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	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75327</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75327"/>
		<updated>2011-02-04T09:11:47Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==Homework 13: Team Problem==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==pH: The Power of Understanding==&lt;br /&gt;
&lt;br /&gt;
The pH scale  measures acidic and basic chemical properties.  The scale ranges from 0  to 14, with a pH of 7 being neutral. Any pH less than 7 is acidic,  whereas a pH of greater than 7 is basic. The pH scale is a logarithmic  scale, which means that each lower numerical value is ten times more  acidic than the next higher numerical value (eg. pH of 3 is ten times  more acidic than a pH of 4, and 100 times (10 times 10) more acidic than  a pH of 5). This property works for basic values as well, with each  higher value being ten times more alkaline, or basic, than the value  below it (eg. pH of 9 is ten times more alkaline than pH of 8).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH  is a measure of the concentration of hydrogen ions.  The definition of  pH is the negative logarithm of a hydrogen ion concentration.  The less  ions that are present, the more alkaline the solution. If more hydrogen  ions are found the more acidic.  To find the pH, we need to find the  logarithm of the H+ (hydrogen) concentration, and then reverse the sign -  the only case in which the sign is not reversed is if the H+  concentration is greater than 1 M (which is highly unusual).&lt;br /&gt;
&lt;br /&gt;
The fact that the pH scale is a logarithmic scale turns out to be very useful when scientists need to calculate very high values of hydrogen ions in solutions. In fact, strongly basic solutions might have, let’s say, 100.000.000.000.000 times less hydrogen ions than strongly basic solutions. It is then obvious that dealing with such large numbers is a source of complications for the scientists that need to analyze this data. A logarithmic scale helps scientists deal with extremely large numbers for pH in a more efficient way, with applications in various fields such as marine biology and oceanography (sea water), medicine (body fluids) and certainly chemistry. &lt;br /&gt;
&lt;br /&gt;
For comparison purposes, the pH scale uses logs because logs can  expand the values between 0 and 1 on a compact scale to make it into a  large scale, this also means that the log scale changes. pH scale uses  logs when the answer or the change in the values depends on the size of  the the change in proportion to the value and not on the absolute size  of change.&lt;br /&gt;
For example, when you graph two acids, one with a  concentration of 0.001 and one with a weaker acid say with pH4 as  opposed to the first, pH3, it would be 0.0001. When you graph this, the  change barely shows and it comes up as the same value, implying that all  neutral and basic values are indistinguishable. Since it is important  to show the difference between the two, a weak and a weaker acid, then  these values are needed, and therefore we use logs to find these values  to plot on our graph. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The  equation of pH is: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;pH = -log [H+]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
eg. what is the pH of a substance with an H+ concentration of 0.001? &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;to find the pH, we first should convert the concentration value into exponential notation:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
0.001M = 10^-3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;we than plug this number into our formula of pH = -log [H+]&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = -log [10^-3]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = 3 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This means that since  pure water is pH 7 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The concentration of hydrogen ions would be 0.0000001, plugged into the pH formula, it would look like pH = -log [10^-7]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:184phscale.gif]]&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;FEEL FREE TO CHECK OUT THE YOUTUBE VIDEO FOR A VISUAL EXPLANATION! &lt;br /&gt;
＊＊＊http://www.youtube.com/watch?v=u837KYKyr9c ＊＊＊&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
 sources: &lt;br /&gt;
 http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;br /&gt;
&lt;br /&gt;
www.science.uwaterloo.ca/~cchieh/cact/c123/ph.html&lt;br /&gt;
&lt;br /&gt;
http://mathforum.org/library/drmath/view/55574.html&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75326</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75326"/>
		<updated>2011-02-04T09:11:20Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==Homework 13: Team Problem==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==pH: The Power of Understanding==&lt;br /&gt;
&lt;br /&gt;
The pH scale  measures acidic and basic chemical properties.  The scale ranges from 0  to 14, with a pH of 7 being neutral. Any pH less than 7 is acidic,  whereas a pH of greater than 7 is basic. The pH scale is a logarithmic  scale, which means that each lower numerical value is ten times more  acidic than the next higher numerical value (eg. pH of 3 is ten times  more acidic than a pH of 4, and 100 times (10 times 10) more acidic than  a pH of 5). This property works for basic values as well, with each  higher value being ten times more alkaline, or basic, than the value  below it (eg. pH of 9 is ten times more alkaline than pH of 8).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH  is a measure of the concentration of hydrogen ions.  The definition of  pH is the negative logarithm of a hydrogen ion concentration.  The less  ions that are present, the more alkaline the solution. If more hydrogen  ions are found the more acidic.  To find the pH, we need to find the  logarithm of the H+ (hydrogen) concentration, and then reverse the sign -  the only case in which the sign is not reversed is if the H+  concentration is greater than 1 M (which is highly unusual).&lt;br /&gt;
&lt;br /&gt;
The fact that the pH scale is a logarithmic scale turns out to be very useful when scientists need to calculate very high values of hydrogen ions in solutions. In fact, strongly basic solutions might have, let’s say, 100.000.000.000.000 times less hydrogen ions than strongly basic solutions. It is then obvious that dealing with such large numbers is a source of complications for the scientists that need to analyze this data. A logarithmic scale helps scientists deal with extremely large numbers for pH in a more efficient way, with applications in various fields such as marine biology and oceanography (sea water), medicine (body fluids) and certainly chemistry. &lt;br /&gt;
&lt;br /&gt;
For comparison purposes, the pH scale uses logs because logs can  expand the values between 0 and 1 on a compact scale to make it into a  large scale, this also means that the log scale changes. pH scale uses  logs when the answer or the change in the values depends on the size of  the the change in proportion to the value and not on the absolute size  of change.&lt;br /&gt;
For example, when you graph two acids, one with a  concentration of 0.001 and one with a weaker acid say with pH4 as  opposed to the first, pH3, it would be 0.0001. When you graph this, the  change barely shows and it comes up as the same value, implying that all  neutral and basic values are indistinguishable. Since it is important  to show the difference between the two, a weak and a weaker acid, then  these values are needed, and therefore we use logs to find these values  to plot on our graph. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The  equation of pH is: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;pH = -log [H+]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
eg. what is the pH of a substance with an H+ concentration of 0.001? &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;to find the pH, we first should convert the concentration value into exponential notation:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
0.001M = 10^-3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;we than plug this number into our formula of pH = -log [H+]&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = -log [10^-3]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = 3 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This means that since  pure water is pH 7 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The concentration of hydrogen ions would be 0.0000001, plugged into the pH formula, it would look like pH = -log [10^-7]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:184phscale.gif]]&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;FEEL FREE TO CHECK OUT THE YOUTUBE VIDEO FOR A VISUAL EXPLANATION! &lt;br /&gt;
＊＊＊http://www.youtube.com/watch?v=u837KYKyr9c ＊＊＊&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 sources: &lt;br /&gt;
 http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;br /&gt;
&lt;br /&gt;
www.science.uwaterloo.ca/~cchieh/cact/c123/ph.html&lt;br /&gt;
&lt;br /&gt;
http://mathforum.org/library/drmath/view/55574.html&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75325</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75325"/>
		<updated>2011-02-04T09:10:14Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==Homework 13: Team Problem==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==pH: The Power of Understanding==&lt;br /&gt;
&lt;br /&gt;
The pH scale  measures acidic and basic chemical properties.  The scale ranges from 0  to 14, with a pH of 7 being neutral. Any pH less than 7 is acidic,  whereas a pH of greater than 7 is basic. The pH scale is a logarithmic  scale, which means that each lower numerical value is ten times more  acidic than the next higher numerical value (eg. pH of 3 is ten times  more acidic than a pH of 4, and 100 times (10 times 10) more acidic than  a pH of 5). This property works for basic values as well, with each  higher value being ten times more alkaline, or basic, than the value  below it (eg. pH of 9 is ten times more alkaline than pH of 8).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH  is a measure of the concentration of hydrogen ions.  The definition of  pH is the negative logarithm of a hydrogen ion concentration.  The less  ions that are present, the more alkaline the solution. If more hydrogen  ions are found the more acidic.  To find the pH, we need to find the  logarithm of the H+ (hydrogen) concentration, and then reverse the sign -  the only case in which the sign is not reversed is if the H+  concentration is greater than 1 M (which is highly unusual).&lt;br /&gt;
&lt;br /&gt;
The fact that the pH scale is a logarithmic scale turns out to be very useful when scientists need to calculate very high values of hydrogen ions in solutions. In fact, strongly basic solutions might have, let’s say, 100.000.000.000.000 times less hydrogen ions than strongly basic solutions. It is then obvious that dealing with such large numbers is a source of complications for the scientists that need to analyze this data. A logarithmic scale helps scientists deal with extremely large numbers for pH in a more efficient way, with applications in various fields such as marine biology and oceanography (sea water), medicine (body fluids) and certainly chemistry. &lt;br /&gt;
&lt;br /&gt;
For comparison purposes, the pH scale uses logs because logs can  expand the values between 0 and 1 on a compact scale to make it into a  large scale, this also means that the log scale changes. pH scale uses  logs when the answer or the change in the values depends on the size of  the the change in proportion to the value and not on the absolute size  of change.&lt;br /&gt;
For example, when you graph two acids, one with a  concentration of 0.001 and one with a weaker acid say with pH4 as  opposed to the first, pH3, it would be 0.0001. When you graph this, the  change barely shows and it comes up as the same value, implying that all  neutral and basic values are indistinguishable. Since it is important  to show the difference between the two, a weak and a weaker acid, then  these values are needed, and therefore we use logs to find these values  to plot on our graph. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The  equation of pH is: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;pH = -log [H+]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
eg. what is the pH of a substance with an H+ concentration of 0.001? &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;to find the pH, we first should convert the concentration value into exponential notation:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
0.001M = 10^-3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;we than plug this number into our formula of pH = -log [H+]&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = -log [10^-3]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = 3 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This means that since  pure water is pH 7 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The concentration of hydrogen ions would be 0.0000001, plugged into the pH formula, it would look like pH = -log [10^-7]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:184phscale.gif]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;＊＊＊http://www.youtube.com/watch?v=u837KYKyr9c＊＊＊&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 sources: &lt;br /&gt;
 http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;br /&gt;
&lt;br /&gt;
www.science.uwaterloo.ca/~cchieh/cact/c123/ph.html&lt;br /&gt;
&lt;br /&gt;
http://mathforum.org/library/drmath/view/55574.html&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75324</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_13&amp;diff=75324"/>
		<updated>2011-02-04T09:07:11Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* pH: The Power of Understanding */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==Homework 13: Team Problem==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==pH: The Power of Understanding==&lt;br /&gt;
&lt;br /&gt;
The pH scale  measures acidic and basic chemical properties.  The scale ranges from 0  to 14, with a pH of 7 being neutral. Any pH less than 7 is acidic,  whereas a pH of greater than 7 is basic. The pH scale is a logarithmic  scale, which means that each lower numerical value is ten times more  acidic than the next higher numerical value (eg. pH of 3 is ten times  more acidic than a pH of 4, and 100 times (10 times 10) more acidic than  a pH of 5). This property works for basic values as well, with each  higher value being ten times more alkaline, or basic, than the value  below it (eg. pH of 9 is ten times more alkaline than pH of 8).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH  is a measure of the concentration of hydrogen ions.  The definition of  pH is the negative logarithm of a hydrogen ion concentration.  The less  ions that are present, the more alkaline the solution. If more hydrogen  ions are found the more acidic.  To find the pH, we need to find the  logarithm of the H+ (hydrogen) concentration, and then reverse the sign -  the only case in which the sign is not reversed is if the H+  concentration is greater than 1 M (which is highly unusual).&lt;br /&gt;
&lt;br /&gt;
The fact that the pH scale is a logarithmic scale turns out to be very useful when scientists need to calculate very high values of hydrogen ions in solutions. In fact, strongly basic solutions might have, let’s say, 100.000.000.000.000 times less hydrogen ions than strongly basic solutions. It is then obvious that dealing with such large numbers is a source of complications for the scientists that need to analyze this data. A logarithmic scale helps scientists deal with extremely large numbers for pH in a more efficient way, with applications in various fields such as marine biology and oceanography (sea water), medicine (body fluids) and certainly chemistry. &lt;br /&gt;
&lt;br /&gt;
For comparison purposes, the pH scale uses logs because logs can  expand the values between 0 and 1 on a compact scale to make it into a  large scale, this also means that the log scale changes. pH scale uses  logs when the answer or the change in the values depends on the size of  the the change in proportion to the value and not on the absolute size  of change.&lt;br /&gt;
For example, when you graph two acids, one with a  concentration of 0.001 and one with a weaker acid say with pH4 as  opposed to the first, pH3, it would be 0.0001. When you graph this, the  change barely shows and it comes up as the same value, implying that all  neutral and basic values are indistinguishable. Since it is important  to show the difference between the two, a weak and a weaker acid, then  these values are needed, and therefore we use logs to find these values  to plot on our graph. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The  equation of pH is: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;pH = -log [H+]&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
eg. what is the pH of a substance with an H+ concentration of 0.001? &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;to find the pH, we first should convert the concentration value into exponential notation:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
0.001M = 10^-3&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;we than plug this number into our formula of pH = -log [H+]&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = -log [10^-3]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
pH = 3 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This means that since  pure water is pH 7 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The concentration of hydrogen ions would be 0.0000001, plugged into the pH formula, it would look like pH = -log [10^-7]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:184phscale.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 sources: &lt;br /&gt;
 http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;br /&gt;
&lt;br /&gt;
www.science.uwaterloo.ca/~cchieh/cact/c123/ph.html&lt;br /&gt;
&lt;br /&gt;
http://mathforum.org/library/drmath/view/55574.html&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73625</id>
		<title>User:Fiona</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73625"/>
		<updated>2011-01-28T09:43:35Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Fiona Ma - Second year, Faculty of Arts.&lt;br /&gt;
&amp;lt;p&amp;gt;[[HOMEWORK 12 - ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Calculus is used in many fields of study, which is why teachers always tell you to continue math past the point where it becomes mandatory because 83 different fields of work require math 12 and beyond. Calculus focuses on limits, functions, derivatives etc. and is the study of change, or the rate of change. Economics is a field that relies most heavily on calculus besides math, because economics focuses on functional relationships such as different variables, income, elasticities etc. Calculus especially is mostly used to determine marginal revenues and costs which helps to calculate how profit will increase in response to each input of production given. &lt;br /&gt;
&amp;lt;p&amp;gt;Example: Microeconomics focuses on firm production, and firms produce its output so that marginal cost=marginal revenue. To calculate profit which equals to revenue-cost, we want to take the derivative and set it to zero to maximize profits. Therefore, marginal profit=marginal revenue-marginal cost and profit will be maximized when marginal revenue = marginal cost&lt;br /&gt;
&amp;lt;p&amp;gt;http://www.youtube.com/watch?v=LmftTXTE3Fw&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
Also, similar to the example we did in class previously on elasticites, we can see that calculus is also used. To calculate the elasticity, we just have to find the percentage change : Elasticity - (percentage change in X) / (percentage change in Y) = (dX/dY) times (Y/X), where dX/dY is the derivative of X with respect to Y, therefore allowing us to calculate elasticity with calculus. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
[[ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
The Pythagorean theorem is used to calculate the hypotenuse of a right angle triangle. This theorem has been around for centuries and researchers found that it was apparent in many historical references. For example, the Egyptians used this theorem to build their pyramids. Although this theorem wasn’t put into a mathematical formula, the Egyptians had their own measuring system using a rope with 12 knots of equal distance showing their understanding of the 3,4, and 5 right angle triangle. To this day, we still use this theorem/method not only in math classes, but to design and build modern-day structures just like the Egyptians used it to build their pyramids.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is defined as: &#039;&#039;&#039;a²+b²=c²&#039;&#039;&#039;  (with c as the length of the hypotenuse).&lt;br /&gt;
 &lt;br /&gt;
An example of this question would be:&lt;br /&gt;
	Matthew loves to zip line across buildings, he wants to zip line across two identical buildings that are 30 meters apart and he wants to zip line from the 10th floor to the 6th floor of the other building. Matthew knows that each floor is 10 meters high. How long should his zip line rope be?&lt;br /&gt;
&lt;br /&gt;
We can deduct that side A is 30m, and side B is 40m (10(10-6)). Using the Pythagorean theorem, we can solve for side C, which is the length of his zip line. &lt;br /&gt;
			&lt;br /&gt;
				&#039;&#039;&#039;A²+B² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;30²+40² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;900+1600 = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (900+1600)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (2500)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = 50m&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using the Pythagorean theorem, we can conclude that Matthew’s zip line needs to be 50m long.&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
References : &lt;br /&gt;
&amp;lt;p&amp;gt;http://ejad.best.vwh.net/java/pythagoras/history.html&lt;br /&gt;
http://en.wikipedia.org/wiki/Pythagorean_theorem&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73624</id>
		<title>User:Fiona</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73624"/>
		<updated>2011-01-28T09:43:20Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Fiona Ma - Second year, Faculty of Arts.&lt;br /&gt;
&amp;lt;p&amp;gt;[[HOMEWORK 12 - ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Calculus is used in many fields of study, which is why teachers always tell you to continue math past the point where it becomes mandatory because 83 different fields of work require math 12 and beyond. Calculus focuses on limits, functions, derivatives etc. and is the study of change, or the rate of change. Economics is a field that relies most heavily on calculus besides math, because economics focuses on functional relationships such as different variables, income, elasticities etc. Calculus especially is mostly used to determine marginal revenues and costs which helps to calculate how profit will increase in response to each input of production given. &lt;br /&gt;
&amp;lt;p&amp;gt;Example: Microeconomics focuses on firm production, and firms produce its output so that marginal cost=marginal revenue. To calculate profit which equals to revenue-cost, we want to take the derivative and set it to zero to maximize profits. Therefore, marginal profit=marginal revenue-marginal cost and profit will be maximized when marginal revenue = marginal cost&lt;br /&gt;
http://www.youtube.com/watch?v=LmftTXTE3Fw&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
Also, similar to the example we did in class previously on elasticites, we can see that calculus is also used. To calculate the elasticity, we just have to find the percentage change : Elasticity - (percentage change in X) / (percentage change in Y) = (dX/dY) times (Y/X), where dX/dY is the derivative of X with respect to Y, therefore allowing us to calculate elasticity with calculus. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
[[ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
The Pythagorean theorem is used to calculate the hypotenuse of a right angle triangle. This theorem has been around for centuries and researchers found that it was apparent in many historical references. For example, the Egyptians used this theorem to build their pyramids. Although this theorem wasn’t put into a mathematical formula, the Egyptians had their own measuring system using a rope with 12 knots of equal distance showing their understanding of the 3,4, and 5 right angle triangle. To this day, we still use this theorem/method not only in math classes, but to design and build modern-day structures just like the Egyptians used it to build their pyramids.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is defined as: &#039;&#039;&#039;a²+b²=c²&#039;&#039;&#039;  (with c as the length of the hypotenuse).&lt;br /&gt;
 &lt;br /&gt;
An example of this question would be:&lt;br /&gt;
	Matthew loves to zip line across buildings, he wants to zip line across two identical buildings that are 30 meters apart and he wants to zip line from the 10th floor to the 6th floor of the other building. Matthew knows that each floor is 10 meters high. How long should his zip line rope be?&lt;br /&gt;
&lt;br /&gt;
We can deduct that side A is 30m, and side B is 40m (10(10-6)). Using the Pythagorean theorem, we can solve for side C, which is the length of his zip line. &lt;br /&gt;
			&lt;br /&gt;
				&#039;&#039;&#039;A²+B² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;30²+40² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;900+1600 = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (900+1600)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (2500)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = 50m&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using the Pythagorean theorem, we can conclude that Matthew’s zip line needs to be 50m long.&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
References : &lt;br /&gt;
&amp;lt;p&amp;gt;http://ejad.best.vwh.net/java/pythagoras/history.html&lt;br /&gt;
http://en.wikipedia.org/wiki/Pythagorean_theorem&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73617</id>
		<title>User:Fiona</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73617"/>
		<updated>2011-01-28T09:30:27Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Fiona Ma - Second year, Faculty of Arts.&lt;br /&gt;
&amp;lt;p&amp;gt;[[HOMEWORK 12 - ESSAY]]&lt;br /&gt;
&lt;br /&gt;
Calculus is used in many fields of study, which is why teachers always tell you to continue math past the point where it becomes mandatory because 83 different fields of work require math 12 and beyond. Economics is a field that relies most heavily on calculus besides math, because economics focuses on functional relationships such as different variables, income, elasticities etc. Calculus especially is mostly used to determine marginal revenues and costs which helps to calculate how profit will increase in response to each input of production given. &lt;br /&gt;
&amp;lt;p&amp;gt;Example: Microeconomics focuses on firm production, and firms produce its output so that marginal cost=marginal revenue. To calculate profit which equals to revenue-cost, we want to take the derivative and set it to zero to maximize profits. Therefore, marginal profit=marginal revenue-marginal cost and profit will be maximized when marginal revenue = marginal cost&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
[[ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
The Pythagorean theorem is used to calculate the hypotenuse of a right angle triangle. This theorem has been around for centuries and researchers found that it was apparent in many historical references. For example, the Egyptians used this theorem to build their pyramids. Although this theorem wasn’t put into a mathematical formula, the Egyptians had their own measuring system using a rope with 12 knots of equal distance showing their understanding of the 3,4, and 5 right angle triangle. To this day, we still use this theorem/method not only in math classes, but to design and build modern-day structures just like the Egyptians used it to build their pyramids.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is defined as: &#039;&#039;&#039;a²+b²=c²&#039;&#039;&#039;  (with c as the length of the hypotenuse).&lt;br /&gt;
 &lt;br /&gt;
An example of this question would be:&lt;br /&gt;
	Matthew loves to zip line across buildings, he wants to zip line across two identical buildings that are 30 meters apart and he wants to zip line from the 10th floor to the 6th floor of the other building. Matthew knows that each floor is 10 meters high. How long should his zip line rope be?&lt;br /&gt;
&lt;br /&gt;
We can deduct that side A is 30m, and side B is 40m (10(10-6)). Using the Pythagorean theorem, we can solve for side C, which is the length of his zip line. &lt;br /&gt;
			&lt;br /&gt;
				&#039;&#039;&#039;A²+B² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;30²+40² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;900+1600 = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (900+1600)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (2500)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = 50m&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using the Pythagorean theorem, we can conclude that Matthew’s zip line needs to be 50m long.&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
References : &lt;br /&gt;
&amp;lt;p&amp;gt;http://ejad.best.vwh.net/java/pythagoras/history.html&lt;br /&gt;
http://en.wikipedia.org/wiki/Pythagorean_theorem&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73616</id>
		<title>User:Fiona</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73616"/>
		<updated>2011-01-28T09:30:06Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Fiona Ma - Second year, Faculty of Arts.&lt;br /&gt;
[[HOMEWORK 12 - ESSAY]]&lt;br /&gt;
&lt;br /&gt;
Calculus is used in many fields of study, which is why teachers always tell you to continue math past the point where it becomes mandatory because 83 different fields of work require math 12 and beyond. Economics is a field that relies most heavily on calculus besides math, because economics focuses on functional relationships such as different variables, income, elasticities etc. Calculus especially is mostly used to determine marginal revenues and costs which helps to calculate how profit will increase in response to each input of production given. &lt;br /&gt;
&amp;lt;p&amp;gt;Example: Microeconomics focuses on firm production, and firms produce its output so that marginal cost=marginal revenue. To calculate profit which equals to revenue-cost, we want to take the derivative and set it to zero to maximize profits. Therefore, marginal profit=marginal revenue-marginal cost and profit will be maximized when marginal revenue = marginal cost&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
[[ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is used to calculate the hypotenuse of a right angle triangle. This theorem has been around for centuries and researchers found that it was apparent in many historical references. For example, the Egyptians used this theorem to build their pyramids. Although this theorem wasn’t put into a mathematical formula, the Egyptians had their own measuring system using a rope with 12 knots of equal distance showing their understanding of the 3,4, and 5 right angle triangle. To this day, we still use this theorem/method not only in math classes, but to design and build modern-day structures just like the Egyptians used it to build their pyramids.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is defined as: &#039;&#039;&#039;a²+b²=c²&#039;&#039;&#039;  (with c as the length of the hypotenuse).&lt;br /&gt;
 &lt;br /&gt;
An example of this question would be:&lt;br /&gt;
	Matthew loves to zip line across buildings, he wants to zip line across two identical buildings that are 30 meters apart and he wants to zip line from the 10th floor to the 6th floor of the other building. Matthew knows that each floor is 10 meters high. How long should his zip line rope be?&lt;br /&gt;
&lt;br /&gt;
We can deduct that side A is 30m, and side B is 40m (10(10-6)). Using the Pythagorean theorem, we can solve for side C, which is the length of his zip line. &lt;br /&gt;
			&lt;br /&gt;
				&#039;&#039;&#039;A²+B² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;30²+40² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;900+1600 = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (900+1600)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (2500)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = 50m&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using the Pythagorean theorem, we can conclude that Matthew’s zip line needs to be 50m long.&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
References : &lt;br /&gt;
&amp;lt;p&amp;gt;http://ejad.best.vwh.net/java/pythagoras/history.html&lt;br /&gt;
http://en.wikipedia.org/wiki/Pythagorean_theorem&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73614</id>
		<title>User:Fiona</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=73614"/>
		<updated>2011-01-28T09:29:35Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Fiona Ma - Second year, Faculty of Arts.&lt;br /&gt;
[[HOMEWORK 12 - ESSAY]]&lt;br /&gt;
&lt;br /&gt;
Calculus is used in many fields of study, which is why teachers always tell you to continue math past the point where it becomes mandatory because 83 different fields of work require math 12 and beyond. Economics is a field that relies most heavily on calculus besides math, because economics focuses on functional relationships such as different variables, income, elasticities etc. Calculus especially is mostly used to determine marginal revenues and costs which helps to calculate how profit will increase in response to each input of production given. &lt;br /&gt;
&amp;lt;p&amp;gt;Example: Microeconomics focuses on firm production, and firms produce its output so that marginal cost=marginal revenue. To calculate profit which equals to revenue-cost, we want to take the derivative and set it to zero to maximize profits. Therefore, marginal profit=marginal revenue-marginal cost and profit will be maximized when marginal revenue = marginal cost&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is used to calculate the hypotenuse of a right angle triangle. This theorem has been around for centuries and researchers found that it was apparent in many historical references. For example, the Egyptians used this theorem to build their pyramids. Although this theorem wasn’t put into a mathematical formula, the Egyptians had their own measuring system using a rope with 12 knots of equal distance showing their understanding of the 3,4, and 5 right angle triangle. To this day, we still use this theorem/method not only in math classes, but to design and build modern-day structures just like the Egyptians used it to build their pyramids.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is defined as: &#039;&#039;&#039;a²+b²=c²&#039;&#039;&#039;  (with c as the length of the hypotenuse).&lt;br /&gt;
 &lt;br /&gt;
An example of this question would be:&lt;br /&gt;
	Matthew loves to zip line across buildings, he wants to zip line across two identical buildings that are 30 meters apart and he wants to zip line from the 10th floor to the 6th floor of the other building. Matthew knows that each floor is 10 meters high. How long should his zip line rope be?&lt;br /&gt;
&lt;br /&gt;
We can deduct that side A is 30m, and side B is 40m (10(10-6)). Using the Pythagorean theorem, we can solve for side C, which is the length of his zip line. &lt;br /&gt;
			&lt;br /&gt;
				&#039;&#039;&#039;A²+B² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;30²+40² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;900+1600 = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (900+1600)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (2500)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = 50m&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using the Pythagorean theorem, we can conclude that Matthew’s zip line needs to be 50m long.&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
References : &lt;br /&gt;
&amp;lt;p&amp;gt;http://ejad.best.vwh.net/java/pythagoras/history.html&lt;br /&gt;
http://en.wikipedia.org/wiki/Pythagorean_theorem&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70883</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70883"/>
		<updated>2011-01-19T07:38:47Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
C= mx+b&lt;br /&gt;
&lt;br /&gt;
where C = total cost&lt;br /&gt;
     mx = variable cost &lt;br /&gt;
     b  = fixed cost &lt;br /&gt;
     m  = marginal cost &lt;br /&gt;
     x  = quantity produced&lt;br /&gt;
&lt;br /&gt;
$100= $7(20units) + b&lt;br /&gt;
&lt;br /&gt;
b = $100 - $140&lt;br /&gt;
&lt;br /&gt;
b = -$40  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Describe your model. What does your model predict for a production of 150 items? According to your model, what happens to the average cost per item as production levels increase?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model predicts the total cost for producing x number of flags given the marginal cost ($7) and a fixed cost (-$40). In the model above, when producing 20 units, the cost of producing a single unit is $5. The reason that this number is lower than the marginal cost is because of the negative fixed cost. One reasoning for the negative fixed cost could be government subsidies for the production of flags. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For the production of 150 items our model predicts a total cost of $1010.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(150units) + (-$40) &lt;br /&gt;
&lt;br /&gt;
C = $1010&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $1010/300 = $6.73&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
According to our model, as the production level increases, the average cost per item increases  &lt;br /&gt;
&lt;br /&gt;
examples: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(300units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $2060&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/300 = $6.87 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(1000units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $6960&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/1000 = $6.96&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost remains constant as production increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
An example would b: TC=3Q&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Q would be quantity and 3 is the marginal cost&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Because in economics, AVC = TC/Q ( average cost = total cost / quantity)&lt;br /&gt;
&amp;lt;p&amp;gt;If you make the cost function 3Q, then that means the average cost will be 3, which is constant no matter what your Q value is.&amp;lt;/p&amp;gt; &lt;br /&gt;
 &lt;br /&gt;
&amp;lt;p&amp;gt;* there is no fixed cost or &#039;b&#039; in the y=mx+b formula *&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost increases as production increases&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;p&amp;gt;An example would be: TC=100+3Q&lt;br /&gt;
&amp;lt;p&amp;gt;The average cost remains constant but now there is a fixed cost added or &#039;b&#039; because as the quantity increase, the fixed cost is shared by an increasing number, therefore the average cost diminishes.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost increases as production increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;An example would be: TC=Q^2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Because AVC=TC/Q, and TC is Q^2, then AVC would then equal Q^2/Q which would make AVC=Q. Therefore as quantity increases average cost increases. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models (not necessarily linear) for which you get other behaviours such as: The average cost remains constant as production increases. The average cost diminishes as production increases. The average cost increases as production increases. You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost. Any other interesting properties that you can think of and create a model for. Bonus points can be obtained for very interesting ideas.&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70871</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70871"/>
		<updated>2011-01-19T07:27:40Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
C= mx+b&lt;br /&gt;
&lt;br /&gt;
where C = total cost&lt;br /&gt;
     mx = variable cost &lt;br /&gt;
     b  = fixed cost &lt;br /&gt;
     m  = marginal cost &lt;br /&gt;
     x  = quantity produced&lt;br /&gt;
&lt;br /&gt;
$100= $7(20units) + b&lt;br /&gt;
&lt;br /&gt;
b = $100 - $140&lt;br /&gt;
&lt;br /&gt;
b = -$40  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Describe your model. What does your model predict for a production of 150 items? According to your model, what happens to the average cost per item as production levels increase?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model predicts the total cost for producing x number of flags given the marginal cost ($7) and a fixed cost (-$40). In the model above, when producing 20 units, the cost of producing a single unit is $5. The reason that this number is lower than the marginal cost is because of the negative fixed cost. One reasoning for the negative fixed cost could be government subsidies for the production of flags. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For the production of 150 items our model predicts a total cost of $1010.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(150units) + (-$40) &lt;br /&gt;
&lt;br /&gt;
C = $1010&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $1010/300 = $6.73&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
According to our model, as the production level increases, the average cost per item increases  &lt;br /&gt;
&lt;br /&gt;
examples: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(300units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $2060&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/300 = $6.87 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(1000units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $6960&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/1000 = $6.96&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost remains constant as production increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
An example would b: Y=3Q&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;q would be quantity and 3 is the marginal cost&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Because in economics, AVC = TC/Q ( average cost = total cost / quantity)&lt;br /&gt;
&amp;lt;p&amp;gt;If you make the cost function 3q, then that means the average cost will be 3, which is constant no matter what your q value is.&amp;lt;/p&amp;gt; &lt;br /&gt;
 &lt;br /&gt;
&amp;lt;p&amp;gt;* there is no fixed cost or &#039;b&#039; in the y=mx+b formula *&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost increases as production increases&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;p&amp;gt;An example would be: Y=100+3Q&lt;br /&gt;
&amp;lt;p&amp;gt;The average cost remains constant but now there is a fixed cost added or &#039;b&#039; because as the quantity increase, the fixed cost is shared by an increasing number, therefore the average cost diminishes.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models (not necessarily linear) for which you get other behaviours such as: The average cost remains constant as production increases. The average cost diminishes as production increases. The average cost increases as production increases. You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost. Any other interesting properties that you can think of and create a model for. Bonus points can be obtained for very interesting ideas.&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70870</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70870"/>
		<updated>2011-01-19T07:27:19Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
C= mx+b&lt;br /&gt;
&lt;br /&gt;
where C = total cost&lt;br /&gt;
     mx = variable cost &lt;br /&gt;
     b  = fixed cost &lt;br /&gt;
     m  = marginal cost &lt;br /&gt;
     x  = quantity produced&lt;br /&gt;
&lt;br /&gt;
$100= $7(20units) + b&lt;br /&gt;
&lt;br /&gt;
b = $100 - $140&lt;br /&gt;
&lt;br /&gt;
b = -$40  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Describe your model. What does your model predict for a production of 150 items? According to your model, what happens to the average cost per item as production levels increase?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model predicts the total cost for producing x number of flags given the marginal cost ($7) and a fixed cost (-$40). In the model above, when producing 20 units, the cost of producing a single unit is $5. The reason that this number is lower than the marginal cost is because of the negative fixed cost. One reasoning for the negative fixed cost could be government subsidies for the production of flags. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For the production of 150 items our model predicts a total cost of $1010.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(150units) + (-$40) &lt;br /&gt;
&lt;br /&gt;
C = $1010&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $1010/300 = $6.73&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
According to our model, as the production level increases, the average cost per item increases  &lt;br /&gt;
&lt;br /&gt;
examples: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(300units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $2060&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/300 = $6.87 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(1000units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $6960&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/1000 = $6.96&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost remains constant as production increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
An example would b: Y=3Q&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;q would be quantity and 3 is the marginal cost&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Because in economics, AVC = TC/Q ( average cost = total cost / quantity)&lt;br /&gt;
&amp;lt;p&amp;gt;If you make the cost function 3q, then that means the average cost will be 3, which is constant no matter what your q value is.&amp;lt;/p&amp;gt; &lt;br /&gt;
 &lt;br /&gt;
&amp;lt;p&amp;gt;* there is no fixed cost or &#039;b&#039; in the y=mx+b formula *&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost increases as production increases&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;p&amp;gt;An example would be: Y=100+3Q&lt;br /&gt;
&amp;lt;p&amp;gt;The average cost remains constant but now there is a fixed cost added or &#039;b&#039; because as the quantity increase, the fixed cost is shared by an increasing number, therefore the average cost diminishes.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models (not necessarily linear) for which you get other behaviours such as: The average cost remains constant as production increases. The average cost diminishes as production increases. The average cost increases as production increases. You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost. Any other interesting properties that you can think of and create a model for. Bonus points can be obtained for very interesting ideas.&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70868</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework_11&amp;diff=70868"/>
		<updated>2011-01-19T07:23:54Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
C= mx+b&lt;br /&gt;
&lt;br /&gt;
where C = total cost&lt;br /&gt;
     mx = variable cost &lt;br /&gt;
     b  = fixed cost &lt;br /&gt;
     m  = marginal cost &lt;br /&gt;
     x  = quantity produced&lt;br /&gt;
&lt;br /&gt;
$100= $7(20units) + b&lt;br /&gt;
&lt;br /&gt;
b = $100 - $140&lt;br /&gt;
&lt;br /&gt;
b = -$40  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Describe your model. What does your model predict for a production of 150 items? According to your model, what happens to the average cost per item as production levels increase?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Our model predicts the total cost for producing x number of flags given the marginal cost ($7) and a fixed cost (-$40). In the model above, when producing 20 units, the cost of producing a single unit is $5. The reason that this number is lower than the marginal cost is because of the negative fixed cost. One reasoning for the negative fixed cost could be government subsidies for the production of flags. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For the production of 150 items our model predicts a total cost of $1010.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(150units) + (-$40) &lt;br /&gt;
&lt;br /&gt;
C = $1010&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $1010/300 = $6.73&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
According to our model, as the production level increases, the average cost per item increases  &lt;br /&gt;
&lt;br /&gt;
examples: &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(300units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $2060&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/300 = $6.87 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = $7(1000units) + (-$40)&lt;br /&gt;
&lt;br /&gt;
C = $6960&lt;br /&gt;
&lt;br /&gt;
The cost per unit in this example is $2060/1000 = $6.96&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The average cost remains constant as production increases&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
An example would be y=3q&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;q would be quantity and 3 is the marginal cost&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Because in economics, AVC = TC/Q ( average cost = total cost / quantity)&lt;br /&gt;
&amp;lt;p&amp;gt;If you make the cost function 3q, then that means the average cost will be 3, which is constant no matter what your q value is.&amp;lt;/p&amp;gt; &lt;br /&gt;
 &lt;br /&gt;
&amp;lt;p&amp;gt;* there is no fixed cost or &#039;b&#039; in the y=mx+b formula *&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Finally, find some other models (not necessarily linear) for which you get other behaviours such as: The average cost remains constant as production increases. The average cost diminishes as production increases. The average cost increases as production increases. You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost. Any other interesting properties that you can think of and create a model for. Bonus points can be obtained for very interesting ideas.&#039;&#039;&#039;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn&amp;diff=69972</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn&amp;diff=69972"/>
		<updated>2011-01-15T21:52:40Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Contact */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Solothurn&lt;br /&gt;
| member 1 = [[User:CurtisDoucette|Curtis Doucette]]&lt;br /&gt;
| member 2 = Fiona Ma&lt;br /&gt;
| member 3 = Gaia Silvestri&lt;br /&gt;
| member 4 = Shannon Lee&lt;br /&gt;
}}&lt;br /&gt;
= Contact =&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Curtis Doucette&#039;&#039;&#039;&lt;br /&gt;
cell: 778-855-2355 email: &amp;lt;curtismdoucette@gmail.com&amp;gt; skype: curtismdoucette&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Fiona Ma&#039;&#039;&#039; cell: 778 688 8537 email: &amp;lt;fionama@live.com&amp;gt; skype: fionama-&amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn&amp;diff=69971</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Solothurn</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Solothurn&amp;diff=69971"/>
		<updated>2011-01-15T21:52:15Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Contact */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Solothurn&lt;br /&gt;
| member 1 = [[User:CurtisDoucette|Curtis Doucette]]&lt;br /&gt;
| member 2 = Fiona Ma&lt;br /&gt;
| member 3 = Gaia Silvestri&lt;br /&gt;
| member 4 = Shannon Lee&lt;br /&gt;
}}&lt;br /&gt;
= Contact =&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Curtis Doucette&#039;&#039;&#039;&lt;br /&gt;
cell: 778-855-2355 email: &amp;lt;curtismdoucette@gmail.com&amp;gt; skype: curtismdoucette&lt;br /&gt;
&#039;&#039;&#039;Fiona Ma&#039;&#039;&#039; &lt;br /&gt;
cell: 778 688 8537 email: &amp;lt;fionama@live.com&amp;gt; skype: fionama-&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=64592</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=64592"/>
		<updated>2010-12-02T06:49:16Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Composition of Functions - Translate/Scale/Reflect Graphs */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;= Constructing new functions by using elementary operations =&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Addition of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=N89Cuq92e6M&lt;br /&gt;
&lt;br /&gt;
== Subtraction of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=ZeW7X-ZTz2o&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Multiplication of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=ayFKhoAkuEk&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Division Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=Wy5dOXWfaUs&lt;br /&gt;
&lt;br /&gt;
== Composition of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=VavmbMh0HOI&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Composition of Functions - Translate/Scale/Reflect Graphs==&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;br /&gt;
===&amp;lt;p&amp;gt;How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;===&lt;br /&gt;
&amp;lt;p&amp;gt;There are three ways in which to change a graph, the most common terms known are: Translating, Stretching/Compressing, and Reflecting. These are the different changes in how a function could change, according to how the x or y of a function is changed. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Translating:&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are four different ways in which to translate a graph. Translating a graph simply means relocating or moving the graph while keeping its properties the same (eg. shape and size). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The first two ways in which to translate a graph is simply left and right.&amp;lt;/p&amp;gt; &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;1) To translate the graph left, we simply &#039;&#039;add&#039;&#039; a number to x.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;p&amp;gt;For example :   f(x) = x² → f(x) = (x+2)²  	- here we have added 2 to x&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the x² function has shifted 2 units to the left.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be  (-2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) To translate the graph right, we simply &#039;&#039;subtract&#039;&#039; a number from x. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = (x-2)²  	- here we have subtracted 2 from x&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the function x²  has shifted 2 units to the right. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be (2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Another way to translate a graph is translating it up or down.&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;3) To translate the graph up, we simply &#039;&#039;add&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt; &lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x²  → f(x) = x²+2  	- here we have added 2 to f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y- values of the function x² has shifted 2 units up.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,2).&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;4) To translate the graph down, we simply &#039;&#039;subtract&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = x²-2        - here we have subtracted 2 from f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y-values of the function x² has shifted 2 units down. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,-2).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Stretching/Compressing:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are 2 different methods in which we can scale a graph, first we can compress a graph and second we can stretch a graph. Both methods can be done vertically and horizontally. In other words, we can have vertical compression, horizontal compression, vertical stretch, and horizontal stretch.&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;1) Vertical Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch vertically, we multiply a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).   Think of this as pulling a rubber band upwards by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;f(x)   ⇒   f(x) = x²  → f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become increased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2,4), we would simply take the x-value of this point (2), and stick it in the new function where we would first square it 2 and then multiply it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 3, the new values would be (2,12). &amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress vertically, we divide by a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).&lt;br /&gt;
Think of this as pushing a rubber band into itself by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = (f(x))/&#039;&#039;&#039;k&#039;&#039;&#039;   ⇒  f(x) = x² → f(x) = (x²)/&#039;&#039;&#039;k&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become decreased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (4,2), we would simply take the x-value of this point (4), and stick it in the new function where we would first square it by 2 and then divide it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new values would be (4,8). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;2) Horizontal Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch horizontally, we divide a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(x/&#039;&#039;&#039;k&#039;&#039;&#039;)  ⇒  f(x) = x² → f(x) = (x/&#039;&#039;&#039;k&#039;&#039;&#039;)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes divided by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2, 3), we would simply take the x-value of this point (2), and stick it into the new function where we would first divide it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is ½, the new point would be (2,1). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress horizontally, we multiply a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(&#039;&#039;&#039;k&#039;&#039;&#039;x)  ⇒  f(x) = x² → f(x) = (&#039;&#039;&#039;k&#039;&#039;&#039;x)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes multiplied by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (3,7), we would simply take the x-value of this point (3), and stick it into the new function where we would first multiply it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new point would be (36,7).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Reflecting&#039;&#039;&#039;:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;By reflecting a graph we are simply flipping it along the x-axis or the y-axis. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt; 1) Reflecting along the y-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the y-axis we place a &#039;&#039;negative&#039;&#039; in front of the x only. NOT the whole function f(x).&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = f(-x)  ⇒  f(x) = x² → f(x) = (-x)²&lt;br /&gt;
&amp;lt;p&amp;gt;To reflect the graph on the y-axis we simply take the whole graph and flip it along the y-axis, this means that all the values of x becomes the negative of itself while all the y-values stay the same. If we were to pick a point on a graph say, (9,9), in order to reflect it along the y-axis we simply take the negative of the original x-value so that the new point would be (-9,9). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) Reflecting along the x-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the x-axis we place a &#039;&#039;negative&#039;&#039; in front of the function f(x).&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = -f(x)  ⇒  f(x) = x² → f(x) = -x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect the graph on the x-axis we simply take the whole graph and flip it along the x-axis this means that all the values of y becomes the negative of itself while all the x-values stay the same. If we were to pick a point on a graph say (8,5), in order to reflect it along the x-axis, we simply take the negative of the original y-value so that the new point would be (8,-5).&amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=64591</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=64591"/>
		<updated>2010-12-02T06:48:54Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Composition of Functions - Translate/Scale/Reflect Graphs */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;= Constructing new functions by using elementary operations =&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Addition of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=N89Cuq92e6M&lt;br /&gt;
&lt;br /&gt;
== Subtraction of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=ZeW7X-ZTz2o&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Multiplication of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=ayFKhoAkuEk&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Division Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=Wy5dOXWfaUs&lt;br /&gt;
&lt;br /&gt;
== Composition of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=VavmbMh0HOI&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Composition of Functions - Translate/Scale/Reflect Graphs==&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;===How to Translate, Scale and Reflect Graphs=== &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are three ways in which to change a graph, the most common terms known are: Translating, Stretching/Compressing, and Reflecting. These are the different changes in how a function could change, according to how the x or y of a function is changed. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Translating:&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are four different ways in which to translate a graph. Translating a graph simply means relocating or moving the graph while keeping its properties the same (eg. shape and size). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The first two ways in which to translate a graph is simply left and right.&amp;lt;/p&amp;gt; &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;1) To translate the graph left, we simply &#039;&#039;add&#039;&#039; a number to x.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;p&amp;gt;For example :   f(x) = x² → f(x) = (x+2)²  	- here we have added 2 to x&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the x² function has shifted 2 units to the left.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be  (-2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) To translate the graph right, we simply &#039;&#039;subtract&#039;&#039; a number from x. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = (x-2)²  	- here we have subtracted 2 from x&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the function x²  has shifted 2 units to the right. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be (2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Another way to translate a graph is translating it up or down.&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;3) To translate the graph up, we simply &#039;&#039;add&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt; &lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x²  → f(x) = x²+2  	- here we have added 2 to f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y- values of the function x² has shifted 2 units up.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,2).&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;4) To translate the graph down, we simply &#039;&#039;subtract&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = x²-2        - here we have subtracted 2 from f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y-values of the function x² has shifted 2 units down. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,-2).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Stretching/Compressing:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are 2 different methods in which we can scale a graph, first we can compress a graph and second we can stretch a graph. Both methods can be done vertically and horizontally. In other words, we can have vertical compression, horizontal compression, vertical stretch, and horizontal stretch.&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;1) Vertical Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch vertically, we multiply a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).   Think of this as pulling a rubber band upwards by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;f(x)   ⇒   f(x) = x²  → f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become increased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2,4), we would simply take the x-value of this point (2), and stick it in the new function where we would first square it 2 and then multiply it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 3, the new values would be (2,12). &amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress vertically, we divide by a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).&lt;br /&gt;
Think of this as pushing a rubber band into itself by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = (f(x))/&#039;&#039;&#039;k&#039;&#039;&#039;   ⇒  f(x) = x² → f(x) = (x²)/&#039;&#039;&#039;k&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become decreased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (4,2), we would simply take the x-value of this point (4), and stick it in the new function where we would first square it by 2 and then divide it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new values would be (4,8). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;2) Horizontal Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch horizontally, we divide a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(x/&#039;&#039;&#039;k&#039;&#039;&#039;)  ⇒  f(x) = x² → f(x) = (x/&#039;&#039;&#039;k&#039;&#039;&#039;)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes divided by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2, 3), we would simply take the x-value of this point (2), and stick it into the new function where we would first divide it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is ½, the new point would be (2,1). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress horizontally, we multiply a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(&#039;&#039;&#039;k&#039;&#039;&#039;x)  ⇒  f(x) = x² → f(x) = (&#039;&#039;&#039;k&#039;&#039;&#039;x)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes multiplied by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (3,7), we would simply take the x-value of this point (3), and stick it into the new function where we would first multiply it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new point would be (36,7).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Reflecting&#039;&#039;&#039;:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;By reflecting a graph we are simply flipping it along the x-axis or the y-axis. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt; 1) Reflecting along the y-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the y-axis we place a &#039;&#039;negative&#039;&#039; in front of the x only. NOT the whole function f(x).&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = f(-x)  ⇒  f(x) = x² → f(x) = (-x)²&lt;br /&gt;
&amp;lt;p&amp;gt;To reflect the graph on the y-axis we simply take the whole graph and flip it along the y-axis, this means that all the values of x becomes the negative of itself while all the y-values stay the same. If we were to pick a point on a graph say, (9,9), in order to reflect it along the y-axis we simply take the negative of the original x-value so that the new point would be (-9,9). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) Reflecting along the x-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the x-axis we place a &#039;&#039;negative&#039;&#039; in front of the function f(x).&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = -f(x)  ⇒  f(x) = x² → f(x) = -x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect the graph on the x-axis we simply take the whole graph and flip it along the x-axis this means that all the values of y becomes the negative of itself while all the x-values stay the same. If we were to pick a point on a graph say (8,5), in order to reflect it along the x-axis, we simply take the negative of the original y-value so that the new point would be (8,-5).&amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=64590</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=64590"/>
		<updated>2010-12-02T06:48:14Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Part 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;= Constructing new functions by using elementary operations =&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Addition of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=N89Cuq92e6M&lt;br /&gt;
&lt;br /&gt;
== Subtraction of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=ZeW7X-ZTz2o&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Multiplication of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=ayFKhoAkuEk&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Division Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=Wy5dOXWfaUs&lt;br /&gt;
&lt;br /&gt;
== Composition of Functions ==&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=VavmbMh0HOI&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Composition of Functions - Translate/Scale/Reflect Graphs==&lt;br /&gt;
&amp;lt;p&amp;gt;How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are three ways in which to change a graph, the most common terms known are: Translating, Stretching/Compressing, and Reflecting. These are the different changes in how a function could change, according to how the x or y of a function is changed. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Translating:&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are four different ways in which to translate a graph. Translating a graph simply means relocating or moving the graph while keeping its properties the same (eg. shape and size). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The first two ways in which to translate a graph is simply left and right.&amp;lt;/p&amp;gt; &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;1) To translate the graph left, we simply &#039;&#039;add&#039;&#039; a number to x.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;p&amp;gt;For example :   f(x) = x² → f(x) = (x+2)²  	- here we have added 2 to x&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the x² function has shifted 2 units to the left.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be  (-2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) To translate the graph right, we simply &#039;&#039;subtract&#039;&#039; a number from x. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = (x-2)²  	- here we have subtracted 2 from x&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the function x²  has shifted 2 units to the right. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be (2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Another way to translate a graph is translating it up or down.&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;3) To translate the graph up, we simply &#039;&#039;add&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt; &lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x²  → f(x) = x²+2  	- here we have added 2 to f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y- values of the function x² has shifted 2 units up.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,2).&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;4) To translate the graph down, we simply &#039;&#039;subtract&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = x²-2        - here we have subtracted 2 from f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y-values of the function x² has shifted 2 units down. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,-2).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Stretching/Compressing:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are 2 different methods in which we can scale a graph, first we can compress a graph and second we can stretch a graph. Both methods can be done vertically and horizontally. In other words, we can have vertical compression, horizontal compression, vertical stretch, and horizontal stretch.&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;1) Vertical Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch vertically, we multiply a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).   Think of this as pulling a rubber band upwards by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;f(x)   ⇒   f(x) = x²  → f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become increased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2,4), we would simply take the x-value of this point (2), and stick it in the new function where we would first square it 2 and then multiply it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 3, the new values would be (2,12). &amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress vertically, we divide by a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).&lt;br /&gt;
Think of this as pushing a rubber band into itself by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = (f(x))/&#039;&#039;&#039;k&#039;&#039;&#039;   ⇒  f(x) = x² → f(x) = (x²)/&#039;&#039;&#039;k&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become decreased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (4,2), we would simply take the x-value of this point (4), and stick it in the new function where we would first square it by 2 and then divide it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new values would be (4,8). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;2) Horizontal Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch horizontally, we divide a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(x/&#039;&#039;&#039;k&#039;&#039;&#039;)  ⇒  f(x) = x² → f(x) = (x/&#039;&#039;&#039;k&#039;&#039;&#039;)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes divided by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2, 3), we would simply take the x-value of this point (2), and stick it into the new function where we would first divide it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is ½, the new point would be (2,1). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress horizontally, we multiply a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(&#039;&#039;&#039;k&#039;&#039;&#039;x)  ⇒  f(x) = x² → f(x) = (&#039;&#039;&#039;k&#039;&#039;&#039;x)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes multiplied by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (3,7), we would simply take the x-value of this point (3), and stick it into the new function where we would first multiply it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new point would be (36,7).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Reflecting&#039;&#039;&#039;:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;By reflecting a graph we are simply flipping it along the x-axis or the y-axis. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt; 1) Reflecting along the y-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the y-axis we place a &#039;&#039;negative&#039;&#039; in front of the x only. NOT the whole function f(x).&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = f(-x)  ⇒  f(x) = x² → f(x) = (-x)²&lt;br /&gt;
&amp;lt;p&amp;gt;To reflect the graph on the y-axis we simply take the whole graph and flip it along the y-axis, this means that all the values of x becomes the negative of itself while all the y-values stay the same. If we were to pick a point on a graph say, (9,9), in order to reflect it along the y-axis we simply take the negative of the original x-value so that the new point would be (-9,9). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) Reflecting along the x-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the x-axis we place a &#039;&#039;negative&#039;&#039; in front of the function f(x).&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = -f(x)  ⇒  f(x) = x² → f(x) = -x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect the graph on the x-axis we simply take the whole graph and flip it along the x-axis this means that all the values of y becomes the negative of itself while all the x-values stay the same. If we were to pick a point on a graph say (8,5), in order to reflect it along the x-axis, we simply take the negative of the original y-value so that the new point would be (8,-5).&amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63720</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63720"/>
		<updated>2010-11-29T13:48:19Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Part 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Part 3 ===&lt;br /&gt;
&amp;lt;p&amp;gt;How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are three ways in which to change a graph, the most common terms known are: Translating, Stretching/Compressing, and Reflecting. These are the different changes in how a function could change, according to how the x or y of a function is changed. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Translating:&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are four different ways in which to translate a graph. Translating a graph simply means relocating or moving the graph while keeping its properties the same (eg. shape and size). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The first two ways in which to translate a graph is simply left and right.&amp;lt;/p&amp;gt; &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;1) To translate the graph left, we simply &#039;&#039;add&#039;&#039; a number to x.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;p&amp;gt;For example :   f(x) = x² → f(x) = (x+2)²  	- here we have added 2 to x&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the x² function has shifted 2 units to the left.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be  (-2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) To translate the graph right, we simply &#039;&#039;subtract&#039;&#039; a number from x. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = (x-2)²  	- here we have subtracted 2 from x&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the function x²  has shifted 2 units to the right. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be (2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Another way to translate a graph is translating it up or down.&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;3) To translate the graph up, we simply &#039;&#039;add&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt; &lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x²  → f(x) = x²+2  	- here we have added 2 to f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y- values of the function x² has shifted 2 units up.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,2).&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;4) To translate the graph down, we simply &#039;&#039;subtract&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = x²-2        - here we have subtracted 2 from f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y-values of the function x² has shifted 2 units down. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,-2).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Stretching/Compressing:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are 2 different methods in which we can scale a graph, first we can compress a graph and second we can stretch a graph. Both methods can be done vertically and horizontally. In other words, we can have vertical compression, horizontal compression, vertical stretch, and horizontal stretch.&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;1) Vertical Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch vertically, we multiply a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).   Think of this as pulling a rubber band upwards by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;f(x)   ⇒   f(x) = x²  → f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become increased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2,4), we would simply take the x-value of this point (2), and stick it in the new function where we would first square it 2 and then multiply it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 3, the new values would be (2,12). &amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress vertically, we divide by a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).&lt;br /&gt;
Think of this as pushing a rubber band into itself by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = (f(x))/&#039;&#039;&#039;k&#039;&#039;&#039;   ⇒  f(x) = x² → f(x) = (x²)/&#039;&#039;&#039;k&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become decreased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (4,2), we would simply take the x-value of this point (4), and stick it in the new function where we would first square it by 2 and then divide it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new values would be (4,8). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;2) Horizontal Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch horizontally, we divide a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(x/&#039;&#039;&#039;k&#039;&#039;&#039;)  ⇒  f(x) = x² → f(x) = (x/&#039;&#039;&#039;k&#039;&#039;&#039;)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes divided by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2, 3), we would simply take the x-value of this point (2), and stick it into the new function where we would first divide it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is ½, the new point would be (2,1). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress horizontally, we multiply a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(&#039;&#039;&#039;k&#039;&#039;&#039;x)  ⇒  f(x) = x² → f(x) = (&#039;&#039;&#039;k&#039;&#039;&#039;x)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes multiplied by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (3,7), we would simply take the x-value of this point (3), and stick it into the new function where we would first multiply it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new point would be (36,7).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Reflecting&#039;&#039;&#039;:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;By reflecting a graph we are simply flipping it along the x-axis or the y-axis. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt; 1) Reflecting along the y-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the y-axis we place a &#039;&#039;negative&#039;&#039; in front of the x only. NOT the whole function f(x).&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = f(-x)  ⇒  f(x) = x² → f(x) = (-x)²&lt;br /&gt;
&amp;lt;p&amp;gt;To reflect the graph on the y-axis we simply take the whole graph and flip it along the y-axis, this means that all the values of x becomes the negative of itself while all the y-values stay the same. If we were to pick a point on a graph say, (9,9), in order to reflect it along the y-axis we simply take the negative of the original x-value so that the new point would be (-9,9). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) Reflecting along the x-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the x-axis we place a &#039;&#039;negative&#039;&#039; in front of the function f(x).&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = -f(x)  ⇒  f(x) = x² → f(x) = -x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect the graph on the x-axis we simply take the whole graph and flip it along the x-axis this means that all the values of y becomes the negative of itself while all the x-values stay the same. If we were to pick a point on a graph say (8,5), in order to reflect it along the x-axis, we simply take the negative of the original y-value so that the new point would be (8,-5).&amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63719</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63719"/>
		<updated>2010-11-29T13:46:49Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Part 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Part 3 ===&lt;br /&gt;
&amp;lt;p&amp;gt;How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are three ways in which to change a graph, the most common terms known are: Translating, Stretching/Compressing, and Reflecting. These are the different changes in how a function could change, according to how the x or y of a function is changed. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Translating:&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are four different ways in which to translate a graph. Translating a graph simply means relocating or moving the graph while keeping its properties the same (eg. shape and size). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The first two ways in which to translate a graph is simply left and right.&amp;lt;/p&amp;gt; &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;1) To translate the graph left, we simply &#039;&#039;add&#039;&#039; a number to x.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;p&amp;gt;For example :   f(x) = x² → f(x) = (x+2)²  	- here we have added 2 to x&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the x² function has shifted 2 units to the left.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be  (-2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) To translate the graph right, we simply &#039;&#039;subtract&#039;&#039; a number from x. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = (x-2)²  	- here we have subtracted 2 from x&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the function x²  has shifted 2 units to the right. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be (2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Another way to translate a graph is translating it up or down.&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;3) To translate the graph up, we simply &#039;&#039;add&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt; &lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x²  → f(x) = x²+2  	- here we have added 2 to f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y- values of the function x² has shifted 2 units up.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,2).&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;4) To translate the graph down, we simply &#039;&#039;subtract&#039;&#039; a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = x²-2        - here we have subtracted 2 from f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y-values of the function x² has shifted 2 units down. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,-2).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Stretching/Compressing:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are 2 different methods in which we can scale a graph, first we can compress a graph and second we can stretch a graph. Both methods can be done vertically and horizontally. In other words, we can have vertical compression, horizontal compression, vertical stretch, and horizontal stretch.&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;1) Vertical Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch vertically, we multiply a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).   Think of this as pulling a rubber band upwards by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;f(x)   ⇒   f(x) = x²  → f(x) = &#039;&#039;&#039;k&#039;&#039;&#039;x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become increased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2,4), we would simply take the x-value of this point (2), and stick it in the new function where we would first square it 2 and then multiply it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 3, the new values would be (2,12). &amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress vertically, we divide by a factor of &#039;&#039;&#039;k&#039;&#039;&#039; to the entire function f(x).&lt;br /&gt;
Think of this as pushing a rubber band into itself by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = (f(x))/&#039;&#039;&#039;k&#039;&#039;&#039;   ⇒  f(x) = x² → f(x) = (x²)/&#039;&#039;&#039;k&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become decreased by a factor of &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (4,2), we would simply take the x-value of this point (4), and stick it in the new function where we would first square it by 2 and then divide it by &#039;&#039;&#039;k&#039;&#039;&#039;. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is 2, the new values would be (4,8). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;2) Horizontal Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch horizontally, we divide a value of &#039;&#039;&#039;k&#039;&#039;&#039; to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(x/&#039;&#039;&#039;k&#039;&#039;&#039;)  ⇒  f(x) = x² → f(x) = (x/&#039;&#039;&#039;k&#039;&#039;&#039;)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes divided by &#039;&#039;&#039;k&#039;&#039;&#039;. If we make a table of values and find that a point on the original graph of x² was (2, 3), we would simply take the x-value of this point (2), and stick it into the new function where we would first divide it to &#039;&#039;&#039;k&#039;&#039;&#039;. and then square it. This would enable us to find the new value of y for the new function. Given that &#039;&#039;&#039;k&#039;&#039;&#039;’s value is ½, the new point would be (2,1). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress horizontally, we multiply a value of k to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(kx)  ⇒  f(x) = x² → f(x) = (kx)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes multiplied by k. If we make a table of values and find that a point on the original graph of x² was (3,7), we would simply take the x-value of this point (3), and stick it into the new function where we would first multiply it to k. and then square it. This would enable us to find the new value of y for the new function. Given that k’s value is 2, the new point would be (36,7).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Reflecting&#039;&#039;&#039;:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;By reflecting a graph we are simply flipping it along the x-axis or the y-axis. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt; 1) Reflecting along the y-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the y-axis we add a negative in front of the x only. NOT the whole function f(x).&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = f(-x)  ⇒  f(x) = x² → f(x) = (-x)²&lt;br /&gt;
&amp;lt;p&amp;gt;To reflect the graph on the y-axis we simply take the whole graph and flip it along the y-axis, this means that all the values of x becomes the negative of itself while all the y-values stay the same. If we were to pick a point on a graph say, (9,9), in order to reflect it along the y-axis we simply take the negative of the original x-value so that the new point would be (-9,9). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) Reflecting along the x-axis&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect along the x-axis we add a negative in front of the function f(x).&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt; For example : f(x) = -f(x)  ⇒  f(x) = x² → f(x) = -x²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt; To reflect the graph on the x-axis we simply take the whole graph and flip it along the x-axis this means that all the values of y becomes the negative of itself while all the x-values stay the same. If we were to pick a point on a graph say (8,5), in order to reflect it along the x-axis, we simply take the negative of the original y-value so that the new point would be (8,-5).&amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63718</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63718"/>
		<updated>2010-11-29T13:30:41Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Part 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Part 3 ===&lt;br /&gt;
&amp;lt;p&amp;gt;How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are three ways in which to change a graph, the most common terms known are: Translating, Stretching/Compressing, and Reflecting. These are the different changes in how a function could change, according to how the x or y of a function is changed. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Translating:&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are four different ways in which to translate a graph. Translating a graph simply means relocating or moving the graph while keeping its properties the same (eg. shape and size). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The first two ways in which to translate a graph is simply left and right.&amp;lt;/p&amp;gt; &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;1) To translate the graph left, we simply add a number to x.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;p&amp;gt;For example :   f(x) = x² → f(x) = (x+2)²  	- here we have added 2 to x&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the x² function has shifted 2 units to the left.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be  (-2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) To translate the graph right, we simply subtract a number from x. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = (x-2)²  	- here we have subtracted 2 from x&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the function x²  has shifted 2 units to the right. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be (2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Another way to translate a graph is translating it up or down.&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;3) To translate the graph up, we simply add a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt; &lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x²  → f(x) = x²+2  	- here we have added 2 to f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y- values of the function x² has shifted 2 units up.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,2).&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;4) To translate the graph down, we simply subtract a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = x²-2        - here we have subtracted 2 from f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y-values of the function x² has shifted 2 units down. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,-2).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Stretching/Compressing:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are 2 different methods in which we can scale a graph, first we can compress a graph and second we can stretch a graph. Both methods can be done vertically and horizontally. In other words, we can have vertical compression, horizontal compression, vertical stretch, and horizontal stretch.&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;1) Vertical Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch vertically, we multiply a factor of k to the entire function f(x).   Think of this as pulling a rubber band upwards by a factor of k. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = kf(x)   ⇒   f(x) = x²  → f(x) = kx²&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become increased by a factor of k. If we make a table of values and find that a point on the original graph of x² was (2,4), we would simply take the x-value of this point (2), and stick it in the new function where we would first square it 2 and then multiply it by k. This would enable us to find the new value of y for the new function. Given that k’s value is 3, the new values would be (2,12). &amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress vertically, we divide by a factor of k to the entire function f(x).&lt;br /&gt;
Think of this as pushing a rubber band into itself by a factor of k. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = (f(x))/k   ⇒  f(x) = x² → f(x) = (x²)/k&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become decreased by a factor of k. If we make a table of values and find that a point on the original graph of x² was (4,2), we would simply take the x-value of this point (4), and stick it in the new function where we would first square it by 2 and then divide it by k. This would enable us to find the new value of y for the new function. Given that k’s value is 2, the new values would be (4,8). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;2) Horizontal Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch horizontally, we divide a value of k to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(x/k)  ⇒  f(x) = x² → f(x) = (x/k)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes divided by k. If we make a table of values and find that a point on the original graph of x² was (2, 3), we would simply take the x-value of this point (2), and stick it into the new function where we would first divide it to k. and then square it. This would enable us to find the new value of y for the new function. Given that k’s value is ½, the new point would be (2,1). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress horizontally, we multiply a value of k to x only. NOT the entire function f(x). &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = f(kx)  ⇒  f(x) = x² → f(x) = (kx)²&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values change as the original values of x becomes multiplied by k. If we make a table of values and find that a point on the original graph of x² was (3,7), we would simply take the x-value of this point (3), and stick it into the new function where we would first multiply it to k. and then square it. This would enable us to find the new value of y for the new function. Given that k’s value is 2, the new point would be (36,7).&amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63717</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63717"/>
		<updated>2010-11-29T13:29:00Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Part 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Part 3 ===&lt;br /&gt;
&amp;lt;p&amp;gt;How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are three ways in which to change a graph, the most common terms known are: Translating, Stretching/Compressing, and Reflecting. These are the different changes in how a function could change, according to how the x or y of a function is changed. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Translating:&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;There are four different ways in which to translate a graph. Translating a graph simply means relocating or moving the graph while keeping its properties the same (eg. shape and size). &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The first two ways in which to translate a graph is simply left and right.&amp;lt;/p&amp;gt; &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;1) To translate the graph left, we simply add a number to x.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;p&amp;gt;For example :   f(x) = x² → f(x) = (x+2)²  	- here we have added 2 to x&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the x² function has shifted 2 units to the left.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be  (-2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;p&amp;gt;2) To translate the graph right, we simply subtract a number from x. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = (x-2)²  	- here we have subtracted 2 from x&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the x-values of the function x²  has shifted 2 units to the right. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be (2,0).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Another way to translate a graph is translating it up or down.&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;3) To translate the graph up, we simply add a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt; &lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x²  → f(x) = x²+2  	- here we have added 2 to f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y- values of the function x² has shifted 2 units up.&lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,2).&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;4) To translate the graph down, we simply subtract a number to the whole function of f(x), not just x.&amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = x² → f(x) = x²-2        - here we have subtracted 2 from f(x)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;In this example, all the y-values of the function x² has shifted 2 units down. &lt;br /&gt;
If the origin of the original function was at (0,0), then the new origin of the function would be at (0,-2).&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Stretching/Compressing:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;There are 2 different methods in which we can scale a graph, first we can compress a graph and second we can stretch a graph. Both methods can be done vertically and horizontally. In other words, we can have vertical compression, horizontal compression, vertical stretch, and horizontal stretch.&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;1) Vertical Stretching and Compressing&amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;a) To stretch vertically, we multiply a factor of k to the entire function f(x).   Think of this as pulling a rubber band upwards by a factor of k. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = kf(x)   ⇒   f(x) = x²  → f(x) = kx²&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become increased by a factor of k. If we make a table of values and find that a point on the original graph of x² was (2,4), we would simply take the x-value of this point (2), and stick it in the new function where we would first square it 2 and then multiply it by k. This would enable us to find the new value of y for the new function. Given that k’s value is 3, the new values would be (2,12). &amp;lt;/p&amp;gt;&lt;br /&gt;
:&amp;lt;p&amp;gt;b) To compress vertically, we divide by a factor of k to the entire function f(x).&lt;br /&gt;
Think of this as pushing a rubber band into itself by a factor of k. &amp;lt;/p&amp;gt;&lt;br /&gt;
::&amp;lt;p&amp;gt;For example : f(x) = (f(x))/k   ⇒  f(x) = x² → f(x) = (x²)/k&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;All the y-values become decreased by a factor of k. If we make a table of values and find that a point on the original graph of x² was (4,2), we would simply take the x-value of this point (4), and stick it in the new function where we would first square it by 2 and then divide it by k. This would enable us to find the new value of y for the new function. Given that k’s value is 2, the new values would be (4,8). &amp;lt;/p&amp;gt;&lt;br /&gt;
2) Horizontal Stretching and Compressing.&lt;br /&gt;
a) To stretch horizontally, we divide a value of k to x only. NOT the entire function &lt;br /&gt;
      f(x). &lt;br /&gt;
	For example : f(x) = f(x/k)  ⇒  f(x) = x² → f(x) = (x/k)²&lt;br /&gt;
All the y-values change as the original values of x becomes divided by k. If we make a table of values and find that a point on the original graph of x² was (2, 3), we would simply take the x-value of this point (2), and stick it into the new function where we would first divide it to k. and then square it. This would enable us to find the new value of y for the new function. Given that k’s value is ½, the new point would be (2,1). &lt;br /&gt;
b) To compress horizontally, we multiply a value of k to x only. NOT the entire &lt;br /&gt;
      function f(x). &lt;br /&gt;
	For example : f(x) = f(kx)  ⇒  f(x) = x² → f(x) = (kx)²&lt;br /&gt;
All the y-values change as the original values of x becomes multiplied by k. If we make a table of values and find that a point on the original graph of x² was (3,7), we would simply take the x-value of this point (3), and stick it into the new function where we would first multiply it to k. and then square it. This would enable us to find the new value of y for the new function. Given that k’s value is 2, the new point would be (36,7).&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63712</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63712"/>
		<updated>2010-11-29T12:09:25Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Part 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Part 3 ===&lt;br /&gt;
&amp;lt;p&amp;gt;How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63711</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63711"/>
		<updated>2010-11-29T12:09:14Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Part 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Part 3 ===&lt;br /&gt;
&amp;lt;p.How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63710</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/BasicSkills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/BasicSkills&amp;diff=63710"/>
		<updated>2010-11-29T12:08:50Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* How the graphs of functions change under different elementary operations of functions */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=How the graphs of functions change under different elementary operations of functions=&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;How the graphs of functions change when subjected to the operations of 1) Addition/Subtraction of functions and 2) Multiplication/Division of functions&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) First, we look at what happens during addition and subtraction of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Now to start off we can look at an easy example: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 3x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 4x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 7x &#039;&#039;&#039;&amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We know from the coefficient of 7x which is 7 that this graph has become steeper; it basically got a steeper slope. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let&#039;s try adding quadratic functions.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; and g(x) = 3&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(f+g)(x) = 4&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;+5x &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
We know from the manipulations we can do to the graph of a function that having a higher number as the coefficient of a squared variable gives a tighter parabola so the graph then becomes skinnier. The 5x component also has an effect, but is less pronounced. The focus is on the leading coefficient (the coefficient of the highest degree component, in this case 4x^2).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
From here we can then say that adding functions serves to &#039;squeeze&#039; the function as it increases the leading coefficient (unless of course the functions are both constants, in which case the horizontal line simply changes its vertical position).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For subtraction, the same concept applies, only the coefficients are usually lowered (the word &#039;usually&#039; is used here because subtracting a negative number is effectively adding it, thus increasing the coefficient). Two things happen here, one is that the function is sort of reflected because the coefficient is negative (i.e. positive quadratic = parabola concave upwards and negative quadratic = concave downwards) and two is that again as the absolute value of the coefficient increases, the function becomes skinnier. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Note: If the functions added/subtracted are of different degrees, then only the lower degree components have a small effect on the resulting function (i.e. the resulting function is simply shifted, with no bearing on its steepness/width).&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) Next, we look at what happens during multiplication and division of functions.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To better understand what happens to a functions graph, let&#039;s recall what happens when we multiply or divide functions.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x) = 2x &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x) = 3x&amp;lt;math&amp;gt;^2&amp;lt;/math&amp;gt;&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;f(x)g(x) = 6&amp;lt;math&amp;gt;x^3&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;g(x)/f(x) = &amp;lt;math&amp;gt;3/2x&amp;lt;/math&amp;gt; &#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
What do we notice here? One is that the leading coefficient changes, though we cannot generalise about its behaviour, we can say that how wide/skinny a function is is affected by multiplication/division of functions. Next, and more importantly is that the degree of the function changes. What this means is that for these functions and higher degree ones, the number of critical points is either increased or decreased in the process of multiplication and division. In other words, the bigger the degree of the function, the more critical points there are and vice versa. This is shown very evidently on its graph by how many times we see a plateau phase or a point on the domain where the tangent line to the curve has a slope of zero. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In general multiplying functions yields higher degree functions and thus more critical points and dividing yields lower degree functions with less critical points, but beware that there are many examples which do not follow this &amp;quot;rule of thumb&amp;quot; if you will.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Part 3 ===&lt;br /&gt;
How to Translate, Scale and Reflect Graphs &amp;lt;/p&amp;gt;&lt;br /&gt;
Link to video : http://www.youtube.com/profile?user=Math110Group13#grid/uploads &amp;lt;/p&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/Homework_7&amp;diff=61107</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13/Homework 7</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13/Homework_7&amp;diff=61107"/>
		<updated>2010-11-12T23:04:41Z</updated>

		<summary type="html">&lt;p&gt;Fiona: Created page with &amp;#039;==Basic Skills Suggestion==   &amp;lt;p&amp;gt;A while ago the prof suggested that we create a section in the wiki regarding the basic skills where each person would offer something that they …&amp;#039;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==Basic Skills Suggestion==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;A while ago the prof suggested that we create a section in the wiki regarding the basic skills where each person would offer something that they know or have a problem with and people would gradually add to it. The idea is to accumulate these basic skills by ourselves through a combined class effort.&lt;br /&gt;
Anyways since a lot of people seemed to have a problem with writing out things on the wiki or spending that much time in general on the wiki, our group thought of making a youtube channel.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Pros:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Most people will have a camera/video editing system built in into their computer&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- OR most people own a digital camera that can take 5 min clips&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- No written explanation required&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Verbal explanation is easier/requires less work&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Visual and verbal explanation go hand in hand&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Easier to show examples/visuals through video&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Allows for more examples to be shown, and someone to talk through it step by step&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt; We propose to make a video study guide for composition of functions that will include a short tutorial video for each of the following topics:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;1. A short explanation of the theory behind composition of functions and why it works the way it does (including examples)&lt;br /&gt;
&amp;lt;p&amp;gt;2. Step by step process for approaching different types of problems including some helpful tips.&lt;br /&gt;
&amp;lt;p&amp;gt;3. Examples worked out from start to finish&lt;br /&gt;
&amp;lt;p&amp;gt;4. Suggestions for useful practice problems to work on&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=61106</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=61106"/>
		<updated>2010-11-12T23:04:25Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Homework #7 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #7=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_7&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #5=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_5&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_4&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_3&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=61104</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=61104"/>
		<updated>2010-11-12T23:03:51Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Homework #7 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #7=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_6&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #5=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_5&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_4&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_3&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=61103</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=61103"/>
		<updated>2010-11-12T23:03:30Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Homework #7 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #7=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_7&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #5=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_5&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_4&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_3&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=60973</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=60973"/>
		<updated>2010-11-12T06:53:24Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Homework #6 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #7=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_6&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #5=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_5&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_4&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_3&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=60723</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=60723"/>
		<updated>2010-11-11T22:03:38Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Basic Skills Suggestion */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
=Homework 6=&lt;br /&gt;
&lt;br /&gt;
==Basic Skills Suggestion==&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;A while ago the prof suggested that we create a section in the wiki regarding the basic skills where each person would offer something that they know or have a problem with and people would gradually add to it. The idea is to accumulate these basic skills by ourselves through a combined class effort.&lt;br /&gt;
Anyways since a lot of people seemed to have a problem with writing out things on the wiki or spending that much time in general on the wiki, our group thought of making a youtube channel.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Pros:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Most people will have a camera/video editing system built in into their computer&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- OR most people own a digital camera that can take 5 min clips&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- No written explanation required&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Verbal explanation is easier/requires less work&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Visual and verbal explanation go hand in hand&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Easier to show examples/visuals through video&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;- Allows for more examples to be shown, and someone to talk through it step by step&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=Homework #5=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_5&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_4&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_3&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=60722</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=60722"/>
		<updated>2010-11-11T22:01:50Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
=Homework 6=&lt;br /&gt;
&lt;br /&gt;
==Basic Skills Suggestion==&lt;br /&gt;
&lt;br /&gt;
A while ago the prof suggested that we create a section in the wiki regarding the basic skills where each person would offer something that they know or have a problem with and people would gradually add to it. The idea is to accumulate these basic skills by ourselves through a combined class effort.&lt;br /&gt;
Anyways since a lot of people seemed to have a problem with writing out things on the wiki or spending that much time in general on the wiki, our group thought of making a youtube channel.&lt;br /&gt;
&lt;br /&gt;
Pros:&lt;br /&gt;
- Most people will have a camera/video editing system built in into their computer&lt;br /&gt;
- OR most people own a digital camera that can take 5 min clips&lt;br /&gt;
- No written explanation required&lt;br /&gt;
- Verbal explanation is easier/requires less work &lt;br /&gt;
- Visual and verbal explanation go hand in hand&lt;br /&gt;
- Easier to show examples/visuals through video&lt;br /&gt;
- Allows for more examples to be shown, and someone to talk through it step by step&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Homework #5=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_5&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_4&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_13/Homework_3&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=57018</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=57018"/>
		<updated>2010-10-21T05:35:50Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan&#039;s mother?&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
- Tosh owns a cat,&amp;lt;br&amp;gt;&lt;br /&gt;
- Bianca owns a frog that she loves,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jaela owns a parrot which keeps calling her &amp;quot;darling, darling&amp;quot;,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jun owns a snake, don&#039;t mess with him,&amp;lt;br&amp;gt;&lt;br /&gt;
- Suzan is the name of the frog,&amp;lt;br&amp;gt;&lt;br /&gt;
- The cat is named Jun,&amp;lt;br&amp;gt;&lt;br /&gt;
- The name by which they call the turtle is the name of the woman whose pet is Tosh,&amp;lt;br&amp;gt;&lt;br /&gt;
- Finally, Suzan&#039;s mother&#039;s pet is Bianca.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Tosh ---&amp;gt; Cat (Jun)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Bianca ---&amp;gt; Frog (Suzan)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♂Jun ---&amp;gt; Snake (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement &amp;quot;Bianca owns a frog that &#039;&#039;&#039;she&#039;&#039;&#039; loves&amp;quot; and we know that Jaela is also female from the statement &amp;quot;Jaela owns a parrot which keeps calling &#039;&#039;&#039;her&#039;&#039;&#039; &#039;darling, darling&#039;&amp;quot;. We also know that Jun is male from the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot;, with him referring to the owner and not the pet, otherwise it would be saying don&#039;t mess with &#039;&#039;&#039;it&#039;&#039;&#039;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can also assume that Suzan&#039;s pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan&#039;s pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle&#039;s name needs to be Suzan which is all very confusing and counter-intuitive. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (Tosh)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This piece of information also allows us to name Suzan&#039;s pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (Jaela)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That only leaves the last part of the problem, which would be Jun&#039;s pet:&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♂Jun ---&amp;gt; Snake (Bianca)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan&#039;s &#039;&#039;&#039;mother&#039;&#039;&#039; cannot possibly be male unless the label &amp;quot;mother&amp;quot; is semantically misleading (i.e. some awkward nickname, etc.).&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There are then two possible solutions after this error. The first is that the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot; has the pronoun &#039;him&#039; refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca, as snakes DO have genders). This solution ends up with Jun being Suzan&#039;s mother and the pet Bianca being her pet. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan&#039;s pet to be Tosh and lastly having Jaela&#039;s pet to be Bianca (and thus having Jaela as Suzan&#039;s mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution. Though again the logic may fit, since there can be any number of unmentioned pets, it is not realistically possible to know who Suzan&#039;s mother is with the given information (for this second solution).&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Though the answer really depends on whether we interpreted the information given correctly, the most plausible solution would probably be that Jun is the mother and she owns a snake pet called Bianca.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;First, simplify the information:&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Bohao, Stewart, Dylan, Tim and Chan &lt;br /&gt;
&lt;br /&gt;
2 are left-handed, 3 are right-handed &lt;br /&gt;
&lt;br /&gt;
Bohnao and Dylan = Same Hand &lt;br /&gt;
&lt;br /&gt;
Tim and Chan = Different Hand&lt;br /&gt;
&lt;br /&gt;
2 are over 2m, 3 are under 2m &lt;br /&gt;
&lt;br /&gt;
Steward and Chan = Same Height &lt;br /&gt;
&lt;br /&gt;
Tim and Dylan = Different Height &lt;br /&gt;
&lt;br /&gt;
Need to find: player that is over 2 meters tall and left-handed&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw a diagram to help solve the first part of the problem (Left/Right Handed).&amp;lt;/u&amp;gt; &lt;br /&gt;
[[Image:Group Project 4 Question 2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project 4 Question 2p2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now, using these two diagrams find a player that is over 2 meters tall and left-handed&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
By process of elimination we know that it can&#039;t be: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao&#039;&#039;&#039; (right-handed)&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Dylan&#039;&#039;&#039; (right-handed) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Stewart&#039;&#039;&#039; (Under 2 meters) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Chan&#039;&#039;&#039;  (Under 2 meters)&lt;br /&gt;
&lt;br /&gt;
Therefore it COULD only be &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  Therefore the player that is over 2 meters tall and left-handed is &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
First we have to look at each piece of information that is given to us.&lt;br /&gt;
We know that there are 6 players, and that each player plays each other once in 5 days:&lt;br /&gt;
 - this means that there are 3 pairs playing against each other everyday&lt;br /&gt;
We are given enough information to determine a set of pairs for each day&lt;br /&gt;
 - using this information, we can determine the other four players and who they play during the 5 days.&lt;br /&gt;
&lt;br /&gt;
First we will group all the pairs that we know has played during the five days&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: CP&#039;&#039;&#039; We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: CJ&#039;&#039;&#039; (L, P, S, and F)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: JL&#039;&#039;&#039; (F, P, L, and S)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: PS&#039;&#039;&#039; (F, J, L, and C)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: F ?&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 1: &amp;quot;CP&amp;quot;  &amp;quot;JS&amp;quot;  &amp;quot;LF&amp;quot;&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: &amp;quot;CJ&amp;quot; &amp;quot;LP&amp;quot; &amp;quot;SF&amp;quot;&#039;&#039;&#039; (We cannot group F with L again because they have played each other on the first day)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: &amp;quot;JL&amp;quot;  &amp;quot;FP&amp;quot;  &amp;quot;LS&amp;quot;&#039;&#039;&#039; (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: &amp;quot;PS&amp;quot; &amp;quot;FJ&amp;quot; &amp;quot;LC&amp;quot;&#039;&#039;&#039;(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once)&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The grouping for the following day is as follows:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: &amp;quot;PS&amp;quot; &amp;lt;u&amp;gt; &amp;quot;FJ&amp;quot; &amp;lt;/u&amp;gt; and &amp;quot;LC&amp;quot;&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Therefore, we can come to the conclusion that &amp;lt;u&amp;gt;Fernanda played Janet&amp;lt;/u&amp;gt; on the last day. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
===Question 16===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Trying to figure out how many S (sons) and D (daughters) so I tried to represent the information in an expression.&lt;br /&gt;
&lt;br /&gt;
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and&lt;br /&gt;
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)&lt;br /&gt;
&lt;br /&gt;
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).&lt;br /&gt;
&lt;br /&gt;
===Question 17===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
If the scale read too low then we know that D’s real weight would be &amp;gt;60 kg and S’s real weight would be &amp;gt;50kg and so we would expect their combined weight to be &amp;gt;than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is &amp;lt;60 and S’s real weight is &amp;lt;50 then we would expect the outcome to be &amp;lt;their combined weight (110kg).&lt;br /&gt;
&lt;br /&gt;
===Question 18===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135. &lt;br /&gt;
&lt;br /&gt;
===Question 19===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;One morning each member of Angela&#039;s family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Angela’s cup can be expressed as:&lt;br /&gt;
¼ Total Milk + 1/6 Total Coffee = 8 oz&lt;br /&gt;
We can also say that:&lt;br /&gt;
Total Milk + Total Coffee/8  = # of family members  (because they each had an 8 oz cup)&lt;br /&gt;
&lt;br /&gt;
If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:&lt;br /&gt;
&lt;br /&gt;
8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn&#039;t really count as a family) that this works with.&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30) &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question #5===&lt;br /&gt;
&lt;br /&gt;
Answer: Homer sleeps late on Tuesday!&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Here, no pair has woken homer together for more than one day and they have not been quiet for three consecutive days.&lt;br /&gt;
&lt;br /&gt;
Here is how I solved this problem.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) I created a chart for Saturday throughout Sunday (similar to the one above) and filled in the provided information&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) For the salesman to be noisy for at least once in three consecutive days, he must have had to be noisy on Friday&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;3)Now, we need one day for Homer to rest undisturbed on Tuesday or Thursday. But for the dog and the construction to be noisy at least once in their three consecutive days, we can place the both of them on Thursday to satisfy the requirements. We can then determine that Homer had a happy sleep on Tuesday!&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56207</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56207"/>
		<updated>2010-10-19T22:56:43Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan&#039;s mother?&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
- Tosh owns a cat,&amp;lt;br&amp;gt;&lt;br /&gt;
- Bianca owns a frog that she loves,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jaela owns a parrot which keeps calling her &amp;quot;darling, darling&amp;quot;,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jun owns a snake, don&#039;t mess with him,&amp;lt;br&amp;gt;&lt;br /&gt;
- Suzan is the name of the frog,&amp;lt;br&amp;gt;&lt;br /&gt;
- The cat is named Jun,&amp;lt;br&amp;gt;&lt;br /&gt;
- The name by which they call the turtle is the name of the woman whose pet is Tosh,&amp;lt;br&amp;gt;&lt;br /&gt;
- Finally, Suzan&#039;s mother&#039;s pet is Bianca.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Tosh ---&amp;gt; Cat (Jun)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Bianca ---&amp;gt; Frog (Suzan)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♂Jun ---&amp;gt; Snake (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement &amp;quot;Bianca owns a frog that &#039;&#039;&#039;she&#039;&#039;&#039; loves&amp;quot; and we know that Jaela is also female from the statement &amp;quot;Jaela owns a parrot which keeps calling &#039;&#039;&#039;her&#039;&#039;&#039; &#039;darling, darling&#039;&amp;quot;. We also know that Jun is male from the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot;, with him referring to the owner and not the pet, otherwise it would be saying don&#039;t mess with &#039;&#039;&#039;it&#039;&#039;&#039;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can also assume that Suzan&#039;s pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan&#039;s pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle&#039;s name needs to be Suzan which is all very confusing and counter-intuitive. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (Tosh)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This piece of information also allows us to name Suzan&#039;s pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (Jaela)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That only leaves the last part of the problem, which would be Jun&#039;s pet:&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♂Jun ---&amp;gt; Snake (Bianca)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan&#039;s &#039;&#039;&#039;mother&#039;&#039;&#039; cannot possibly be male unless the label &amp;quot;mother&amp;quot; is semantically misleading (i.e. nickname, etc.).&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There are then two possible solutions after this error. The first is that the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot; has the pronoun &#039;him&#039; refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca). This solution ends up with Jun being Suzan&#039;s mother and the pet Bianca being her pet. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan&#039;s pet to be Tosh and lastly having Jaela&#039;s pet to be Bianca (and thus having Jaela as Suzan&#039;s mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution, and a more likely one then the previously aforementioned. In this case, it is not realistically possible to know who Suzan&#039;s mother is with the given information.&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;First, simplify the information:&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Bohao, Stewart, Dylan, Tim and Chan &lt;br /&gt;
&lt;br /&gt;
2 are left-handed, 3 are right-handed &lt;br /&gt;
&lt;br /&gt;
Bohnao and Dylan = Same Hand &lt;br /&gt;
&lt;br /&gt;
Tim and Chan = Different Hand&lt;br /&gt;
&lt;br /&gt;
2 are over 2m, 3 are under 2m &lt;br /&gt;
&lt;br /&gt;
Steward and Chan = Same Height &lt;br /&gt;
&lt;br /&gt;
Tim and Dylan = Different Height &lt;br /&gt;
&lt;br /&gt;
Need to find: player that is over 2 meters tall and left-handed&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw a diagram to help solve the first part of the problem (Left/Right Handed).&amp;lt;/u&amp;gt; &lt;br /&gt;
[[Image:Group Project 4 Question 2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project 4 Question 2p2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now, using these two diagrams find a player that is over 2 meters tall and left-handed&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
By process of elimination we know that it can&#039;t be: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao&#039;&#039;&#039; (right-handed)&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Dylan&#039;&#039;&#039; (right-handed) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Stewart&#039;&#039;&#039; (Under 2 meters) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Chan&#039;&#039;&#039;  (Under 2 meters)&lt;br /&gt;
&lt;br /&gt;
Therefore it COULD only be &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  Therefore the player that is over 2 meters tall and left-handed is &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
First we have to look at each piece of information that is given to us.&lt;br /&gt;
We know that there are 6 players, and that each player plays each other once in 5 days:&lt;br /&gt;
 - this means that there are 3 pairs playing against each other everyday&lt;br /&gt;
We are given enough information to determine a set of pairs for each day&lt;br /&gt;
 - using this information, we can determine the other four players and who they play during the 5 days.&lt;br /&gt;
&lt;br /&gt;
First we will group all the pairs that we know has played during the five days&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: CP&#039;&#039;&#039; We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: CJ&#039;&#039;&#039; (L, P, S, and F)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: JL&#039;&#039;&#039; (F, P, L, and S)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: PS&#039;&#039;&#039; (F, J, L, and C)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: F ?&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 1: &amp;quot;CP&amp;quot;  &amp;quot;JS&amp;quot;  &amp;quot;LF&amp;quot;&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: &amp;quot;CJ&amp;quot; &amp;quot;LP&amp;quot; &amp;quot;SF&amp;quot;&#039;&#039;&#039; (We cannot group F with L again because they have played each other on the first day)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: &amp;quot;JL&amp;quot;  &amp;quot;FP&amp;quot;  &amp;quot;LS&amp;quot;&#039;&#039;&#039; (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: &amp;quot;PS&amp;quot; &amp;quot;FJ&amp;quot; &amp;quot;LC&amp;quot;&#039;&#039;&#039;(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once)&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The grouping for the following day is as follows:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: &amp;quot;PS&amp;quot; &amp;lt;u&amp;gt; &amp;quot;FJ&amp;quot; &amp;lt;/u&amp;gt; and &amp;quot;LC&amp;quot;&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Therefore, we can come to the conclusion that &amp;lt;u&amp;gt;Fernanda played Janet&amp;lt;/u&amp;gt; on the last day. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
===Question 16===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Trying to figure out how many S (sons) and D (daughters) so I tried to represent the information in an expression.&lt;br /&gt;
&lt;br /&gt;
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and&lt;br /&gt;
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)&lt;br /&gt;
&lt;br /&gt;
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).&lt;br /&gt;
&lt;br /&gt;
===Question 17===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
If the scale read too low then we know that D’s real weight would be &amp;gt;60 kg and S’s real weight would be &amp;gt;50kg and so we would expect their combined weight to be &amp;gt;than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is &amp;lt;60 and S’s real weight is &amp;lt;50 then we would expect the outcome to be &amp;lt;their combined weight (110kg).&lt;br /&gt;
&lt;br /&gt;
===Question 18===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135. &lt;br /&gt;
&lt;br /&gt;
===Question 19===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;One morning each member of Angela&#039;s family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Angela’s cup can be expressed as:&lt;br /&gt;
¼ Total Milk + 1/6 Total Coffee = 8 oz&lt;br /&gt;
We can also say that:&lt;br /&gt;
Total Milk + Total Coffee/8  = # of family members  (because they each had an 8 oz cup)&lt;br /&gt;
&lt;br /&gt;
If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:&lt;br /&gt;
&lt;br /&gt;
8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn&#039;t really count as a family) that this works with.&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30) &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question #5===&lt;br /&gt;
&lt;br /&gt;
Answer: Homer sleeps late on Tuesday!&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Here, no pair has woken homer together for more than one day and they have not been quiet for three consecutive days.&lt;br /&gt;
&lt;br /&gt;
Here is how I solved this problem.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) I created a chart for Saturday throughout Sunday (similar to the one above) and filled in the provided information&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) For the salesman to be noisy for at least once in three consecutive days, he must have had to be noisy on Friday&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;3)Now, we need one day for Homer to rest undisturbed on Tuesday or Thursday. But for the dog and the construction to be noisy at least once in their three consecutive days, we can place the both of them on Thursday to satisfy the requirements. We can then determine that Homer had a happy sleep on Tuesday!&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56206</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56206"/>
		<updated>2010-10-19T22:56:13Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan&#039;s mother?&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
- Tosh owns a cat,&amp;lt;br&amp;gt;&lt;br /&gt;
- Bianca owns a frog that she loves,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jaela owns a parrot which keeps calling her &amp;quot;darling, darling&amp;quot;,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jun owns a snake, don&#039;t mess with him,&amp;lt;br&amp;gt;&lt;br /&gt;
- Suzan is the name of the frog,&amp;lt;br&amp;gt;&lt;br /&gt;
- The cat is named Jun,&amp;lt;br&amp;gt;&lt;br /&gt;
- The name by which they call the turtle is the name of the woman whose pet is Tosh,&amp;lt;br&amp;gt;&lt;br /&gt;
- Finally, Suzan&#039;s mother&#039;s pet is Bianca.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Tosh ---&amp;gt; Cat (Jun)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Bianca ---&amp;gt; Frog (Suzan)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♂Jun ---&amp;gt; Snake (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement &amp;quot;Bianca owns a frog that &#039;&#039;&#039;she&#039;&#039;&#039; loves&amp;quot; and we know that Jaela is also female from the statement &amp;quot;Jaela owns a parrot which keeps calling &#039;&#039;&#039;her&#039;&#039;&#039; &#039;darling, darling&#039;&amp;quot;. We also know that Jun is male from the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot;, with him referring to the owner and not the pet, otherwise it would be saying don&#039;t mess with &#039;&#039;&#039;it&#039;&#039;&#039;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can also assume that Suzan&#039;s pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan&#039;s pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle&#039;s name needs to be Suzan which is all very confusing and counter-intuitive. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (Tosh)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This piece of information also allows us to name Suzan&#039;s pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (Jaela)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That only leaves the last part of the problem, which would be Jun&#039;s pet:&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♂Jun ---&amp;gt; Snake (Bianca)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan&#039;s &#039;&#039;&#039;mother&#039;&#039;&#039; cannot possibly be male unless the label &amp;quot;mother&amp;quot; is semantically misleading (i.e. nickname, etc.).&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There are then two possible solutions after this error. The first is that the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot; has the pronoun &#039;him&#039; refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca). This solution ends up with Jun being Suzan&#039;s mother and the pet Bianca being her pet. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan&#039;s pet to be Tosh and lastly having Jaela&#039;s pet to be Bianca (and thus having Jaela as Suzan&#039;s mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution, and a more likely one then the previously aforementioned. In this case, it is not realistically possible to know who Suzan&#039;s mother is with the given information.&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;First, simplify the information:&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Bohao, Stewart, Dylan, Tim and Chan &lt;br /&gt;
&lt;br /&gt;
2 are left-handed, 3 are right-handed &lt;br /&gt;
&lt;br /&gt;
Bohnao and Dylan = Same Hand &lt;br /&gt;
&lt;br /&gt;
Tim and Chan = Different Hand&lt;br /&gt;
&lt;br /&gt;
2 are over 2m, 3 are under 2m &lt;br /&gt;
&lt;br /&gt;
Steward and Chan = Same Height &lt;br /&gt;
&lt;br /&gt;
Tim and Dylan = Different Height &lt;br /&gt;
&lt;br /&gt;
Need to find: player that is over 2 meters tall and left-handed&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw a diagram to help solve the first part of the problem (Left/Right Handed).&amp;lt;/u&amp;gt; &lt;br /&gt;
[[Image:Group Project 4 Question 2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project 4 Question 2p2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now, using these two diagrams find a player that is over 2 meters tall and left-handed&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
By process of elimination we know that it can&#039;t be: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao&#039;&#039;&#039; (right-handed)&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Dylan&#039;&#039;&#039; (right-handed) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Stewart&#039;&#039;&#039; (Under 2 meters) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Chan&#039;&#039;&#039;  (Under 2 meters)&lt;br /&gt;
&lt;br /&gt;
Therefore it COULD only be &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  Therefore the player that is over 2 meters tall and left-handed is &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
First we have to look at each piece of information that is given to us.&lt;br /&gt;
We know that there are 6 players, and that each player plays each other once in 5 days:&lt;br /&gt;
 - this means that there are 3 pairs playing against each other everyday&lt;br /&gt;
We are given enough information to determine a set of pairs for each day&lt;br /&gt;
 - using this information, we can determine the other four players and who they play during the 5 days.&lt;br /&gt;
&lt;br /&gt;
First we will group all the pairs that we know has played during the five days&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: CP&#039;&#039;&#039; We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: CJ&#039;&#039;&#039; (L, P, S, and F)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: JL&#039;&#039;&#039; (F, P, L, and S)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: PS&#039;&#039;&#039; (F, J, L, and C)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: F ?&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 1: &amp;quot;CP&amp;quot;  &amp;quot;JS&amp;quot;  &amp;quot;LF&amp;quot;&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: &amp;quot;CJ&amp;quot; &amp;quot;LP&amp;quot; &amp;quot;SF&amp;quot;&#039;&#039;&#039; (We cannot group F with L again because they have played each other on the first day)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: &amp;quot;JL&amp;quot;  &amp;quot;FP&amp;quot;  &amp;quot;LS&amp;quot;&#039;&#039;&#039; (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: &amp;quot;PS&amp;quot; &amp;quot;FJ&amp;quot; &amp;quot;LC&amp;quot;&#039;&#039;&#039;(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once)&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The grouping for the following day is as follows:&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: &amp;quot;PS&amp;quot; &amp;lt;u&amp;gt; &amp;quot;FJ&amp;quot; &amp;lt;/u&amp;gt; and &amp;quot;LC&amp;quot;&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Therefore, we can come to the conclusion that Fernanda played Janet on the last day. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
===Question 16===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Trying to figure out how many S (sons) and D (daughters) so I tried to represent the information in an expression.&lt;br /&gt;
&lt;br /&gt;
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and&lt;br /&gt;
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)&lt;br /&gt;
&lt;br /&gt;
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).&lt;br /&gt;
&lt;br /&gt;
===Question 17===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
If the scale read too low then we know that D’s real weight would be &amp;gt;60 kg and S’s real weight would be &amp;gt;50kg and so we would expect their combined weight to be &amp;gt;than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is &amp;lt;60 and S’s real weight is &amp;lt;50 then we would expect the outcome to be &amp;lt;their combined weight (110kg).&lt;br /&gt;
&lt;br /&gt;
===Question 18===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135. &lt;br /&gt;
&lt;br /&gt;
===Question 19===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;One morning each member of Angela&#039;s family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Angela’s cup can be expressed as:&lt;br /&gt;
¼ Total Milk + 1/6 Total Coffee = 8 oz&lt;br /&gt;
We can also say that:&lt;br /&gt;
Total Milk + Total Coffee/8  = # of family members  (because they each had an 8 oz cup)&lt;br /&gt;
&lt;br /&gt;
If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:&lt;br /&gt;
&lt;br /&gt;
8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn&#039;t really count as a family) that this works with.&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30) &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question #5===&lt;br /&gt;
&lt;br /&gt;
Answer: Homer sleeps late on Tuesday!&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Here, no pair has woken homer together for more than one day and they have not been quiet for three consecutive days.&lt;br /&gt;
&lt;br /&gt;
Here is how I solved this problem.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) I created a chart for Saturday throughout Sunday (similar to the one above) and filled in the provided information&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) For the salesman to be noisy for at least once in three consecutive days, he must have had to be noisy on Friday&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;3)Now, we need one day for Homer to rest undisturbed on Tuesday or Thursday. But for the dog and the construction to be noisy at least once in their three consecutive days, we can place the both of them on Thursday to satisfy the requirements. We can then determine that Homer had a happy sleep on Tuesday!&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56205</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56205"/>
		<updated>2010-10-19T22:55:23Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan&#039;s mother?&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
- Tosh owns a cat,&amp;lt;br&amp;gt;&lt;br /&gt;
- Bianca owns a frog that she loves,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jaela owns a parrot which keeps calling her &amp;quot;darling, darling&amp;quot;,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jun owns a snake, don&#039;t mess with him,&amp;lt;br&amp;gt;&lt;br /&gt;
- Suzan is the name of the frog,&amp;lt;br&amp;gt;&lt;br /&gt;
- The cat is named Jun,&amp;lt;br&amp;gt;&lt;br /&gt;
- The name by which they call the turtle is the name of the woman whose pet is Tosh,&amp;lt;br&amp;gt;&lt;br /&gt;
- Finally, Suzan&#039;s mother&#039;s pet is Bianca.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Tosh ---&amp;gt; Cat (Jun)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Bianca ---&amp;gt; Frog (Suzan)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♂Jun ---&amp;gt; Snake (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement &amp;quot;Bianca owns a frog that &#039;&#039;&#039;she&#039;&#039;&#039; loves&amp;quot; and we know that Jaela is also female from the statement &amp;quot;Jaela owns a parrot which keeps calling &#039;&#039;&#039;her&#039;&#039;&#039; &#039;darling, darling&#039;&amp;quot;. We also know that Jun is male from the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot;, with him referring to the owner and not the pet, otherwise it would be saying don&#039;t mess with &#039;&#039;&#039;it&#039;&#039;&#039;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can also assume that Suzan&#039;s pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan&#039;s pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle&#039;s name needs to be Suzan which is all very confusing and counter-intuitive. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (Tosh)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This piece of information also allows us to name Suzan&#039;s pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (Jaela)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That only leaves the last part of the problem, which would be Jun&#039;s pet:&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♂Jun ---&amp;gt; Snake (Bianca)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan&#039;s &#039;&#039;&#039;mother&#039;&#039;&#039; cannot possibly be male unless the label &amp;quot;mother&amp;quot; is semantically misleading (i.e. nickname, etc.).&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There are then two possible solutions after this error. The first is that the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot; has the pronoun &#039;him&#039; refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca). This solution ends up with Jun being Suzan&#039;s mother and the pet Bianca being her pet. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan&#039;s pet to be Tosh and lastly having Jaela&#039;s pet to be Bianca (and thus having Jaela as Suzan&#039;s mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution, and a more likely one then the previously aforementioned. In this case, it is not realistically possible to know who Suzan&#039;s mother is with the given information.&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;First, simplify the information:&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Bohao, Stewart, Dylan, Tim and Chan &lt;br /&gt;
&lt;br /&gt;
2 are left-handed, 3 are right-handed &lt;br /&gt;
&lt;br /&gt;
Bohnao and Dylan = Same Hand &lt;br /&gt;
&lt;br /&gt;
Tim and Chan = Different Hand&lt;br /&gt;
&lt;br /&gt;
2 are over 2m, 3 are under 2m &lt;br /&gt;
&lt;br /&gt;
Steward and Chan = Same Height &lt;br /&gt;
&lt;br /&gt;
Tim and Dylan = Different Height &lt;br /&gt;
&lt;br /&gt;
Need to find: player that is over 2 meters tall and left-handed&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw a diagram to help solve the first part of the problem (Left/Right Handed).&amp;lt;/u&amp;gt; &lt;br /&gt;
[[Image:Group Project 4 Question 2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project 4 Question 2p2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now, using these two diagrams find a player that is over 2 meters tall and left-handed&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
By process of elimination we know that it can&#039;t be: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao&#039;&#039;&#039; (right-handed)&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Dylan&#039;&#039;&#039; (right-handed) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Stewart&#039;&#039;&#039; (Under 2 meters) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Chan&#039;&#039;&#039;  (Under 2 meters)&lt;br /&gt;
&lt;br /&gt;
Therefore it COULD only be &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  Therefore the player that is over 2 meters tall and left-handed is &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?&lt;br /&gt;
&lt;br /&gt;
First we have to look at each piece of information that is given to us.&lt;br /&gt;
We know that there are 6 players, and that each player plays each other once in 5 days:&lt;br /&gt;
 - this means that there are 3 pairs playing against each other everyday&lt;br /&gt;
We are given enough information to determine a set of pairs for each day&lt;br /&gt;
 - using this information, we can determine the other four players and who they play during the 5 days.&lt;br /&gt;
&lt;br /&gt;
First we will group all the pairs that we know has played during the five days&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: CP&#039;&#039;&#039; We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: CJ&#039;&#039;&#039; (L, P, S, and F)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: JL&#039;&#039;&#039; (F, P, L, and S)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: PS&#039;&#039;&#039; (F, J, L, and C)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: F ?&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 1: &amp;quot;CP&amp;quot;  &amp;quot;JS&amp;quot;  &amp;quot;LF&amp;quot;&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;Day 2: &amp;quot;CJ&amp;quot; &amp;quot;LP&amp;quot; &amp;quot;SF&amp;quot;&#039;&#039;&#039; (We cannot group F with L again because they have played each other on the first day)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: &amp;quot;JL&amp;quot;  &amp;quot;FP&amp;quot;  &amp;quot;LS&amp;quot;&#039;&#039;&#039; (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: &amp;quot;PS&amp;quot; &amp;quot;FJ&amp;quot; &amp;quot;LC&amp;quot;&#039;&#039;&#039;(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once)&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;The grouping for the following day is as follows:&lt;br /&gt;
&#039;&#039;&#039;Day 5: &amp;quot;PS&amp;quot; &amp;lt;u&amp;gt; &amp;quot;FJ&amp;quot; &amp;lt;/u&amp;gt; and &amp;quot;LC&amp;quot;&#039;&#039;&#039;&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.&amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;p&amp;gt;Therefore, we can come to the conclusion that Fernanda played Janet on the last day. &amp;lt;/p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
===Question 16===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Trying to figure out how many S (sons) and D (daughters) so I tried to represent the information in an expression.&lt;br /&gt;
&lt;br /&gt;
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and&lt;br /&gt;
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)&lt;br /&gt;
&lt;br /&gt;
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).&lt;br /&gt;
&lt;br /&gt;
===Question 17===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
If the scale read too low then we know that D’s real weight would be &amp;gt;60 kg and S’s real weight would be &amp;gt;50kg and so we would expect their combined weight to be &amp;gt;than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is &amp;lt;60 and S’s real weight is &amp;lt;50 then we would expect the outcome to be &amp;lt;their combined weight (110kg).&lt;br /&gt;
&lt;br /&gt;
===Question 18===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135. &lt;br /&gt;
&lt;br /&gt;
===Question 19===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;One morning each member of Angela&#039;s family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Angela’s cup can be expressed as:&lt;br /&gt;
¼ Total Milk + 1/6 Total Coffee = 8 oz&lt;br /&gt;
We can also say that:&lt;br /&gt;
Total Milk + Total Coffee/8  = # of family members  (because they each had an 8 oz cup)&lt;br /&gt;
&lt;br /&gt;
If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:&lt;br /&gt;
&lt;br /&gt;
8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn&#039;t really count as a family) that this works with.&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30) &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question #5===&lt;br /&gt;
&lt;br /&gt;
Answer: Homer sleeps late on Tuesday!&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Here, no pair has woken homer together for more than one day and they have not been quiet for three consecutive days.&lt;br /&gt;
&lt;br /&gt;
Here is how I solved this problem.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) I created a chart for Saturday throughout Sunday (similar to the one above) and filled in the provided information&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) For the salesman to be noisy for at least once in three consecutive days, he must have had to be noisy on Friday&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;3)Now, we need one day for Homer to rest undisturbed on Tuesday or Thursday. But for the dog and the construction to be noisy at least once in their three consecutive days, we can place the both of them on Thursday to satisfy the requirements. We can then determine that Homer had a happy sleep on Tuesday!&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56204</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56204"/>
		<updated>2010-10-19T22:53:55Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Question 3 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan&#039;s mother?&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
- Tosh owns a cat,&amp;lt;br&amp;gt;&lt;br /&gt;
- Bianca owns a frog that she loves,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jaela owns a parrot which keeps calling her &amp;quot;darling, darling&amp;quot;,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jun owns a snake, don&#039;t mess with him,&amp;lt;br&amp;gt;&lt;br /&gt;
- Suzan is the name of the frog,&amp;lt;br&amp;gt;&lt;br /&gt;
- The cat is named Jun,&amp;lt;br&amp;gt;&lt;br /&gt;
- The name by which they call the turtle is the name of the woman whose pet is Tosh,&amp;lt;br&amp;gt;&lt;br /&gt;
- Finally, Suzan&#039;s mother&#039;s pet is Bianca.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Tosh ---&amp;gt; Cat (Jun)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Bianca ---&amp;gt; Frog (Suzan)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♂Jun ---&amp;gt; Snake (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement &amp;quot;Bianca owns a frog that &#039;&#039;&#039;she&#039;&#039;&#039; loves&amp;quot; and we know that Jaela is also female from the statement &amp;quot;Jaela owns a parrot which keeps calling &#039;&#039;&#039;her&#039;&#039;&#039; &#039;darling, darling&#039;&amp;quot;. We also know that Jun is male from the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot;, with him referring to the owner and not the pet, otherwise it would be saying don&#039;t mess with &#039;&#039;&#039;it&#039;&#039;&#039;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can also assume that Suzan&#039;s pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan&#039;s pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle&#039;s name needs to be Suzan which is all very confusing and counter-intuitive. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (Tosh)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This piece of information also allows us to name Suzan&#039;s pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (Jaela)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That only leaves the last part of the problem, which would be Jun&#039;s pet:&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♂Jun ---&amp;gt; Snake (Bianca)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan&#039;s &#039;&#039;&#039;mother&#039;&#039;&#039; cannot possibly be male unless the label &amp;quot;mother&amp;quot; is semantically misleading (i.e. nickname, etc.).&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There are then two possible solutions after this error. The first is that the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot; has the pronoun &#039;him&#039; refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca). This solution ends up with Jun being Suzan&#039;s mother and the pet Bianca being her pet. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan&#039;s pet to be Tosh and lastly having Jaela&#039;s pet to be Bianca (and thus having Jaela as Suzan&#039;s mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution, and a more likely one then the previously aforementioned. In this case, it is not realistically possible to know who Suzan&#039;s mother is with the given information.&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;First, simplify the information:&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Bohao, Stewart, Dylan, Tim and Chan &lt;br /&gt;
&lt;br /&gt;
2 are left-handed, 3 are right-handed &lt;br /&gt;
&lt;br /&gt;
Bohnao and Dylan = Same Hand &lt;br /&gt;
&lt;br /&gt;
Tim and Chan = Different Hand&lt;br /&gt;
&lt;br /&gt;
2 are over 2m, 3 are under 2m &lt;br /&gt;
&lt;br /&gt;
Steward and Chan = Same Height &lt;br /&gt;
&lt;br /&gt;
Tim and Dylan = Different Height &lt;br /&gt;
&lt;br /&gt;
Need to find: player that is over 2 meters tall and left-handed&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw a diagram to help solve the first part of the problem (Left/Right Handed).&amp;lt;/u&amp;gt; &lt;br /&gt;
[[Image:Group Project 4 Question 2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project 4 Question 2p2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now, using these two diagrams find a player that is over 2 meters tall and left-handed&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
By process of elimination we know that it can&#039;t be: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao&#039;&#039;&#039; (right-handed)&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Dylan&#039;&#039;&#039; (right-handed) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Stewart&#039;&#039;&#039; (Under 2 meters) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Chan&#039;&#039;&#039;  (Under 2 meters)&lt;br /&gt;
&lt;br /&gt;
Therefore it COULD only be &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  Therefore the player that is over 2 meters tall and left-handed is &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?&lt;br /&gt;
&lt;br /&gt;
First we have to look at each piece of information that is given to us.&lt;br /&gt;
We know that there are 6 players, and that each player plays each other once in 5 days:&lt;br /&gt;
 - this means that there are 3 pairs playing against each other everyday&lt;br /&gt;
We are given enough information to determine a set of pairs for each day&lt;br /&gt;
 - using this information, we can determine the other four players and who they play during the 5 days.&lt;br /&gt;
&lt;br /&gt;
First we will group all the pairs that we know has played during the five days&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: CP&#039;&#039;&#039; We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 2: CJ&#039;&#039;&#039; (L, P, S, and F)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 3: JL&#039;&#039;&#039; (F, P, L, and S)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 4: PS&#039;&#039;&#039; (F, J, L, and C)&amp;lt;/p&amp;gt;&lt;br /&gt;
&amp;lt;p&amp;gt;&#039;&#039;&#039;Day 5: F ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: &amp;quot;CP&amp;quot;  &amp;quot;JS&amp;quot;  &amp;quot;LF&amp;quot;&lt;br /&gt;
Day 2: &amp;quot;CJ&amp;quot; &amp;quot;LP&amp;quot; &amp;quot;SF&amp;quot;&#039;&#039;&#039; (We cannot group F with L again because they have played each other on the first day)&lt;br /&gt;
&#039;&#039;&#039;Day 3: &amp;quot;JL&amp;quot;  &amp;quot;FP&amp;quot;  &amp;quot;LS&amp;quot;&#039;&#039;&#039; (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)&lt;br /&gt;
&#039;&#039;&#039;Day 4: &amp;quot;PS&amp;quot; &amp;quot;FJ&amp;quot; &amp;quot;LC&amp;quot;&#039;&#039;&#039;(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once) &lt;br /&gt;
The grouping for the following day is as follows:&lt;br /&gt;
&#039;&#039;&#039;Day 5: &amp;quot;PS&amp;quot; &amp;lt;u&amp;gt; &amp;quot;FJ&amp;quot; &amp;lt;/u&amp;gt; and &amp;quot;LC&amp;quot;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.&lt;br /&gt;
&lt;br /&gt;
Therefore, we can come to the conclusion that Fernanda played Janet on the last day. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
===Question 16===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Trying to figure out how many S (sons) and D (daughters) so I tried to represent the information in an expression.&lt;br /&gt;
&lt;br /&gt;
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and&lt;br /&gt;
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)&lt;br /&gt;
&lt;br /&gt;
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).&lt;br /&gt;
&lt;br /&gt;
===Question 17===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
If the scale read too low then we know that D’s real weight would be &amp;gt;60 kg and S’s real weight would be &amp;gt;50kg and so we would expect their combined weight to be &amp;gt;than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is &amp;lt;60 and S’s real weight is &amp;lt;50 then we would expect the outcome to be &amp;lt;their combined weight (110kg).&lt;br /&gt;
&lt;br /&gt;
===Question 18===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135. &lt;br /&gt;
&lt;br /&gt;
===Question 19===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;One morning each member of Angela&#039;s family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Angela’s cup can be expressed as:&lt;br /&gt;
¼ Total Milk + 1/6 Total Coffee = 8 oz&lt;br /&gt;
We can also say that:&lt;br /&gt;
Total Milk + Total Coffee/8  = # of family members  (because they each had an 8 oz cup)&lt;br /&gt;
&lt;br /&gt;
If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:&lt;br /&gt;
&lt;br /&gt;
8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn&#039;t really count as a family) that this works with.&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30) &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question #5===&lt;br /&gt;
&lt;br /&gt;
Answer: Homer sleeps late on Tuesday!&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Here, no pair has woken homer together for more than one day and they have not been quiet for three consecutive days.&lt;br /&gt;
&lt;br /&gt;
Here is how I solved this problem.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) I created a chart for Saturday throughout Sunday (similar to the one above) and filled in the provided information&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) For the salesman to be noisy for at least once in three consecutive days, he must have had to be noisy on Friday&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;3)Now, we need one day for Homer to rest undisturbed on Tuesday or Thursday. But for the dog and the construction to be noisy at least once in their three consecutive days, we can place the both of them on Thursday to satisfy the requirements. We can then determine that Homer had a happy sleep on Tuesday!&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56203</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=56203"/>
		<updated>2010-10-19T22:53:04Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Homework #4 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 13&lt;br /&gt;
| member 1 = [[User:VictoriaBass|Victoria Bass]]&lt;br /&gt;
| member 2 = [[User:MiguelCaruncho|Miguel Caruncho]]&lt;br /&gt;
| member 3 = [[User:Fiona|Fiona Ma]]&lt;br /&gt;
| member 4 = [[User:AdamsNguyen|Adam Nguyen]]&lt;br /&gt;
| member 5 = [[User:UnaVuckovic|Una Vuckovic]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
===Contact Information===&lt;br /&gt;
* Victoria Bass | email: bassvm@interchange.ubc.ca|&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic |&lt;br /&gt;
&lt;br /&gt;
I think you should add that above information to your user profiles instea. --  [[User:DavidKohler|DavidKohler]]]&lt;br /&gt;
----&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan&#039;s mother?&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
- Tosh owns a cat,&amp;lt;br&amp;gt;&lt;br /&gt;
- Bianca owns a frog that she loves,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jaela owns a parrot which keeps calling her &amp;quot;darling, darling&amp;quot;,&amp;lt;br&amp;gt;&lt;br /&gt;
- Jun owns a snake, don&#039;t mess with him,&amp;lt;br&amp;gt;&lt;br /&gt;
- Suzan is the name of the frog,&amp;lt;br&amp;gt;&lt;br /&gt;
- The cat is named Jun,&amp;lt;br&amp;gt;&lt;br /&gt;
- The name by which they call the turtle is the name of the woman whose pet is Tosh,&amp;lt;br&amp;gt;&lt;br /&gt;
- Finally, Suzan&#039;s mother&#039;s pet is Bianca.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Tosh ---&amp;gt; Cat (Jun)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Bianca ---&amp;gt; Frog (Suzan)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♂Jun ---&amp;gt; Snake (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (No name) &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement &amp;quot;Bianca owns a frog that &#039;&#039;&#039;she&#039;&#039;&#039; loves&amp;quot; and we know that Jaela is also female from the statement &amp;quot;Jaela owns a parrot which keeps calling &#039;&#039;&#039;her&#039;&#039;&#039; &#039;darling, darling&#039;&amp;quot;. We also know that Jun is male from the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot;, with him referring to the owner and not the pet, otherwise it would be saying don&#039;t mess with &#039;&#039;&#039;it&#039;&#039;&#039;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can also assume that Suzan&#039;s pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets.&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan&#039;s pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle&#039;s name needs to be Suzan which is all very confusing and counter-intuitive. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Jaela ---&amp;gt; Parrot (Tosh)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This piece of information also allows us to name Suzan&#039;s pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♀Suzan ---&amp;gt; Turtle (Jaela)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That only leaves the last part of the problem, which would be Jun&#039;s pet:&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
♂Jun ---&amp;gt; Snake (Bianca)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan&#039;s &#039;&#039;&#039;mother&#039;&#039;&#039; cannot possibly be male unless the label &amp;quot;mother&amp;quot; is semantically misleading (i.e. nickname, etc.).&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
There are then two possible solutions after this error. The first is that the statement &amp;quot;Jun owns a snake, don&#039;t mess with &#039;&#039;&#039;him&#039;&#039;&#039;&amp;quot; has the pronoun &#039;him&#039; refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca). This solution ends up with Jun being Suzan&#039;s mother and the pet Bianca being her pet. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan&#039;s pet to be Tosh and lastly having Jaela&#039;s pet to be Bianca (and thus having Jaela as Suzan&#039;s mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution, and a more likely one then the previously aforementioned. In this case, it is not realistically possible to know who Suzan&#039;s mother is with the given information.&lt;br /&gt;
&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;First, simplify the information:&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
Bohao, Stewart, Dylan, Tim and Chan &lt;br /&gt;
&lt;br /&gt;
2 are left-handed, 3 are right-handed &lt;br /&gt;
&lt;br /&gt;
Bohnao and Dylan = Same Hand &lt;br /&gt;
&lt;br /&gt;
Tim and Chan = Different Hand&lt;br /&gt;
&lt;br /&gt;
2 are over 2m, 3 are under 2m &lt;br /&gt;
&lt;br /&gt;
Steward and Chan = Same Height &lt;br /&gt;
&lt;br /&gt;
Tim and Dylan = Different Height &lt;br /&gt;
&lt;br /&gt;
Need to find: player that is over 2 meters tall and left-handed&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw a diagram to help solve the first part of the problem (Left/Right Handed).&amp;lt;/u&amp;gt; &lt;br /&gt;
[[Image:Group Project 4 Question 2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project 4 Question 2p2.JPG|center]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;u&amp;gt;Now, using these two diagrams find a player that is over 2 meters tall and left-handed&amp;lt;/u&amp;gt; &lt;br /&gt;
&lt;br /&gt;
By process of elimination we know that it can&#039;t be: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Bohao&#039;&#039;&#039; (right-handed)&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Dylan&#039;&#039;&#039; (right-handed) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Stewart&#039;&#039;&#039; (Under 2 meters) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Chan&#039;&#039;&#039;  (Under 2 meters)&lt;br /&gt;
&lt;br /&gt;
Therefore it COULD only be &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  Therefore the player that is over 2 meters tall and left-handed is &#039;&#039;&#039;TIM&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?&lt;br /&gt;
&lt;br /&gt;
First we have to look at each piece of information that is given to us.&lt;br /&gt;
We know that there are 6 players, and that each player plays each other once in 5 days:&lt;br /&gt;
 - this means that there are 3 pairs playing against each other everyday&lt;br /&gt;
We are given enough information to determine a set of pairs for each day&lt;br /&gt;
 - using this information, we can determine the other four players and who they play during the 5 days.&lt;br /&gt;
&lt;br /&gt;
First we will group all the pairs that we know has played during the five days&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: CP&#039;&#039;&#039; We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)&lt;br /&gt;
&#039;&#039;&#039;Day 2: CJ&#039;&#039;&#039; (L, P, S, and F)&lt;br /&gt;
&#039;&#039;&#039;Day 3: JL&#039;&#039;&#039; (F, P, L, and S)&lt;br /&gt;
&#039;&#039;&#039;Day 4: PS&#039;&#039;&#039; (F, J, L, and C)&lt;br /&gt;
&#039;&#039;&#039;Day 5: F ?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Day 1: &amp;quot;CP&amp;quot;  &amp;quot;JS&amp;quot;  &amp;quot;LF&amp;quot;&lt;br /&gt;
Day 2: &amp;quot;CJ&amp;quot; &amp;quot;LP&amp;quot; &amp;quot;SF&amp;quot;&#039;&#039;&#039; (We cannot group F with L again because they have played each other on the first day)&lt;br /&gt;
&#039;&#039;&#039;Day 3: &amp;quot;JL&amp;quot;  &amp;quot;FP&amp;quot;  &amp;quot;LS&amp;quot;&#039;&#039;&#039; (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)&lt;br /&gt;
&#039;&#039;&#039;Day 4: &amp;quot;PS&amp;quot; &amp;quot;FJ&amp;quot; &amp;quot;LC&amp;quot;&#039;&#039;&#039;(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once) &lt;br /&gt;
The grouping for the following day is as follows:&lt;br /&gt;
&#039;&#039;&#039;Day 5: &amp;quot;PS&amp;quot; &amp;lt;u&amp;gt; &amp;quot;FJ&amp;quot; &amp;lt;/u&amp;gt; and &amp;quot;LC&amp;quot;&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.&lt;br /&gt;
&lt;br /&gt;
Therefore, we can come to the conclusion that Fernanda played Janet on the last day. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
=Homework #3=&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
===Question 16===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Trying to figure out how many S (sons) and D (daughters) so I tried to represent the information in an expression.&lt;br /&gt;
&lt;br /&gt;
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and&lt;br /&gt;
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)&lt;br /&gt;
&lt;br /&gt;
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).&lt;br /&gt;
&lt;br /&gt;
===Question 17===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain.&amp;quot;&lt;br /&gt;
&lt;br /&gt;
If the scale read too low then we know that D’s real weight would be &amp;gt;60 kg and S’s real weight would be &amp;gt;50kg and so we would expect their combined weight to be &amp;gt;than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is &amp;lt;60 and S’s real weight is &amp;lt;50 then we would expect the outcome to be &amp;lt;their combined weight (110kg).&lt;br /&gt;
&lt;br /&gt;
===Question 18===&lt;br /&gt;
&lt;br /&gt;
&amp;quot; Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135. &lt;br /&gt;
&lt;br /&gt;
===Question 19===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;One morning each member of Angela&#039;s family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Angela’s cup can be expressed as:&lt;br /&gt;
¼ Total Milk + 1/6 Total Coffee = 8 oz&lt;br /&gt;
We can also say that:&lt;br /&gt;
Total Milk + Total Coffee/8  = # of family members  (because they each had an 8 oz cup)&lt;br /&gt;
&lt;br /&gt;
If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:&lt;br /&gt;
&lt;br /&gt;
8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn&#039;t really count as a family) that this works with.&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&amp;quot;Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30) &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;br /&gt;
&lt;br /&gt;
=Homework #4=&lt;br /&gt;
&lt;br /&gt;
===Question #5===&lt;br /&gt;
&lt;br /&gt;
Answer: Homer sleeps late on Tuesday!&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Here, no pair has woken homer together for more than one day and they have not been quiet for three consecutive days.&lt;br /&gt;
&lt;br /&gt;
Here is how I solved this problem.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1) I created a chart for Saturday throughout Sunday (similar to the one above) and filled in the provided information&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;2) For the salesman to be noisy for at least once in three consecutive days, he must have had to be noisy on Friday&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;3)Now, we need one day for Homer to rest undisturbed on Tuesday or Thursday. But for the dog and the construction to be noisy at least once in their three consecutive days, we can place the both of them on Thursday to satisfy the requirements. We can then determine that Homer had a happy sleep on Tuesday!&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
Sat - Salesman&amp;lt;br&amp;gt;&lt;br /&gt;
Sun - Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Mon - Salesman and Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Tuesday - &#039;&#039;&#039;Homer sleeps late!&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Wednesday - Salesman and Dog&amp;lt;br&amp;gt;&lt;br /&gt;
Thursday - &#039;&#039;&#039;Construction and Dog&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Friday - &#039;&#039;&#039;Salesman&#039;&#039;&#039;&amp;lt;br&amp;gt;&lt;br /&gt;
Saturday - Construction&amp;lt;br&amp;gt;&lt;br /&gt;
Sunday - Dog&amp;lt;br&amp;gt;&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=54313</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=54313"/>
		<updated>2010-10-13T05:58:06Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Question 14 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Victoria Bass&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Fiona Ma&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic | email: v_una@hotmail.com | cell: 778 829 8862&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
=Problems= &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=54311</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=54311"/>
		<updated>2010-10-13T05:57:40Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Question 14 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Victoria Bass&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Fiona Ma&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic | email: v_una@hotmail.com | cell: 778 829 8862&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
=Problems= &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i).  First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii).  Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.&lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=54309</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_13&amp;diff=54309"/>
		<updated>2010-10-13T05:56:56Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Victoria Bass&lt;br /&gt;
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242&lt;br /&gt;
* Fiona Ma&lt;br /&gt;
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987&lt;br /&gt;
* Una Vuckovic | email: v_una@hotmail.com | cell: 778 829 8862&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don&#039;t have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!&lt;br /&gt;
&lt;br /&gt;
ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
=Problems= &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 1===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain. &lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_1.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min = 1hour&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;60min + 20min = 80min&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1hour + 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
  Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same &lt;br /&gt;
  as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because... &lt;br /&gt;
   &#039;&#039;&#039;1hour and 20min = 80min&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 2===&lt;br /&gt;
&lt;br /&gt;
A lady did not have her driver&#039;s license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_2.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The simple explanation to this question is that the &#039;&#039;&#039;lady is a pedestrian&#039;&#039;&#039;. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
     Therefore the witness policeman did not stop her because she did NOT break the law.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
===Question 3===&lt;br /&gt;
&lt;br /&gt;
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain. &lt;br /&gt;
[[Image:Group Project question -3.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The problem can be solved by taking out the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. If an APPLE is pulled&lt;br /&gt;
out of Box #1, you know that the contents of &#039;&#039;&#039;Box #1&#039;&#039;&#039; is just APPLES. This is because the two options &lt;br /&gt;
(labelled in blue) are either just APPLES or just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Given that &#039;&#039;&#039;Box #1&#039;&#039;&#039; has only APPLES you know that &#039;&#039;&#039;Box #3&#039;&#039;&#039; has APPLES &amp;amp; ORANGES. This is because the two&lt;br /&gt;
options are just APPLES or APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #1&#039;&#039;&#039; already contains just APPLES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
By process of elimination &#039;&#039;&#039;Box #2&#039;&#039;&#039; contains just oranges. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The same logic applies if an ORANGE is initially pulled out of &#039;&#039;&#039;Box #1&#039;&#039;&#039;. Then &#039;&#039;&#039;Box #2&#039;&#039;&#039; would contain &lt;br /&gt;
APPLES &amp;amp; ORANGES and &#039;&#039;&#039;Box #3&#039;&#039;&#039; would contain just ORANGES. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
       Therefore the solution can be solved by opening &#039;&#039;&#039;Box #1&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 4===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am the brother of the blind fiddler, but brothers I have none. How can this be?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group_Project_question_4.JPG‎]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    The key to this question is to avoid gender bias. The logical solution is that the boy is the brother &lt;br /&gt;
                       of a blind fiddler, who is his sister, therefore he has no brothers.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 5===&lt;br /&gt;
&lt;br /&gt;
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Group Project question 5.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
    Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.&lt;br /&gt;
&lt;br /&gt;
===Question 6===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&amp;lt;br&amp;gt;If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 7===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i) pair of the same colour&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii) a pair with different colours?&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.wholesalefootballkits.com/images/navyblue_socks.jpg&lt;br /&gt;
&lt;br /&gt;
===Question 8===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.&#039;&#039;&#039; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If Reuben&#039;s birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Dec 30th 2010 - 20 years old &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 - turns 21 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. &amp;lt;br&amp;gt;&lt;br /&gt;
Dec 31st 2011 (same year) - turns 22 &amp;lt;br&amp;gt;&lt;br /&gt;
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Question 9===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.&lt;br /&gt;
&lt;br /&gt;
===Question 10===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;10. Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i and ii&#039;&#039;&#039;)&amp;lt;br&amp;gt;&amp;lt;br&amp;gt; It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.&lt;br /&gt;
&lt;br /&gt;
===Question 11===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;11. A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?&lt;br /&gt;
&lt;br /&gt;
First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible. &lt;br /&gt;
&lt;br /&gt;
===Question 12===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;12. A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man&#039;s Bronx girlfriend left him because he rarely visited. Give a (logical) explanation. &lt;br /&gt;
&lt;br /&gt;
Because each of the trains arrive 10 min apart, we can come up with a train schedule:&lt;br /&gt;
Bronx train arrives at 10:10, 10:20, 10:30 etc.&lt;br /&gt;
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.&lt;br /&gt;
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 13===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;13. If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)? &lt;br /&gt;
&lt;br /&gt;
For each 5 seconds the clock chimes 5 times with equal lengths of space in between. &lt;br /&gt;
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5. &lt;br /&gt;
5 chimes / 4 spaces = 1.25seconds&lt;br /&gt;
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes. &lt;br /&gt;
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 14===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;14. One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;i). First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.&lt;br /&gt;
AB = ABCD, ABDC&lt;br /&gt;
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up &lt;br /&gt;
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;ii). Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly. &lt;br /&gt;
&lt;br /&gt;
===Question 15===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;15. Alex says to you, “I&#039;ll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet? &lt;br /&gt;
&lt;br /&gt;
Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards. &lt;br /&gt;
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).&lt;br /&gt;
&lt;br /&gt;
-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards&lt;br /&gt;
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.&lt;br /&gt;
&lt;br /&gt;
We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 21===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?&#039;&#039;&#039; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First we have to recognise Sven&#039;s place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We also know that Sven came in before Dan, who is 10th. Thus giving the definition: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Sven &amp;lt; 10 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Total = 2(Sven) - 1 &amp;gt; 16 &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 22===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement &amp;quot;every rainy afternoon was preceded by a sunny morning&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement &amp;quot;when it rained in the morning, the afternoon was sunny&amp;quot;. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let z be the number of days where it didn&#039;t rain at all. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From these definitions we can come up with equations to represent these pieces of information.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For one we can say that &amp;lt;math&amp;gt; x + 7 = 11 &amp;lt;/math&amp;gt; since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next we can say that &amp;lt;math&amp;gt; y + z = 13 &amp;lt;/math&amp;gt; since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly we can say that &amp;lt;math&amp;gt; x + y = 13 &amp;lt;/math&amp;gt; since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then we can simply solve the system of equations. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x = 13 - y &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; x + z = 11 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 13 - y + z = 11 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt; y + z = 12 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 12 - 11 = y + z - (13 - y + z)&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = y + z - 13 + y - z&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 1 = 2y - 13&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; y = 7 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + x = 13&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; x = 6&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; y + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; 7 + z = 12&amp;lt;/math&amp;gt; --&amp;gt; &amp;lt;math&amp;gt; z = 5&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus &amp;lt;math&amp;gt; x + y + z = 7 + 6 + 5 = 18 &amp;lt;/math&amp;gt;, with 18 being the length of the entire vacation.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 23===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today&#039;s date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula&#039;s children? Explain.&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To start off, you can easily list down all of the 3-set factors of 36 which are: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
{ ( 36, 1, 1) &amp;lt;br&amp;gt;&lt;br /&gt;
(18, 2, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(12, 3, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 4, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(9, 2, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 6, 1)&amp;lt;br&amp;gt;&lt;br /&gt;
(6, 3, 2)&amp;lt;br&amp;gt;&lt;br /&gt;
(4, 3, 3) } &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The next clue is that the sum of these ages would equal today&#039;s date, thus the sum of their ages cannot exceed 31. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Adding all the ages of the 3-set factors would yield the following: &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
36 + 1 + 1 = 38 &amp;lt;br&amp;gt;&lt;br /&gt;
18 + 2 + 1 = 21 &amp;lt;br&amp;gt;&lt;br /&gt;
12 + 3 + 1 = 16&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 4 + 1 = 14&amp;lt;br&amp;gt;&lt;br /&gt;
9 + 2 + 2 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 6 + 1 = 13&amp;lt;br&amp;gt;&lt;br /&gt;
6 + 3 + 2 = 11&amp;lt;br&amp;gt;&lt;br /&gt;
4 + 3 + 3 = 10&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like &amp;quot;my oldest children have red hair&amp;quot;. Thus leaving only the set of (9, 2, 2) left.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 24===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?&#039;&#039;&#039; &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((6-t)/6)&amp;lt;/math&amp;gt;      with t being the number of hours elapsed since the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The second candle&#039;s equation is as follows: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;L=((3-t)/3)&amp;lt;/math&amp;gt;      with t again being the number of hours after the candle was lit &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;((6-1)/6) = 5/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-1)/6) = 2/3&amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;((6-2)/6) = 4/6&amp;lt;/math&amp;gt; &amp;lt;----&amp;gt; &amp;lt;math&amp;gt;((3-2)/6) = 1/3&amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; 4/6 = 2/3&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; 2/3 = 2 (1/3)&amp;lt;/math&amp;gt; with 1/3 being the length of the second candle. &amp;lt;br&amp;gt; &amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus giving the answer as after 2 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Question 25===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.&#039;&#039;&#039;&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since when L is at half its length, namely &amp;lt;math&amp;gt; (1/2)L&amp;lt;/math&amp;gt;, it equals 4 or 4 hours after 0, then we can put up the equation: &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (1/2) L = 4 &amp;lt;/math&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt; L = 8 &amp;lt;/math&amp;gt; &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
L then equals 8.&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Homework/2&amp;diff=51575</id>
		<title>Course:MATH110/Archive/2010-2011/003/Homework/2</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Homework/2&amp;diff=51575"/>
		<updated>2010-10-02T05:55:22Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Need some help for the homework? Ask your questions here. --[[User:DavidKohler|DavidKohler]]&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
help! how do you do this problem:&lt;br /&gt;
The domain of the function h(x)=(x+11)^2(2x-12)^{1/4} is &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
thanks!&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Lectures/Lecture_7&amp;diff=48874</id>
		<title>Course:MATH110/Archive/2010-2011/003/Lectures/Lecture 7</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Lectures/Lecture_7&amp;diff=48874"/>
		<updated>2010-09-22T06:12:00Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Your questions for Lecture 7, September 22 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Your questions for Lecture 7, September 22=&lt;br /&gt;
&lt;br /&gt;
Please add here any questions or topics that you would like to discuss in class. Make sure to separate each questions by a horizontal line (just type &amp;lt;nowiki&amp;gt;----&amp;lt;/nowiki&amp;gt; in between them)&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
I was just looking through the sample for the basic math skills test, and was wondering if we would have anything about composite functions on the test. I was wondering how one would go about solving a question along these lines: &lt;br /&gt;
&lt;br /&gt;
Given the composite function &amp;lt;math&amp;gt;f(g(x)) = 1/(x2/4)&amp;lt;/math&amp;gt;, find f(x) and g(x).&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
--[[User:BenJeffery|BenJeffery]] 03:53, 22 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
There were two WebWorks questions that I didn&#039;t know how to solve,&lt;br /&gt;
&lt;br /&gt;
The first one is : P(x)= &amp;lt;math&amp;gt;x^{4/3}&amp;lt;/math&amp;gt; - &amp;lt;math&amp;gt;7x^{2/3}&amp;lt;/math&amp;gt;+6 ; solve for x&lt;br /&gt;
&lt;br /&gt;
The second is : &amp;lt;math&amp;gt; x^2 &amp;lt;/math&amp;gt; / x+13=14 ; solve for x.&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Lectures/Lecture_7&amp;diff=48873</id>
		<title>Course:MATH110/Archive/2010-2011/003/Lectures/Lecture 7</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Lectures/Lecture_7&amp;diff=48873"/>
		<updated>2010-09-22T06:11:35Z</updated>

		<summary type="html">&lt;p&gt;Fiona: /* Your questions for Lecture 7, September 22 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Your questions for Lecture 7, September 22=&lt;br /&gt;
&lt;br /&gt;
Please add here any questions or topics that you would like to discuss in class. Make sure to separate each questions by a horizontal line (just type &amp;lt;nowiki&amp;gt;----&amp;lt;/nowiki&amp;gt; in between them)&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
I was just looking through the sample for the basic math skills test, and was wondering if we would have anything about composite functions on the test. I was wondering how one would go about solving a question along these lines: &lt;br /&gt;
&lt;br /&gt;
Given the composite function &amp;lt;math&amp;gt;f(g(x)) = 1/(x2/4)&amp;lt;/math&amp;gt;, find f(x) and g(x).&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
--[[User:BenJeffery|BenJeffery]] 03:53, 22 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
There were two WebWorks questions that I didn&#039;t know how to solve,&lt;br /&gt;
&lt;br /&gt;
The first one is : P(x)= &amp;lt;math&amp;gt;x^{4/3}&amp;lt;/math&amp;gt; - &amp;lt;math&amp;gt;7x^{2/3}&amp;lt;/math&amp;gt;+6&lt;br /&gt;
&lt;br /&gt;
The second is : &amp;lt;math&amp;gt; x^2 &amp;lt;/math&amp;gt; / x+13=14 ; solve for x.&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48872</id>
		<title>Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48872"/>
		<updated>2010-09-22T06:11:10Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Question 17==&lt;br /&gt;
&lt;br /&gt;
Hi everyone! how do i submit the answer to question 17n if it has a square root on it????&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Put sqrt right before the rest...For instance, sqrt(3V/h) for &amp;lt;math&amp;gt;\sqrt{3V\over h}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
I&#039;m still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
Try to use exponent notation instead of the square root. So write &amp;lt;math&amp;gt;x^{1/2}&amp;lt;/math&amp;gt; instead of &amp;lt;math&amp;gt;\sqrt{x}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide &amp;lt;math&amp;gt;\frac {x^{1}}{2}&amp;lt;/math&amp;gt;. Took me a few tries to figure this one out.&lt;br /&gt;
&lt;br /&gt;
[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
==Question 26==&lt;br /&gt;
&lt;br /&gt;
Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{6\over x-1}-{6\over x}\geqq 1&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I still believe the correct answer is &amp;lt;math&amp;gt;(-\infty, -2]\cup[3,\infty)&amp;lt;/math&amp;gt; but that and a myriad of other solutions I tried were apparently all wrong. I&#039;d be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]&lt;br /&gt;
&lt;br /&gt;
==Question 15==&lt;br /&gt;
&lt;br /&gt;
Hi guys,&lt;br /&gt;
&lt;br /&gt;
I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1). &lt;br /&gt;
&lt;br /&gt;
I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.&lt;br /&gt;
&lt;br /&gt;
[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]&lt;br /&gt;
&lt;br /&gt;
===Re: Question 15===&lt;br /&gt;
&lt;br /&gt;
Hi Arabella,&lt;br /&gt;
&lt;br /&gt;
This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are &#039;&#039;A (7,6) and B (13,0)&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
I didn&#039;t use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you&#039;re used to: &lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
((y2+y1)/(2)) , ((x2+x1)/(2))&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different): &lt;br /&gt;
&lt;br /&gt;
((0+6))/(2)) , ((13+7)/(2)) &lt;br /&gt;
= (y,x)&lt;br /&gt;
Therefore midpoint &#039;&#039;&#039;(x,y) = (10,3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3=1(10) + b&lt;br /&gt;
Now I just isolate b and solve:&lt;br /&gt;
b=7 &lt;br /&gt;
&lt;br /&gt;
Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7&lt;br /&gt;
&lt;br /&gt;
I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)&lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hi Steffany,&lt;br /&gt;
&lt;br /&gt;
Your explanation makes totally sense! I was omitting the midpoint part! &lt;br /&gt;
&lt;br /&gt;
Thank you. Arabella.&lt;br /&gt;
&lt;br /&gt;
==Question 7==&lt;br /&gt;
&lt;br /&gt;
Hello, I&#039;m struggling on question 7 on our first webwork assignment. Here is the question: &lt;br /&gt;
&lt;br /&gt;
Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).&lt;br /&gt;
&lt;br /&gt;
So what I did was that I first used the distance formula for point (0,b) with point (2,2): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(2-0)^2+(2-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and with point (6,-5): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(6-0)^2+(-5-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and then I made them equal to each other to solve for b.&lt;br /&gt;
&lt;br /&gt;
This simplifies to &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;2+(2-b)=6+(-5-b)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b is then found to be &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
But when I plugged in &amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt; into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?&lt;br /&gt;
&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Adams, I think you&#039;ve got the sets of points wrong. The question is supposed to be &amp;quot;Find the point (0, b) on the y-axis that is equidistant from the points &amp;lt;b&amp;gt;(3, 3)&amp;lt;/b&amp;gt; and &amp;lt;b&amp;gt;(5, -4)&amp;lt;/b&amp;gt;. I got the right answer with the method you used, so it&#039;s just a matter of using the correct data!&lt;br /&gt;
&lt;br /&gt;
[[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
The PDF Version that I printed out says (2,2) and (6,-5) :(&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hello guys,&lt;br /&gt;
&lt;br /&gt;
For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3). &lt;br /&gt;
&lt;br /&gt;
I tried Adam&#039;s method but my answer is not a solution!&lt;br /&gt;
&lt;br /&gt;
I initially, I worked as follows though:&lt;br /&gt;
&lt;br /&gt;
-I found the slope m= -3-3/ 4-3= -6/1&lt;br /&gt;
&lt;br /&gt;
-Then I equated using the formula y-b=m(x-0)&lt;br /&gt;
&lt;br /&gt;
for point (3, 3) &lt;br /&gt;
&lt;br /&gt;
3-b=-6(3-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
for point (4, -3)&lt;br /&gt;
&lt;br /&gt;
-3-b=-6(4-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
However, like Adam, the program is not accepting my answer. I&#039;m not sure what I&#039;m doing wrong. Thank you.&lt;br /&gt;
&lt;br /&gt;
Arabella.&lt;br /&gt;
&lt;br /&gt;
===Re: Question 7===&lt;br /&gt;
Hi everyone,&lt;br /&gt;
&lt;br /&gt;
We must all be given different values for the same question. &lt;br /&gt;
&lt;br /&gt;
Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point &#039;&#039;equidistant&#039;&#039; from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying. &lt;br /&gt;
&lt;br /&gt;
Good Luck! &lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steffany, &lt;br /&gt;
&lt;br /&gt;
I am following what you&#039;re saying about using the distance formula but when I went to solve for &#039;b&#039; by putting them equal to one another, I get confused. Can you step me through your simplifying?&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
LaBri Krahn --&lt;br /&gt;
&lt;br /&gt;
Hi,&lt;br /&gt;
&lt;br /&gt;
When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow. &lt;br /&gt;
&lt;br /&gt;
Steffany&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48871</id>
		<title>Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48871"/>
		<updated>2010-09-22T06:09:39Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==WebWorks Questions==&lt;br /&gt;
There were two WebWorks questions that I didn&#039;t know how to solve and am hoping Prof. Kholer will go over in class tomorrow.&lt;br /&gt;
&lt;br /&gt;
The first one is : P(x)= &amp;lt;math&amp;gt;x^{4/3}&amp;lt;/math&amp;gt; - &amp;lt;math&amp;gt;7x^{2/3}&amp;lt;/math&amp;gt;+6&lt;br /&gt;
&lt;br /&gt;
The second is : &amp;lt;math&amp;gt; x^2 &amp;lt;/math&amp;gt; / x+13=14 ; solve for x.&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Question 17==&lt;br /&gt;
&lt;br /&gt;
Hi everyone! how do i submit the answer to question 17n if it has a square root on it????&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Put sqrt right before the rest...For instance, sqrt(3V/h) for &amp;lt;math&amp;gt;\sqrt{3V\over h}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
I&#039;m still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
Try to use exponent notation instead of the square root. So write &amp;lt;math&amp;gt;x^{1/2}&amp;lt;/math&amp;gt; instead of &amp;lt;math&amp;gt;\sqrt{x}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide &amp;lt;math&amp;gt;\frac {x^{1}}{2}&amp;lt;/math&amp;gt;. Took me a few tries to figure this one out.&lt;br /&gt;
&lt;br /&gt;
[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
==Question 26==&lt;br /&gt;
&lt;br /&gt;
Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{6\over x-1}-{6\over x}\geqq 1&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I still believe the correct answer is &amp;lt;math&amp;gt;(-\infty, -2]\cup[3,\infty)&amp;lt;/math&amp;gt; but that and a myriad of other solutions I tried were apparently all wrong. I&#039;d be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]&lt;br /&gt;
&lt;br /&gt;
==Question 15==&lt;br /&gt;
&lt;br /&gt;
Hi guys,&lt;br /&gt;
&lt;br /&gt;
I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1). &lt;br /&gt;
&lt;br /&gt;
I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.&lt;br /&gt;
&lt;br /&gt;
[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]&lt;br /&gt;
&lt;br /&gt;
===Re: Question 15===&lt;br /&gt;
&lt;br /&gt;
Hi Arabella,&lt;br /&gt;
&lt;br /&gt;
This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are &#039;&#039;A (7,6) and B (13,0)&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
I didn&#039;t use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you&#039;re used to: &lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
((y2+y1)/(2)) , ((x2+x1)/(2))&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different): &lt;br /&gt;
&lt;br /&gt;
((0+6))/(2)) , ((13+7)/(2)) &lt;br /&gt;
= (y,x)&lt;br /&gt;
Therefore midpoint &#039;&#039;&#039;(x,y) = (10,3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3=1(10) + b&lt;br /&gt;
Now I just isolate b and solve:&lt;br /&gt;
b=7 &lt;br /&gt;
&lt;br /&gt;
Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7&lt;br /&gt;
&lt;br /&gt;
I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)&lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hi Steffany,&lt;br /&gt;
&lt;br /&gt;
Your explanation makes totally sense! I was omitting the midpoint part! &lt;br /&gt;
&lt;br /&gt;
Thank you. Arabella.&lt;br /&gt;
&lt;br /&gt;
==Question 7==&lt;br /&gt;
&lt;br /&gt;
Hello, I&#039;m struggling on question 7 on our first webwork assignment. Here is the question: &lt;br /&gt;
&lt;br /&gt;
Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).&lt;br /&gt;
&lt;br /&gt;
So what I did was that I first used the distance formula for point (0,b) with point (2,2): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(2-0)^2+(2-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and with point (6,-5): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(6-0)^2+(-5-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and then I made them equal to each other to solve for b.&lt;br /&gt;
&lt;br /&gt;
This simplifies to &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;2+(2-b)=6+(-5-b)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b is then found to be &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
But when I plugged in &amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt; into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?&lt;br /&gt;
&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Adams, I think you&#039;ve got the sets of points wrong. The question is supposed to be &amp;quot;Find the point (0, b) on the y-axis that is equidistant from the points &amp;lt;b&amp;gt;(3, 3)&amp;lt;/b&amp;gt; and &amp;lt;b&amp;gt;(5, -4)&amp;lt;/b&amp;gt;. I got the right answer with the method you used, so it&#039;s just a matter of using the correct data!&lt;br /&gt;
&lt;br /&gt;
[[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
The PDF Version that I printed out says (2,2) and (6,-5) :(&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hello guys,&lt;br /&gt;
&lt;br /&gt;
For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3). &lt;br /&gt;
&lt;br /&gt;
I tried Adam&#039;s method but my answer is not a solution!&lt;br /&gt;
&lt;br /&gt;
I initially, I worked as follows though:&lt;br /&gt;
&lt;br /&gt;
-I found the slope m= -3-3/ 4-3= -6/1&lt;br /&gt;
&lt;br /&gt;
-Then I equated using the formula y-b=m(x-0)&lt;br /&gt;
&lt;br /&gt;
for point (3, 3) &lt;br /&gt;
&lt;br /&gt;
3-b=-6(3-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
for point (4, -3)&lt;br /&gt;
&lt;br /&gt;
-3-b=-6(4-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
However, like Adam, the program is not accepting my answer. I&#039;m not sure what I&#039;m doing wrong. Thank you.&lt;br /&gt;
&lt;br /&gt;
Arabella.&lt;br /&gt;
&lt;br /&gt;
===Re: Question 7===&lt;br /&gt;
Hi everyone,&lt;br /&gt;
&lt;br /&gt;
We must all be given different values for the same question. &lt;br /&gt;
&lt;br /&gt;
Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point &#039;&#039;equidistant&#039;&#039; from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying. &lt;br /&gt;
&lt;br /&gt;
Good Luck! &lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steffany, &lt;br /&gt;
&lt;br /&gt;
I am following what you&#039;re saying about using the distance formula but when I went to solve for &#039;b&#039; by putting them equal to one another, I get confused. Can you step me through your simplifying?&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
LaBri Krahn --&lt;br /&gt;
&lt;br /&gt;
Hi,&lt;br /&gt;
&lt;br /&gt;
When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow. &lt;br /&gt;
&lt;br /&gt;
Steffany&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48870</id>
		<title>Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48870"/>
		<updated>2010-09-22T06:07:41Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==WebWorks Questions==&lt;br /&gt;
There were two WebWorks questions that I didn&#039;t know how to solve and am hoping Prof. Kholer will go over in class tomorrow.&lt;br /&gt;
&lt;br /&gt;
The first one is : P(x)= x^(4/3)-7x^(2/3)+6&lt;br /&gt;
&lt;br /&gt;
The second is : &amp;lt;math&amp;gt; x^2 &amp;lt;/math&amp;gt; / x+13=14 ; solve for x.&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Question 17==&lt;br /&gt;
&lt;br /&gt;
Hi everyone! how do i submit the answer to question 17n if it has a square root on it????&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Put sqrt right before the rest...For instance, sqrt(3V/h) for &amp;lt;math&amp;gt;\sqrt{3V\over h}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
I&#039;m still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
Try to use exponent notation instead of the square root. So write &amp;lt;math&amp;gt;x^{1/2}&amp;lt;/math&amp;gt; instead of &amp;lt;math&amp;gt;\sqrt{x}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide &amp;lt;math&amp;gt;\frac {x^{1}}{2}&amp;lt;/math&amp;gt;. Took me a few tries to figure this one out.&lt;br /&gt;
&lt;br /&gt;
[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
==Question 26==&lt;br /&gt;
&lt;br /&gt;
Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{6\over x-1}-{6\over x}\geqq 1&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I still believe the correct answer is &amp;lt;math&amp;gt;(-\infty, -2]\cup[3,\infty)&amp;lt;/math&amp;gt; but that and a myriad of other solutions I tried were apparently all wrong. I&#039;d be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]&lt;br /&gt;
&lt;br /&gt;
==Question 15==&lt;br /&gt;
&lt;br /&gt;
Hi guys,&lt;br /&gt;
&lt;br /&gt;
I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1). &lt;br /&gt;
&lt;br /&gt;
I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.&lt;br /&gt;
&lt;br /&gt;
[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]&lt;br /&gt;
&lt;br /&gt;
===Re: Question 15===&lt;br /&gt;
&lt;br /&gt;
Hi Arabella,&lt;br /&gt;
&lt;br /&gt;
This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are &#039;&#039;A (7,6) and B (13,0)&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
I didn&#039;t use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you&#039;re used to: &lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
((y2+y1)/(2)) , ((x2+x1)/(2))&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different): &lt;br /&gt;
&lt;br /&gt;
((0+6))/(2)) , ((13+7)/(2)) &lt;br /&gt;
= (y,x)&lt;br /&gt;
Therefore midpoint &#039;&#039;&#039;(x,y) = (10,3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3=1(10) + b&lt;br /&gt;
Now I just isolate b and solve:&lt;br /&gt;
b=7 &lt;br /&gt;
&lt;br /&gt;
Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7&lt;br /&gt;
&lt;br /&gt;
I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)&lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hi Steffany,&lt;br /&gt;
&lt;br /&gt;
Your explanation makes totally sense! I was omitting the midpoint part! &lt;br /&gt;
&lt;br /&gt;
Thank you. Arabella.&lt;br /&gt;
&lt;br /&gt;
==Question 7==&lt;br /&gt;
&lt;br /&gt;
Hello, I&#039;m struggling on question 7 on our first webwork assignment. Here is the question: &lt;br /&gt;
&lt;br /&gt;
Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).&lt;br /&gt;
&lt;br /&gt;
So what I did was that I first used the distance formula for point (0,b) with point (2,2): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(2-0)^2+(2-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and with point (6,-5): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(6-0)^2+(-5-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and then I made them equal to each other to solve for b.&lt;br /&gt;
&lt;br /&gt;
This simplifies to &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;2+(2-b)=6+(-5-b)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b is then found to be &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
But when I plugged in &amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt; into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?&lt;br /&gt;
&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Adams, I think you&#039;ve got the sets of points wrong. The question is supposed to be &amp;quot;Find the point (0, b) on the y-axis that is equidistant from the points &amp;lt;b&amp;gt;(3, 3)&amp;lt;/b&amp;gt; and &amp;lt;b&amp;gt;(5, -4)&amp;lt;/b&amp;gt;. I got the right answer with the method you used, so it&#039;s just a matter of using the correct data!&lt;br /&gt;
&lt;br /&gt;
[[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
The PDF Version that I printed out says (2,2) and (6,-5) :(&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hello guys,&lt;br /&gt;
&lt;br /&gt;
For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3). &lt;br /&gt;
&lt;br /&gt;
I tried Adam&#039;s method but my answer is not a solution!&lt;br /&gt;
&lt;br /&gt;
I initially, I worked as follows though:&lt;br /&gt;
&lt;br /&gt;
-I found the slope m= -3-3/ 4-3= -6/1&lt;br /&gt;
&lt;br /&gt;
-Then I equated using the formula y-b=m(x-0)&lt;br /&gt;
&lt;br /&gt;
for point (3, 3) &lt;br /&gt;
&lt;br /&gt;
3-b=-6(3-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
for point (4, -3)&lt;br /&gt;
&lt;br /&gt;
-3-b=-6(4-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
However, like Adam, the program is not accepting my answer. I&#039;m not sure what I&#039;m doing wrong. Thank you.&lt;br /&gt;
&lt;br /&gt;
Arabella.&lt;br /&gt;
&lt;br /&gt;
===Re: Question 7===&lt;br /&gt;
Hi everyone,&lt;br /&gt;
&lt;br /&gt;
We must all be given different values for the same question. &lt;br /&gt;
&lt;br /&gt;
Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point &#039;&#039;equidistant&#039;&#039; from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying. &lt;br /&gt;
&lt;br /&gt;
Good Luck! &lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steffany, &lt;br /&gt;
&lt;br /&gt;
I am following what you&#039;re saying about using the distance formula but when I went to solve for &#039;b&#039; by putting them equal to one another, I get confused. Can you step me through your simplifying?&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
LaBri Krahn --&lt;br /&gt;
&lt;br /&gt;
Hi,&lt;br /&gt;
&lt;br /&gt;
When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow. &lt;br /&gt;
&lt;br /&gt;
Steffany&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48869</id>
		<title>Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_A1&amp;diff=48869"/>
		<updated>2010-09-22T06:05:35Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==WebWorks Questions==&lt;br /&gt;
There were two WebWorks questions that I didn&#039;t know how to solve and am hoping Prof. Kholer will go over in class tomorrow.&lt;br /&gt;
&lt;br /&gt;
The first one is : P(x)=&amp;lt;math&amp;gt; x^4/3 &amp;lt;/math&amp;gt;-&amp;lt;math&amp;gt; 7x^2/3 &amp;lt;/math&amp;gt;+6&lt;br /&gt;
&lt;br /&gt;
The second is : &amp;lt;math&amp;gt; x^2 &amp;lt;/math&amp;gt;/x+13=14 ; solve for x.&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Question 17==&lt;br /&gt;
&lt;br /&gt;
Hi everyone! how do i submit the answer to question 17n if it has a square root on it????&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Put sqrt right before the rest...For instance, sqrt(3V/h) for &amp;lt;math&amp;gt;\sqrt{3V\over h}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
I&#039;m still hoping someone can answer my question please :( [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
Try to use exponent notation instead of the square root. So write &amp;lt;math&amp;gt;x^{1/2}&amp;lt;/math&amp;gt; instead of &amp;lt;math&amp;gt;\sqrt{x}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
-[[User:DavidKohler|DavidKohler]] 05:58, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide &amp;lt;math&amp;gt;\frac {x^{1}}{2}&amp;lt;/math&amp;gt;. Took me a few tries to figure this one out.&lt;br /&gt;
&lt;br /&gt;
[[User:TrevorShumka|TrevorShumka]] 06:18, 20 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
==Question 26==&lt;br /&gt;
&lt;br /&gt;
Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{6\over x-1}-{6\over x}\geqq 1&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I still believe the correct answer is &amp;lt;math&amp;gt;(-\infty, -2]\cup[3,\infty)&amp;lt;/math&amp;gt; but that and a myriad of other solutions I tried were apparently all wrong. I&#039;d be grateful for suggestions on possible ways to solve this. [[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! [[User:JustineVallieres|JustineVallieres]]&lt;br /&gt;
&lt;br /&gt;
==Question 15==&lt;br /&gt;
&lt;br /&gt;
Hi guys,&lt;br /&gt;
&lt;br /&gt;
I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1). &lt;br /&gt;
&lt;br /&gt;
I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.&lt;br /&gt;
&lt;br /&gt;
[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]&lt;br /&gt;
&lt;br /&gt;
===Re: Question 15===&lt;br /&gt;
&lt;br /&gt;
Hi Arabella,&lt;br /&gt;
&lt;br /&gt;
This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are &#039;&#039;A (7,6) and B (13,0)&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
I didn&#039;t use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you&#039;re used to: &lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
((y2+y1)/(2)) , ((x2+x1)/(2))&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different): &lt;br /&gt;
&lt;br /&gt;
((0+6))/(2)) , ((13+7)/(2)) &lt;br /&gt;
= (y,x)&lt;br /&gt;
Therefore midpoint &#039;&#039;&#039;(x,y) = (10,3)&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
3=1(10) + b&lt;br /&gt;
Now I just isolate b and solve:&lt;br /&gt;
b=7 &lt;br /&gt;
&lt;br /&gt;
Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7&lt;br /&gt;
&lt;br /&gt;
I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):)&lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 22:03, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hi Steffany,&lt;br /&gt;
&lt;br /&gt;
Your explanation makes totally sense! I was omitting the midpoint part! &lt;br /&gt;
&lt;br /&gt;
Thank you. Arabella.&lt;br /&gt;
&lt;br /&gt;
==Question 7==&lt;br /&gt;
&lt;br /&gt;
Hello, I&#039;m struggling on question 7 on our first webwork assignment. Here is the question: &lt;br /&gt;
&lt;br /&gt;
Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).&lt;br /&gt;
&lt;br /&gt;
So what I did was that I first used the distance formula for point (0,b) with point (2,2): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(2-0)^2+(2-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and with point (6,-5): &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\sqrt{(6-0)^2+(-5-b)^2}&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
and then I made them equal to each other to solve for b.&lt;br /&gt;
&lt;br /&gt;
This simplifies to &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;2+(2-b)=6+(-5-b)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
b is then found to be &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
But when I plugged in &amp;lt;math&amp;gt;-4.25&amp;lt;/math&amp;gt; into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?&lt;br /&gt;
&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 02:26, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Adams, I think you&#039;ve got the sets of points wrong. The question is supposed to be &amp;quot;Find the point (0, b) on the y-axis that is equidistant from the points &amp;lt;b&amp;gt;(3, 3)&amp;lt;/b&amp;gt; and &amp;lt;b&amp;gt;(5, -4)&amp;lt;/b&amp;gt;. I got the right answer with the method you used, so it&#039;s just a matter of using the correct data!&lt;br /&gt;
&lt;br /&gt;
[[User:BellaTory|BellaTory]]&lt;br /&gt;
&lt;br /&gt;
The PDF Version that I printed out says (2,2) and (6,-5) :(&lt;br /&gt;
--[[User:AdamsNguyen|AdamsNguyen]] 07:41, 18 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
Hello guys,&lt;br /&gt;
&lt;br /&gt;
For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3). &lt;br /&gt;
&lt;br /&gt;
I tried Adam&#039;s method but my answer is not a solution!&lt;br /&gt;
&lt;br /&gt;
I initially, I worked as follows though:&lt;br /&gt;
&lt;br /&gt;
-I found the slope m= -3-3/ 4-3= -6/1&lt;br /&gt;
&lt;br /&gt;
-Then I equated using the formula y-b=m(x-0)&lt;br /&gt;
&lt;br /&gt;
for point (3, 3) &lt;br /&gt;
&lt;br /&gt;
3-b=-6(3-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
for point (4, -3)&lt;br /&gt;
&lt;br /&gt;
-3-b=-6(4-0)&lt;br /&gt;
&lt;br /&gt;
b=21&lt;br /&gt;
&lt;br /&gt;
However, like Adam, the program is not accepting my answer. I&#039;m not sure what I&#039;m doing wrong. Thank you.&lt;br /&gt;
&lt;br /&gt;
Arabella.&lt;br /&gt;
&lt;br /&gt;
===Re: Question 7===&lt;br /&gt;
Hi everyone,&lt;br /&gt;
&lt;br /&gt;
We must all be given different values for the same question. &lt;br /&gt;
&lt;br /&gt;
Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point &#039;&#039;equidistant&#039;&#039; from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying. &lt;br /&gt;
&lt;br /&gt;
Good Luck! &lt;br /&gt;
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 03:26, 19 September 2010 (UTC)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Steffany, &lt;br /&gt;
&lt;br /&gt;
I am following what you&#039;re saying about using the distance formula but when I went to solve for &#039;b&#039; by putting them equal to one another, I get confused. Can you step me through your simplifying?&lt;br /&gt;
&lt;br /&gt;
Thanks!&lt;br /&gt;
LaBri Krahn --&lt;br /&gt;
&lt;br /&gt;
Hi,&lt;br /&gt;
&lt;br /&gt;
When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow. &lt;br /&gt;
&lt;br /&gt;
Steffany&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=48253</id>
		<title>User:Fiona</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=48253"/>
		<updated>2010-09-20T05:14:36Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Fiona Ma - Second year, Faculty of Arts.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is used to calculate the hypotenuse of a right angle triangle. This theorem has been around for centuries and researchers found that it was apparent in many historical references. For example, the Egyptians used this theorem to build their pyramids. Although this theorem wasn’t put into a mathematical formula, the Egyptians had their own measuring system using a rope with 12 knots of equal distance showing their understanding of the 3,4, and 5 right angle triangle. To this day, we still use this theorem/method not only in math classes, but to design and build modern-day structures just like the Egyptians used it to build their pyramids.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is defined as: &#039;&#039;&#039;a²+b²=c²&#039;&#039;&#039;  (with c as the length of the hypotenuse).&lt;br /&gt;
 &lt;br /&gt;
An example of this question would be:&lt;br /&gt;
	Matthew loves to zip line across buildings, he wants to zip line across two identical buildings that are 30 meters apart and he wants to zip line from the 10th floor to the 6th floor of the other building. Matthew knows that each floor is 10 meters high. How long should his zip line rope be?&lt;br /&gt;
&lt;br /&gt;
We can deduct that side A is 30m, and side B is 40m (10(10-6)). Using the Pythagorean theorem, we can solve for side C, which is the length of his zip line. &lt;br /&gt;
			&lt;br /&gt;
				&#039;&#039;&#039;A²+B² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;30²+40² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;900+1600 = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (900+1600)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (2500)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = 50m&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using the Pythagorean theorem, we can conclude that Matthew’s zip line needs to be 50m long.&lt;br /&gt;
&amp;lt;p&amp;gt;&lt;br /&gt;
References : &lt;br /&gt;
&amp;lt;p&amp;gt;http://ejad.best.vwh.net/java/pythagoras/history.html&lt;br /&gt;
http://en.wikipedia.org/wiki/Pythagorean_theorem&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=48252</id>
		<title>User:Fiona</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Fiona&amp;diff=48252"/>
		<updated>2010-09-20T05:13:10Z</updated>

		<summary type="html">&lt;p&gt;Fiona: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Fiona Ma - Second year, Faculty of Arts.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[ESSAY]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is used to calculate the hypotenuse of a right angle triangle. This theorem has been around for centuries and researchers found that it was apparent in many historical references. For example, the Egyptians used this theorem to build their pyramids. Although this theorem wasn’t put into a mathematical formula, the Egyptians had their own measuring system using a rope with 12 knots of equal distance showing their understanding of the 3,4, and 5 right angle triangle. To this day, we still use this theorem/method not only in math classes, but to design and build modern-day structures just like the Egyptians used it to build their pyramids.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is defined as: &#039;&#039;&#039;a²+b²=c²&#039;&#039;&#039;  (with c as the length of the hypotenuse).&lt;br /&gt;
 &lt;br /&gt;
An example of this question would be:&lt;br /&gt;
	Matthew loves to zip line across buildings, he wants to zip line across two identical buildings that are 30 meters apart and he wants to zip line from the 10th floor to the 6th floor of the other building. Matthew knows that each floor is 10 meters high. How long should his zip line rope be?&lt;br /&gt;
&lt;br /&gt;
We can deduct that side A is 30m, and side B is 40m (10(10-6)). Using the Pythagorean theorem, we can solve for side C, which is the length of his zip line. &lt;br /&gt;
			&lt;br /&gt;
				&#039;&#039;&#039;A²+B² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;30²+40² = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;900+1600 = C²&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (900+1600)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = √ (2500)&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
				&#039;&#039;&#039;C = 50m&#039;&#039;&#039;&amp;lt;p&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Using the Pythagorean theorem, we can conclude that Matthew’s zip line needs to be 50m long.&lt;br /&gt;
&amp;lt;p&amp;gt; ©FionaMa&lt;/div&gt;</summary>
		<author><name>Fiona</name></author>
	</entry>
</feed>