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	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=93121</id>
		<title>User:CatherinePuiLingChen</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=93121"/>
		<updated>2011-05-03T00:12:09Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: Blanked the page&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=73715</id>
		<title>User:CatherinePuiLingChen</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=73715"/>
		<updated>2011-01-28T15:37:20Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hello!&lt;br /&gt;
&lt;br /&gt;
This is Catherine Chen (Pui Ling)&lt;br /&gt;
&lt;br /&gt;
---------------------------------------------------------------------------------------------------------------------------&lt;br /&gt;
Marginalism in Business and Economics&lt;br /&gt;
Calculus is largely about functions and derivatives. In economics, the most important concept is the relationship between variables, often X and Y. In functions, the value of Y changes as the value of X changes, that’s the relationship of a function. Similarly, economics is all about relationships between an independent variable and an dependent variable, for example, price of oil/gallon and the quantity demanded of oil. Taking derivatives of functions allows economists to measure an average change in price of oil in relation to quantity demanded in a certain amount of time (which is actually the definition of demand: the amount of a particular economic good or service that consumers will want to purchase at a given price in a given time. &lt;br /&gt;
Marginalism examines the change in an outcome that results from a single-unit increase in a variable. To put it into a simple question, let’s ask, how much does it cost to produce a single extra unit of my product? Or, How much more revenue can I make from selling one more unit of my product? This concept in business helps companies maximize their profits. Actually, the firm does not make a gain until the marginal revenue exceeds that marginal cost. We can play around with independent variables to determine when our production of good/services will allow us to increase our profits. &lt;br /&gt;
Calculus allows us to use derivative skills to solve problems like: &lt;br /&gt;
Total Cost = C(x)&lt;br /&gt;
Marginal Cost = C’(x)&lt;br /&gt;
Average Cost = c(x)=C(x)/x&lt;br /&gt;
Price function =p(x)&lt;br /&gt;
Revenue function = R(x)= x p(x)&lt;br /&gt;
Marginal  Revenue= R’(x)&lt;br /&gt;
Profit Function = P(x)= R(x)-C(x) (Revenue – Cost)&lt;br /&gt;
Marginal Profit = P’(x)= R’(x)-C’(x) (Marginal Revenue – Marginal Cost)&lt;br /&gt;
 &lt;br /&gt;
Marginal is the key term. Whenever we are looking for Marginal Cost/ Revenue/Proft, we know that we are looking for the instantaneous rate of change, or, the derivative. &lt;br /&gt;
If we want to maximize the profit, then the marginal revenue should equal the marginal cost; If we want to minimize the cost, then the marginal cost should equal the average cost. &lt;br /&gt;
So what exactly is the Marginal cost and Marginal revenue, and how do we compute them mathematically? &lt;br /&gt;
Suppose C(x) is the total cost of manufacturing x units of your company’s product, let’s say, bottle of distilled water. If your company produces 1000 bottles of distilled water, C(1000) would be the cost of producing all 1000 products. The question is, now that you have produced this amount of product, how much would it cost to produce 1 more bottle of water? That’s something that the Marginal cost would be able to tell you. C’(1000) is the marginal cost function, and it is also the derivative. In economics, the derivative of the cost is called the marginal cost.&lt;br /&gt;
To visualize this, we can write:&lt;br /&gt;
 &lt;br /&gt;
C’(1000) = lim h→0 [C(1000+h)-C(1000)]/1 &lt;br /&gt;
C’(1000) = C(1001)-C(1000)&lt;br /&gt;
that&#039;s the cost of producing one more.&lt;br /&gt;
So this all relates to optimization. What does optimization mean? Optimization means finding the maximum of minimum values of a quantity, or finding when these max/mins occur. We want to maximize our profit and minimize our costs. Now we know why we&#039;re learning calculus, it may not be very obvious that we need to use it frequently in introuction to micro/macro economics, but in depth ecnomics is actually mostly calculations and it is important to understand visually (which to me is mathematically) how to obtain each single unit of profit.&lt;br /&gt;
&lt;br /&gt;
-------------------------------------------------------------------------------------------------------------&lt;br /&gt;
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I am trying to love math!&lt;br /&gt;
&lt;br /&gt;
Parabola, some people read it like “Para-ba-la”, some say “Para-BO-la”. In either way, it sounds like a fun word to me and so I thought that the concept would be interesting by its name. &lt;br /&gt;
The parabola in physics was discovered by Galileo in the early 17th century, which is then proven mathematically by Issac Newton. The most well know example of a parabola in physics is “the trajectory of body in motion under the influence of a uniform gravitation field without air friction” (tripatlas.com) I’m not a physics student, nor am I a keen math person, and I never quite understand how they proved it right or wrong, mathematically or physics-wise. But if you tell me some real world examples of parabolae, I would probably get it. So there’s the example of cables on the suspension bridge. The free-hanging cables form curve, and appears in the shape of a parabola.It looks like a parabola that is horizontally stretched on a graph paper. In physics, there is also a principle called the “parabolic reflector” discovered in the 3rd century BC by Archimedes. It is a great idea that s a reflective device can concentrates light or other forms of electromagnetic radiation to a common focal point. We cannot prove whether or not the ancient Syracuse had used the principle to concentrate sunray to set fire on the Romans, but it is really amazing that such concept can be applied in the invention of microwaves and telescopes. After all, math is a bunch of great ideas!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Basic Parabola in equations&lt;br /&gt;
f(x)=x^2&lt;br /&gt;
f(x)=x^3 cubic function&lt;br /&gt;
f(x)=ax^2+bx+c quadratic function&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=73713</id>
		<title>User:CatherinePuiLingChen</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=73713"/>
		<updated>2011-01-28T15:24:27Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hello!&lt;br /&gt;
&lt;br /&gt;
This is Catherine Chen (Pui Ling)&lt;br /&gt;
&lt;br /&gt;
---------------------------------------------------------------------------------------------------------------------------&lt;br /&gt;
Marginalism in Business and Economics&lt;br /&gt;
Calculus is largely about functions and derivatives. In economics, the most important concept is the relationship between variables, often X and Y. In functions, the value of Y changes as the value of X changes, that’s the relationship of a function. Similarly, economics is all about relationships between an independent variable and an dependent variable, for example, price of oil/gallon and the quantity demanded of oil. Taking derivatives of functions allows economists to measure an average change in price of oil in relation to quantity demanded in a certain amount of time (which is actually the definition of demand: the amount of a particular economic good or service that consumers will want to purchase at a given price in a given time. &lt;br /&gt;
Marginalism examines the change in an outcome that results from a single-unit increase in a variable. To put it into a simple question, let’s ask, how much does it cost to produce a single extra unit of my product? Or, How much more revenue can I make from selling one more unit of my product? This concept in business helps companies maximize their profits. Actually, the firm does not make a gain until the marginal revenue exceeds that marginal cost. We can play around with independent variables to determine when our production of good/services will allow us to increase our profits. &lt;br /&gt;
Calculus allows us to use derivative skills to solve problems like: &lt;br /&gt;
Total Cost = C(x)&lt;br /&gt;
Marginal Cost = C’(x)&lt;br /&gt;
Average Cost = c(x)=C(x)/x&lt;br /&gt;
Price function =p(x)&lt;br /&gt;
Revenue function = R(x)= x p(x)&lt;br /&gt;
Marginal  Revenue= R’(x)&lt;br /&gt;
Profit Function = P(x)= R(x)-C(x) (Revenue – Cost)&lt;br /&gt;
Marginal Profit = P’(x)= R’(x)-C’(x) (Marginal Revenue – Marginal Cost)&lt;br /&gt;
 &lt;br /&gt;
Marginal is the key term. Whenever we are looking for Marginal Cost/ Revenue/Proft, we know that we are looking for the instantaneous rate of change, or, the derivative. &lt;br /&gt;
If we want to maximize the profit, then the marginal revenue should equal the marginal cost; If we want to minimize the cost, then the marginal cost should equal the average cost. &lt;br /&gt;
So what exactly is the Marginal cost and Marginal revenue, and how do we compute them mathematically? &lt;br /&gt;
Suppose C(x) is the total cost of manufacturing x units of your company’s product, let’s say, bottle of distilled water. If your company produces 1000 bottles of distilled water, C(1000) would be the cost of producing all 1000 products. The question is, now that you have produced this amount of product, how much would it cost to produce 1 more bottle of water? That’s something that the Marginal cost would be able to tell you. C’(1000) is the marginal cost function, and it is also the derivative. In economics, the derivative of the cost is called the marginal cost.&lt;br /&gt;
To visualize this, we can write:&lt;br /&gt;
 &lt;br /&gt;
C’(1000) = lim h→0 [C(1000+h)-C(1000)]/1 &lt;br /&gt;
C’(1000) = C(1001)-C(1000)&lt;br /&gt;
The cost of producing one more.&lt;br /&gt;
So this all relates to optimization. What does optimization mean? Optimization means finding the maximum of minimum values of a quantity, or finding when these max/mins occur. We want to maximize our profit and minimize our costs. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
I am trying to love math!&lt;br /&gt;
&lt;br /&gt;
Parabola, some people read it like “Para-ba-la”, some say “Para-BO-la”. In either way, it sounds like a fun word to me and so I thought that the concept would be interesting by its name. &lt;br /&gt;
The parabola in physics was discovered by Galileo in the early 17th century, which is then proven mathematically by Issac Newton. The most well know example of a parabola in physics is “the trajectory of body in motion under the influence of a uniform gravitation field without air friction” (tripatlas.com) I’m not a physics student, nor am I a keen math person, and I never quite understand how they proved it right or wrong, mathematically or physics-wise. But if you tell me some real world examples of parabolae, I would probably get it. So there’s the example of cables on the suspension bridge. The free-hanging cables form curve, and appears in the shape of a parabola.It looks like a parabola that is horizontally stretched on a graph paper. In physics, there is also a principle called the “parabolic reflector” discovered in the 3rd century BC by Archimedes. It is a great idea that s a reflective device can concentrates light or other forms of electromagnetic radiation to a common focal point. We cannot prove whether or not the ancient Syracuse had used the principle to concentrate sunray to set fire on the Romans, but it is really amazing that such concept can be applied in the invention of microwaves and telescopes. After all, math is a bunch of great ideas!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Basic Parabola in equations&lt;br /&gt;
f(x)=x^2&lt;br /&gt;
f(x)=x^3 cubic function&lt;br /&gt;
f(x)=ax^2+bx+c quadratic function&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Glarus&amp;diff=71116</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Glarus</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Glarus&amp;diff=71116"/>
		<updated>2011-01-19T21:24:53Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Glarus&lt;br /&gt;
| member 1 = [[User:AndreaMameri|Andrea Mameri]]&lt;br /&gt;
| member 2 = Catherine Chen&lt;br /&gt;
| member 3 = Supreet Saran&lt;br /&gt;
| member 4 = Victoria Bass&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
Hi.... My name is Andrea Mameri&lt;br /&gt;
&lt;br /&gt;
phone number: 604-716-9520&lt;br /&gt;
&lt;br /&gt;
email address: andreamameri@aim.com&lt;br /&gt;
&lt;br /&gt;
skype name: andrea.mameri&lt;br /&gt;
&lt;br /&gt;
You can find me on facebook as: Andrea Mameri&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Supreet Saran  &lt;br /&gt;
&lt;br /&gt;
skype: supreetsaran&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hi everyone. I&#039;m Victoria.  &lt;br /&gt;
&lt;br /&gt;
My email is bassvm@interchange.ubc.ca Yay!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hi! I&#039;m Catherine.&lt;br /&gt;
 &lt;br /&gt;
fb and email: catherine chen (catherine1012_ling@hotmail.com)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Homework Question:&lt;br /&gt;
&lt;br /&gt;
Model: y= 7x-40&lt;br /&gt;
&lt;br /&gt;
The model is specified to be linear so we know that it will follow the formula of y=mx+b. We also know that the marginal cost is $7/unit. Since x is the # of units we know that 7x describes part of our cost. Meaning we now have y=7x+b. When we produce 20 items (when x = 20) our total cost is $100. This gives us 100= 7(20)+b. We can then solve for b and find that b=-40.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For 150 items our model predicts a cost of $1010. We find this by:&lt;br /&gt;
&lt;br /&gt;
y=7(150)-40&lt;br /&gt;
&lt;br /&gt;
y=1010&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This means that the average cost per item increases as production levels increase. We find this through the following logic:&lt;br /&gt;
&lt;br /&gt;
When we produced 20 items our cost was $100 (x=20, y=100) this means that the average cost per item was $5 (100/20). When we produced 150 items our cost was $1010 (x=150, y=1010) this means that the average cost per item was $6.73 (1010/150)  &lt;br /&gt;
&lt;br /&gt;
In consideration of other models, this model would be an example of one where average cost increases as production increases (as we have just shown.)&lt;br /&gt;
&lt;br /&gt;
a) The average cost remains constant as production increases:&lt;br /&gt;
y= 100&lt;br /&gt;
In this case, since there is no x variable, it does not matter how much is produced and the average cost will always be 100. It is a linear function which would be a horizontal line on the function graph.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Glarus&amp;diff=70793</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Glarus</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Glarus&amp;diff=70793"/>
		<updated>2011-01-19T05:11:45Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Glarus&lt;br /&gt;
| member 1 = [[User:AndreaMameri|Andrea Mameri]]&lt;br /&gt;
| member 2 = Catherine Chen&lt;br /&gt;
| member 3 = Supreet Saran&lt;br /&gt;
| member 4 = Victoria Bass&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
Hi.... My name is Andrea Mameri&lt;br /&gt;
&lt;br /&gt;
phone number: 604-716-9520&lt;br /&gt;
&lt;br /&gt;
email address: andreamameri@aim.com&lt;br /&gt;
&lt;br /&gt;
skype name: andrea.mameri&lt;br /&gt;
&lt;br /&gt;
You can find me on facebook as: Andrea Mameri&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hi everyone. I&#039;m Victoria. My email is bassvm@interchange.ubc.ca Yay!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In workshop G.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Hi! I&#039;m Catherine. &lt;br /&gt;
fb and email: catherine chen (catherine1012_ling@hotmail.com)&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum&amp;diff=66502</id>
		<title>Course:MATH110/Archive/2010-2011/003/Math Forum</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum&amp;diff=66502"/>
		<updated>2010-12-13T09:23:22Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Winter Final */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This space is meant to be organized by the students. The instructor will help organize the wiki, but won&#039;t necessarily interfere with the math content.&lt;br /&gt;
&lt;br /&gt;
For example, you could ask more details about the quiz of Lecture 2. Or get some help understanding something from the reading in section 3.1 or 4.2.&lt;br /&gt;
&lt;br /&gt;
==Subpages of the Math Forum==&lt;br /&gt;
&lt;br /&gt;
Once in a while, I&#039;l archive the discussions in subpages, you can find it all on the following pages.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;dpl&amp;gt;&lt;br /&gt;
titlematch={{PAGENAME}}/%&lt;br /&gt;
namespace={{NAMESPACE}}&lt;br /&gt;
shownamespace=false&lt;br /&gt;
&amp;lt;/dpl&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Winter Final==&lt;br /&gt;
&lt;br /&gt;
Is there any practice questions or last years winter final that we could use to study?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
OPTIMIZATION is NOT on the finals, right?&lt;br /&gt;
&lt;br /&gt;
==Nov.19 Problem 1==&lt;br /&gt;
&lt;br /&gt;
Anybody have any ideas to even start finding the proof? Thanks.&amp;lt;br&amp;gt;&lt;br /&gt;
http://wiki.ubc.ca/images/math/c/6/2/c62af4f5040ad1a27d5f1c9e38731786.png&lt;br /&gt;
&lt;br /&gt;
-adam&lt;br /&gt;
&lt;br /&gt;
==Webwork Week 10==&lt;br /&gt;
&lt;br /&gt;
Does anyone have an helpful tips towards solving for the second derivative below:&lt;br /&gt;
 Find y&amp;quot; for y= 2(x^2-6)^3&lt;br /&gt;
&lt;br /&gt;
-Ghita Youssefi&lt;br /&gt;
&lt;br /&gt;
:Did you compute &amp;lt;math&amp;gt;y&#039;&amp;lt;/math&amp;gt; already?&lt;br /&gt;
&lt;br /&gt;
Try expanding what is in the brackets first, that should make it easier to find the first derivative. Hope this helps.&lt;br /&gt;
Steffany&lt;br /&gt;
&lt;br /&gt;
== Rock paper scissors ==&lt;br /&gt;
Does anyone have any helpful tips towards solving the question regarding rock-paper-scissors?&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Homework 4: Problem Solving ==&lt;br /&gt;
Here&#039;s some space to help each other solve these problems. I would appreciate if you kept it at helping each other and not giving away the answer too easily. Struggling and hitting your head against the walls is a healthy way to do mathematics. [[User:DavidKohler|DavidKohler]]&lt;br /&gt;
&lt;br /&gt;
=== Problem 1 ===&lt;br /&gt;
=== Problem 2 ===&lt;br /&gt;
=== Problem 3 ===&lt;br /&gt;
hey guys, ive been giving a shot at question 3 but I have many doubts regarding the wording... in one of the situations, it states that Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married. however, Ed&#039;s sister is engaged to the second baseman. does this mean that I can consider the second baseman as married or as a bachelor?&lt;br /&gt;
&lt;br /&gt;
also, consider these two statements:&lt;br /&gt;
Pascal and Charles each won $20 from the pitcher at a poker game, &lt;br /&gt;
Ed and the outfielders play cards during their free time, &lt;br /&gt;
I figured that this would imply that pascal and charles are both outfielders and that Ed is the pitcher. however, this is not the case.&lt;br /&gt;
The pitcher can&#039;t be Ed since Ed is a bachelor and it states that the pitcher has a wife.&lt;br /&gt;
 &lt;br /&gt;
The reason I also brought up Pascal and Charles is because from the previous statement it would seem as if pascal and charles are part of the outfieldiers. We are also told that &amp;quot;All the battery and infield except Charles, Hassan and Adam are shorter than Sung&amp;quot; nulling the fact that charles is an outfielder...&lt;br /&gt;
&lt;br /&gt;
It also states that Sung is in the process of getting divorced. once again can I consider him a bachelor or married? if I still consider him as married. what is the importance of that phrase in the context of the question?&lt;br /&gt;
I am extremely confused as to how to take on this question.&lt;br /&gt;
&lt;br /&gt;
If someone could look at it id greatly appreciate it. &lt;br /&gt;
Thx&lt;br /&gt;
&lt;br /&gt;
[[User:MarcoGasparian|MarcoGasparian]]&lt;br /&gt;
&lt;br /&gt;
Hey Marco,&lt;br /&gt;
&lt;br /&gt;
I had similar issues with the problem and I still haven&#039;t completed but can&#039;t contribute a bit here.&lt;br /&gt;
&lt;br /&gt;
Logically, being engaged is not being married. So, I treated the second baseman as a bachelor.&lt;br /&gt;
&lt;br /&gt;
You are right; since Ed is a bachelor, he cannot be the pitcher.&lt;br /&gt;
&lt;br /&gt;
I think this statement might be trying to confuse us a bit. Maybe there are, within the team, different subgroups of card-playing buddies and Ed and the outfielders form one subgroup and Pascal, Charles and the pitcher form the second etc....?&lt;br /&gt;
&lt;br /&gt;
Once again, I treated Sung as a married man because technically he is still married. This information is somewhat relevant in the context of this problem because when you are considering solutions, Sung will be part of the married players. I guess they are just twisting the phrase to confuse us!&lt;br /&gt;
&lt;br /&gt;
Hope this helps at least a little! :D&lt;br /&gt;
&lt;br /&gt;
Arabella.&lt;br /&gt;
&lt;br /&gt;
=== Problem 4===&lt;br /&gt;
=== Problem 5===&lt;br /&gt;
&lt;br /&gt;
== A2 Question 10 ==&lt;br /&gt;
&lt;br /&gt;
I gave this question a try but it&#039;s not making any more sense: &lt;br /&gt;
&lt;br /&gt;
&amp;quot;The graph of the function f(x)=-4^x can be obtained from the graph of g(x)=4^x by one of the following actions:&amp;quot;&lt;br /&gt;
&lt;br /&gt;
I don&#039;t know how the changes are reflected on the graph...I&#039;d say the function has shifted vertically. What would the range for this be? I am lost...a hint please, someone? Thanks kindly.&lt;br /&gt;
&lt;br /&gt;
Arabella.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=65918</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=65918"/>
		<updated>2010-12-05T19:44:44Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* =Basic Skills Review - Area and Volume */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 3 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 4 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 5 = &lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
==Basic Skills Review - Area and Volume=&lt;br /&gt;
&lt;br /&gt;
[[File:Math110 01 Area and Volume group project.pdf]]&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
&lt;br /&gt;
5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
&lt;br /&gt;
a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
&lt;br /&gt;
b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
&lt;br /&gt;
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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
&lt;br /&gt;
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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
&lt;br /&gt;
[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Solving problems (2)==&lt;br /&gt;
&lt;br /&gt;
4)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Math chart.png]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Basic Skills Evaluation==&lt;br /&gt;
&lt;br /&gt;
1)Those that pose &#039;&#039;&#039;no problem&#039;&#039;&#039; to anyone in the group. &lt;br /&gt;
&lt;br /&gt;
2.1 Basic functions&lt;br /&gt;
2.2 Properties of functions&lt;br /&gt;
2.3 Equations&lt;br /&gt;
2.5 Composition of functions&lt;br /&gt;
2.8 Intersections of functions&lt;br /&gt;
2.10 Distances and lines&lt;br /&gt;
2.11 Operations on graphs of functions&lt;br /&gt;
2.14 Areas and volumes&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2)Those that &#039;&#039;&#039;some&#039;&#039;&#039; of you have issues with, but not everyone in the group. &lt;br /&gt;
2.4 Inequalities&lt;br /&gt;
2.6 Polynomial long division&lt;br /&gt;
2.7 Graphs of functions&lt;br /&gt;
2.9 Reading graphs of functions&lt;br /&gt;
2.12 Construction of graphs&lt;br /&gt;
2.13 Trigonometry and the Pythagorean theorem&lt;br /&gt;
2.15 Mathematical writing&lt;br /&gt;
&lt;br /&gt;
3)Those that &#039;&#039;&#039;no one&#039;&#039;&#039; in the group &#039;&#039;&#039;knows&#039;&#039;&#039; how to handle.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=65917</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=65917"/>
		<updated>2010-12-05T19:44:26Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 3 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 4 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 5 = &lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
==Basic Skills Review - Area and Volume=&lt;br /&gt;
&lt;br /&gt;
[[File:Math110 01 Area and Volume group project.pdf&lt;br /&gt;
.jpg]]&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
&lt;br /&gt;
5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
&lt;br /&gt;
a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
&lt;br /&gt;
b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
&lt;br /&gt;
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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
&lt;br /&gt;
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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
&lt;br /&gt;
[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
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==Solving problems (2)==&lt;br /&gt;
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4)&lt;br /&gt;
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[[File:Math chart.png]]&lt;br /&gt;
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==Basic Skills Evaluation==&lt;br /&gt;
&lt;br /&gt;
1)Those that pose &#039;&#039;&#039;no problem&#039;&#039;&#039; to anyone in the group. &lt;br /&gt;
&lt;br /&gt;
2.1 Basic functions&lt;br /&gt;
2.2 Properties of functions&lt;br /&gt;
2.3 Equations&lt;br /&gt;
2.5 Composition of functions&lt;br /&gt;
2.8 Intersections of functions&lt;br /&gt;
2.10 Distances and lines&lt;br /&gt;
2.11 Operations on graphs of functions&lt;br /&gt;
2.14 Areas and volumes&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2)Those that &#039;&#039;&#039;some&#039;&#039;&#039; of you have issues with, but not everyone in the group. &lt;br /&gt;
2.4 Inequalities&lt;br /&gt;
2.6 Polynomial long division&lt;br /&gt;
2.7 Graphs of functions&lt;br /&gt;
2.9 Reading graphs of functions&lt;br /&gt;
2.12 Construction of graphs&lt;br /&gt;
2.13 Trigonometry and the Pythagorean theorem&lt;br /&gt;
2.15 Mathematical writing&lt;br /&gt;
&lt;br /&gt;
3)Those that &#039;&#039;&#039;no one&#039;&#039;&#039; in the group &#039;&#039;&#039;knows&#039;&#039;&#039; how to handle.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Area_%26_Volume.pdf&amp;diff=65856</id>
		<title>File:Area &amp; Volume.pdf</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Area_%26_Volume.pdf&amp;diff=65856"/>
		<updated>2010-12-04T05:33:52Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
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&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=63722</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=63722"/>
		<updated>2010-11-29T16:31:39Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 3 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 4 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 5 = &lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
==Basic Skills Review - Area and Volume=&lt;br /&gt;
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[[File:Area &amp;amp; Volume.jpg]]&lt;br /&gt;
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== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
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[[File:P1120678.JPG]]&lt;br /&gt;
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[[Media:P1120679.jpg]]&lt;br /&gt;
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Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
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== Working on Solving Problems ==&lt;br /&gt;
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1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
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5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
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a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
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b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
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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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
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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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
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5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
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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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
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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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
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[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
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&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Solving problems (2)==&lt;br /&gt;
&lt;br /&gt;
4)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Math chart.png]]&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
==Basic Skills Evaluation==&lt;br /&gt;
&lt;br /&gt;
1)Those that pose &#039;&#039;&#039;no problem&#039;&#039;&#039; to anyone in the group. &lt;br /&gt;
&lt;br /&gt;
2.1 Basic functions&lt;br /&gt;
2.2 Properties of functions&lt;br /&gt;
2.3 Equations&lt;br /&gt;
2.5 Composition of functions&lt;br /&gt;
2.8 Intersections of functions&lt;br /&gt;
2.10 Distances and lines&lt;br /&gt;
2.11 Operations on graphs of functions&lt;br /&gt;
2.14 Areas and volumes&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2)Those that &#039;&#039;&#039;some&#039;&#039;&#039; of you have issues with, but not everyone in the group. &lt;br /&gt;
2.4 Inequalities&lt;br /&gt;
2.6 Polynomial long division&lt;br /&gt;
2.7 Graphs of functions&lt;br /&gt;
2.9 Reading graphs of functions&lt;br /&gt;
2.12 Construction of graphs&lt;br /&gt;
2.13 Trigonometry and the Pythagorean theorem&lt;br /&gt;
2.15 Mathematical writing&lt;br /&gt;
&lt;br /&gt;
3)Those that &#039;&#039;&#039;no one&#039;&#039;&#039; in the group &#039;&#039;&#039;knows&#039;&#039;&#039; how to handle.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Basic_Skills&amp;diff=61829</id>
		<title>Course:MATH110/Archive/2010-2011/003/Basic Skills</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Basic_Skills&amp;diff=61829"/>
		<updated>2010-11-17T00:44:53Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Even if mathematics is mainly about ideas and concepts, we need to be able to manipulate some of its elements in order to get the job done. By basic skills, we understand some of the technical skills that are usually addressed in elementary and high school. Math 110 students usually have difficulties with those or have never seen some aspects of it. It is one of the goals of Math 110 to have students close this gap and be able to successfully take the UBC basic skills test. Our goal is to get this done by the end of December so that we can move on to our other goals. The December final exam will address this and test you on those basic skills.&lt;br /&gt;
&lt;br /&gt;
==Basic Skills Learning Guide Project==&lt;br /&gt;
&lt;br /&gt;
So far, the following topics have been claimed. Groups who have topics which partially overlap should communicate with each other to coordinate their work.&lt;br /&gt;
&lt;br /&gt;
Since some learning goals are smaller than some others, just claiming a learning goal isn&#039;t always enough.&lt;br /&gt;
&lt;br /&gt;
* Equations: group 7&lt;br /&gt;
::Linear equations&lt;br /&gt;
::Quadratic equations&lt;br /&gt;
::Tips and tricks for other polynomial equations (degree more than 2)&lt;br /&gt;
::Equations with square roots&lt;br /&gt;
* Inequalities: group 9 and group 10&lt;br /&gt;
::What they are&lt;br /&gt;
::How to represent their solutions&lt;br /&gt;
::How to solve them&lt;br /&gt;
* Operations on functions and graphs: group 13&lt;br /&gt;
::Addition, multiplication and composition of functions&lt;br /&gt;
::Link to operations on graphs&lt;br /&gt;
* Piecewise functions: group 6&lt;br /&gt;
::What is it&lt;br /&gt;
::How to graph it&lt;br /&gt;
::Link to absolute value functions&lt;br /&gt;
* Absolute value functions&lt;br /&gt;
::What is it&lt;br /&gt;
::How to graph it&lt;br /&gt;
::Equations &amp;amp; Inequalities&lt;br /&gt;
* Find intersection of graphs&lt;br /&gt;
::What&#039;s the idea&lt;br /&gt;
::How to do it&lt;br /&gt;
* Trigonometric functions: group 11&lt;br /&gt;
* Exponential functions&lt;br /&gt;
::What they are&lt;br /&gt;
::How to graph them&lt;br /&gt;
::Link to logarithms&lt;br /&gt;
* Logarithmic functions&lt;br /&gt;
::What they are&lt;br /&gt;
::How to graph them&lt;br /&gt;
::Link to exponentials&lt;br /&gt;
* Polynomial long division: group 8&lt;br /&gt;
::How to do it&lt;br /&gt;
::Link with asymptotes&lt;br /&gt;
* Pythagorean theorem: group 17&lt;br /&gt;
::Proofs&lt;br /&gt;
::Relation to distance formula&lt;br /&gt;
* Distances and lines: group 2&lt;br /&gt;
::Distance formula&lt;br /&gt;
::How to find the equation of a line&lt;br /&gt;
::Slopes&lt;br /&gt;
* Areas and Volumes : Group 1&lt;br /&gt;
* Mathematical writing: group 12&lt;br /&gt;
&lt;br /&gt;
==Learning Objectives==&lt;br /&gt;
&lt;br /&gt;
A good grasp of what we call the basic skills is encompassed in the following learning objectives.&lt;br /&gt;
&lt;br /&gt;
===Basic functions===&lt;br /&gt;
to evaluate, simplify and manipulate basic functions which includes:&lt;br /&gt;
* polynomials,&lt;br /&gt;
* radical functions,&lt;br /&gt;
* trigonometric functions,&lt;br /&gt;
* inverse trigonometric functions,&lt;br /&gt;
* exponential functions,&lt;br /&gt;
* logarithmic functions,&lt;br /&gt;
* absolute-valued functions,&lt;br /&gt;
* functions that are constructed by additions, subtractions, multiplications, divisions, exponentiations and/or compositions of the above functions,&lt;br /&gt;
* piecewise functions;&lt;br /&gt;
&lt;br /&gt;
===Properties of functions===&lt;br /&gt;
to find the domain, range and intercepts of a basic function (see above), and the behaviour of such a function at/near the endpoints of the domain;&lt;br /&gt;
&lt;br /&gt;
===Equations===&lt;br /&gt;
to solve linear, quadratic, rational, radical, trigonometric, exponential, logarithmic, and absolute-valued equations;&lt;br /&gt;
===Inequalities===&lt;br /&gt;
to solve linear, quadratic, rational, radical, trigonometric, exponential and logarithmic inequalities;&lt;br /&gt;
===Polynomial long division===&lt;br /&gt;
to perform long divisons of polynomials and write the result out, whether there is a remainder or not;&lt;br /&gt;
===Graphs of functions===&lt;br /&gt;
to relate graphs to simple functions such as linear, quadratic, power, root, reciprocal, absolute-valued, trigonometric, inverse trigonometric, exponential, logarithmic and piecewise functions as well as equations involving circles and ellipses; i.e., plot a graph from a given equation and find the equation from a given graph;&lt;br /&gt;
===Intersections of functions===&lt;br /&gt;
to find intersections of two or more graphs;&lt;br /&gt;
===Reading graphs of functions===&lt;br /&gt;
to find a value of a function from its graph and determine whether a point of given coordinates lies on the graph;&lt;br /&gt;
===Distances and lines===&lt;br /&gt;
to find the distance between two given points and the slope/equation of the line containing two given points;&lt;br /&gt;
===Operations on graphs of functions===&lt;br /&gt;
to construct new functions by using elementary operations such as:&lt;br /&gt;
* addition/subtraction of functions&lt;br /&gt;
* multiplication/division of functions&lt;br /&gt;
* composition of functions&lt;br /&gt;
to identify the various functions that make up a composite function;&amp;lt;br /&amp;gt;&lt;br /&gt;
to translate, scale and reflect graphs;&lt;br /&gt;
===Construction of graphs===&lt;br /&gt;
to construct a graph from a given context and extract information related to a given context from a graph;&lt;br /&gt;
===Trigonometry and the Pythagorean theorem===&lt;br /&gt;
to apply the Pythagorean theorem, write down trigonometric relationships involving the sides and angles of a right triangle, and express proportional relations between similar triangles;&lt;br /&gt;
===Areas and volumes===&lt;br /&gt;
to compute the area of basic 2D shapes, and the surface area and the volume of basic 3D shapes;&lt;br /&gt;
===Mathematical writing===&lt;br /&gt;
to construct neat, logical, understandable explanations and solutions.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58286</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58286"/>
		<updated>2010-10-28T23:42:28Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Basic Skills Evaluation */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:CurtisDoucette|Curtis Doucette]]&lt;br /&gt;
| member 3 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 4 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 5 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
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[[File:P1120678.JPG]]&lt;br /&gt;
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[[Media:P1120679.jpg]]&lt;br /&gt;
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Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
&lt;br /&gt;
5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
&lt;br /&gt;
a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
&lt;br /&gt;
b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
&lt;br /&gt;
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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
&lt;br /&gt;
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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
&lt;br /&gt;
[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Solving problems (2)==&lt;br /&gt;
&lt;br /&gt;
4)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Math chart.png]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Basic Skills Evaluation==&lt;br /&gt;
&lt;br /&gt;
1)Those that pose &#039;&#039;&#039;no problem&#039;&#039;&#039; to anyone in the group. &lt;br /&gt;
&lt;br /&gt;
2.1 Basic functions&lt;br /&gt;
2.2 Properties of functions&lt;br /&gt;
2.3 Equations&lt;br /&gt;
2.5 Composition of functions&lt;br /&gt;
2.8 Intersections of functions&lt;br /&gt;
2.10 Distances and lines&lt;br /&gt;
2.11 Operations on graphs of functions&lt;br /&gt;
2.14 Areas and volumes&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2)Those that &#039;&#039;&#039;some&#039;&#039;&#039; of you have issues with, but not everyone in the group. &lt;br /&gt;
2.4 Inequalities&lt;br /&gt;
2.6 Polynomial long division&lt;br /&gt;
2.7 Graphs of functions&lt;br /&gt;
2.9 Reading graphs of functions&lt;br /&gt;
2.12 Construction of graphs&lt;br /&gt;
2.13 Trigonometry and the Pythagorean theorem&lt;br /&gt;
2.15 Mathematical writing&lt;br /&gt;
&lt;br /&gt;
3)Those that &#039;&#039;&#039;no one&#039;&#039;&#039; in the group &#039;&#039;&#039;knows&#039;&#039;&#039; how to handle.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58281</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58281"/>
		<updated>2010-10-28T23:38:51Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:CurtisDoucette|Curtis Doucette]]&lt;br /&gt;
| member 3 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 4 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 5 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
&lt;br /&gt;
5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
&lt;br /&gt;
a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
&lt;br /&gt;
b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
&lt;br /&gt;
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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
&lt;br /&gt;
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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
&lt;br /&gt;
[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Solving problems (2)==&lt;br /&gt;
&lt;br /&gt;
4)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Math chart.png]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Basic Skills Evaluation==&lt;br /&gt;
&lt;br /&gt;
1)Those that pose &#039;&#039;&#039;no problem&#039;&#039;&#039; to anyone in the group. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2)Those that &#039;&#039;&#039;some&#039;&#039;&#039; of you have issues with, but not everyone in the group. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3)Those that &#039;&#039;&#039;no one&#039;&#039;&#039; in the group &#039;&#039;&#039;knows&#039;&#039;&#039; how to handle.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58279</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58279"/>
		<updated>2010-10-28T23:35:53Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Solving problems (2) */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:CurtisDoucette|Curtis Doucette]]&lt;br /&gt;
| member 3 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 4 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 5 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
&lt;br /&gt;
5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
&lt;br /&gt;
a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
&lt;br /&gt;
b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
&lt;br /&gt;
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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
&lt;br /&gt;
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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
&lt;br /&gt;
[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Solving problems (2)=&lt;br /&gt;
&lt;br /&gt;
4)&lt;br /&gt;
&lt;br /&gt;
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[[File:Math chart.png]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58278</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58278"/>
		<updated>2010-10-28T23:35:20Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:CurtisDoucette|Curtis Doucette]]&lt;br /&gt;
| member 3 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 4 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 5 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
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[[File:P1120678.JPG]]&lt;br /&gt;
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[[Media:P1120679.jpg]]&lt;br /&gt;
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Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
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----&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
&lt;br /&gt;
5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
&lt;br /&gt;
a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
&lt;br /&gt;
b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
&lt;br /&gt;
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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
&lt;br /&gt;
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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
&lt;br /&gt;
[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Solving problems (2)=&lt;br /&gt;
&lt;br /&gt;
4)&lt;br /&gt;
[[File:Math chart.png]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Math_chart.png&amp;diff=58276</id>
		<title>File:Math chart.png</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Math_chart.png&amp;diff=58276"/>
		<updated>2010-10-28T23:34:16Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58274</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=58274"/>
		<updated>2010-10-28T23:26:16Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 1&lt;br /&gt;
| member 1 = [[User:CatherinePuiLingChen|Catherine Chen]]&lt;br /&gt;
| member 2 = [[User:CurtisDoucette|Curtis Doucette]]&lt;br /&gt;
| member 3 = [[User:TanyaJacob|Tanya Jacob]]&lt;br /&gt;
| member 4 = [[User:AlbertKonig|Albert König]]&lt;br /&gt;
| member 5 = [[User:ShaunaMaty|Shauna Maty]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
1)      There is no time difference! It is how the time has been written! One hour consists of 60 minutes. and When we add 20 minutes to this it adds up to 80 minutes bus drive. And this is the same exact amount that the driver needed for returning to the terminal&lt;br /&gt;
&lt;br /&gt;
2)	As the question is not saying at what time and at what place the policeman saw the woman, I conclude that the policeman was not there when the lady broke the law. The policeman only &amp;quot;might have&amp;quot; her driving. So she might be driving the right way at that time, but 5minues ago she was breaking the law in the absence of the law.&lt;br /&gt;
Another conclusion that can be made from this question is that the question is not including cars or any other types of vehicles that are related to an act of crime while driving. Hence we can also conclude that the woman must have been driving a bicycle instead of a car&lt;br /&gt;
&lt;br /&gt;
3)	The probability of labeling Apple and orange box correctly is 100% for people who know what an orange and what an apple looks like. But when we reach box three, it becomes tricky. The reason is that there are two different fruits inside of it and when we choose only one fruit, we will label that box according to the fruit picked. Hence the chance of saying the right name for the last box is 0. Because, if we pick an orange then we label the box as orange-box but in fact it is a orange-apple box. The same procedure happens when we pick apple from that box. The only chance of getting this right is to pick at least 3 different fruits from the third box and when we see that we have picked two different fruits we know that it is a combination.&lt;br /&gt;
&lt;br /&gt;
4)	If we look at brother in the first part of the sentence and then the plural form of brothers in the second half, we can easily say that this blind fiddler has only one brother.&lt;br /&gt;
Looking at it from another point we know that a fiddler is a person who cheats on people mainly for the sake of “robbing” their money. So we can look at this as a gang where one persons say that everyone in the organization is connected to the blind fiddler but none of us inside the organization are connected to each other. It looks like a pyramid, where the tip can be having multiple lines towards the bottom.&lt;br /&gt;
&lt;br /&gt;
5)	From different point of views there different numbers of rotationsa. &lt;br /&gt;
&lt;br /&gt;
a)One way is when the picture on the coin is facing the same direction then it has revolved 2 times. One time at 0 degrees and one time at 180 degrees. &lt;br /&gt;
&lt;br /&gt;
b)If we don’t care about the direction the coin’s picture is looking at we had a 360o rotation about its axis, which means that we had indefinite times of turn, until it reaches its origin.&lt;br /&gt;
&lt;br /&gt;
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;
&lt;br /&gt;
Probability of taking apple of one of three kinds is 1/3 therefore taking 2 of the same kind is 1/3 X1/3&lt;br /&gt;
= 1/9&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
i) Probability of taking out 1 blue sock for example is 1/2, therefore probability of taking a pair of same color is 1/4&lt;br /&gt;
ii) 1/4&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
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?&lt;br /&gt;
&lt;br /&gt;
5 rungs will be showing as the whole rope is 10 foot with 10 rungs one foot apart&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
No it doesn&#039;t as is does not specify whether all 1/2 of the women are chocolate eaters or all 1/2 of the men are chocolate eaters. The number of chocolate eaters can be distributed between all the men and women to make 1/2. &lt;br /&gt;
&lt;br /&gt;
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;
This is not possible. Based on how it is phrased, it has to be either the son or the daughter, because the mother and her OLDER brother are not the same age. Therefore, since the best and worst players are of opposite sex, this cannot be possible.&lt;br /&gt;
&lt;br /&gt;
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;
Because it never specifies what intervals the trains come at, it could be the following: Brooklyn- 11:59, 12:09, 12:19 Bronx- 12:00, 12:10, 12:20. Based on when the man arrives at the train station, he could almost always end up picking Brooklyn because it departs 1 minute early. &lt;br /&gt;
&lt;br /&gt;
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;
Although it seems that it would just take 10 seconds for the clock to strike ten, simply double, this cannot be right because between 5 chimes, there is only 4 intervals of time so, letting C=chimes and I=intervals, it can be said that 5C+4I=5 and then the formula for the second one would be 10C+9I=x. So, each interval for the 5 chimes is equal to 5/4. Since there is 9 intervals when the clock strikes 10, you would have 5/4*9 which equals 11.25 therefore, it takes 11.25 seconds for the clock to strike ten. &lt;br /&gt;
&lt;br /&gt;
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;
there are 6 ways that two of the four babies can be directly tagged. there is no way that three of the four babies can be directly tagged. &lt;br /&gt;
&lt;br /&gt;
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;
No, you should not accept his bet. No matter how many red cards are in the first half, there has to be the exact same of black cards in the second half as there are red cards in the first half. A half of a deck totals to 26 cards and since there are two colors, red and black, the number of red and black cards will be mirrored oppositely. Example: if Alex splits the deck of cards, and we count what we have in the first half, say 20 black cards and 6 red cards, we know without looking that there are going to be 20 red and 6 black in the other half, simply because there are only 2 colors and 26+26=52&lt;br /&gt;
&lt;br /&gt;
[[&#039;&#039;&#039;Curtis: 16-20&#039;&#039;&#039;]]&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;16.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;First we must translate this information into an equation for the daughters and sons. Let S= sisters and let B= brothers then our equation for the daughters is: S-1=B, and for the sons is: S=2(B-1) Next we solve for B by substituting the information we have that S=B+1: B+1=2(B-1), 1=2B-2-B,  3=2B-B, B=3 therefore by substituting B=3 into S-1=B we get: S-1=3 so S=4. We can then see that there are 4 sisters and 3 brothers.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;17.&#039;&#039;&#039; 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.&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Again, we start with equations. Let D= Dan&#039;s weight, let S= Sarah&#039;s weight and let x= the amount the scale is off by. Then our equations will be D+x=60, S+x=50, and D+S+x=105. Then we can do some simple algebra and substitution to get D=60-x, S=50-x, and (60-x)+(50-x)+x=105. Finally, we can solve for x: -2x+x=105-60-50, -x=-5, x=5. So, the scale is adding 5 kilograms.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;18.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This time we can simply write an equation for the problem letting x= the number of pennies in the jar: 2(2(2x/3)/3)/3=40 and then by reversing this operation we get: x=3(3(3(40)/2)/2)/2 which is really terrible to look at so it can also be viewed as x=40(3/2)^3 therefore x=135. The number of pennies that was in the jar to begin with is 135.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;19.&#039;&#039;&#039; 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? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Once more, you guessed it! We are going to write an equation. Let M= total milk consumed by Angela&#039;s family in the morning and C= total coffee consumed by Angela&#039;s family in the morning and x= the number of members in Angela&#039;s family. Our equation will be (M/4 + C/6)x= M+C. Regrouping, we get 2C(6-x)=3M(x-4). Since both C and M are positive quantities, both (6-x), and (x-4) are also positive, which is only possible when x = 5. Therefore, Anglela&#039;s family has 5 members in it.&lt;br /&gt;
&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;20.&#039;&#039;&#039; 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?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;This one is pretty easy since every hour each clock moves 5 minutes away from the other. ie the gap between them is increased by 10 minutes each hour. 60(minutes in an hour)/10(minutes clocks move apart)=6 so, after 6 hours the clocks will be an hour apart. Therefore, the clocks will be 6 hours apart at 6 am.&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;br /&gt;
[[Link title]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Solving problems (2)=&lt;br /&gt;
&lt;br /&gt;
4)&lt;br /&gt;
&lt;br /&gt;
[[File:Example.jpg]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=54095</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=54095"/>
		<updated>2010-10-12T22:30:48Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Working on Solving Problems */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
It is impossible to determine the ages of Paula&#039;s children. The first piece of information only gives possible combinations that adds up/ multiplies up to 36. We don&#039;t know the date of today, we only know that the sum cannot be larger than 31, and their ages has to be smaller than 10 for each child because their product cannot exceed 36. &lt;br /&gt;
The second piece of information&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=54059</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=54059"/>
		<updated>2010-10-12T21:39:51Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Working on Solving Problems */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Sven is the median of the sequence. Dan is the 10th and Lars is the 16th, so there must be at least 16 runners in order to have a 16th placement. Since 16 is an even number the isn&#039;t an exact median in the sequence. So 17, the next number would be reasonable. The median would be 9. Sven is placed exactly the 9th, which is the middle among all 17 runners, faster than Dan and Lars.&lt;br /&gt;
&lt;br /&gt;
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? &lt;br /&gt;
&lt;br /&gt;
Let the length of the candle be 12cm. For the candle that takes 6 hours to burn out, we call it (a), for the other that takes 4 hours to burn out, we call it (b). With the length of 12 cm, we can calculate the rate of burning. For (a), the rate is 2cm/hr, for (b), the rate is 4cm/hr.&lt;br /&gt;
&lt;br /&gt;
After an hour, (a) would be 10cm while (b) would be 8cm. After two hours, (a) would then be 8cm while (b) would be 4cm. This is when the length are exactly twice. So it takes two hours to have one candle exactly twice as long as the other candle.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=52438</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=52438"/>
		<updated>2010-10-06T08:29:45Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Working on Solving Problems ==&lt;br /&gt;
&lt;br /&gt;
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. &lt;br /&gt;
Let&#039;s say Reuben&#039;s birthday is on Dec 12, two days on Dec 10 he was 20 years old. On Dec 12 he is 21 years old. The next Dec12 he would be 22 years old. Later in Dec 13 the next year he would become 23 years old.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:P1120679.jpg&amp;diff=51787</id>
		<title>File:P1120679.jpg</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:P1120679.jpg&amp;diff=51787"/>
		<updated>2010-10-04T08:42:27Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51786</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51786"/>
		<updated>2010-10-04T08:42:13Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;br /&gt;
&lt;br /&gt;
[[Media:P1120679.jpg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Unable to show steps on wiki. Hand written work will be submitted.&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51785</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51785"/>
		<updated>2010-10-04T08:37:11Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Deriving to the area of the pentagon using squares */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51784</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51784"/>
		<updated>2010-10-04T08:36:39Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Deriving to the area of the pentagon using squares */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:File-P1120678.JPG&amp;diff=51783</id>
		<title>File:File-P1120678.JPG</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:File-P1120678.JPG&amp;diff=51783"/>
		<updated>2010-10-04T08:35:41Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:--File.File.jpg&amp;diff=51778</id>
		<title>File:--File.File.jpg</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:--File.File.jpg&amp;diff=51778"/>
		<updated>2010-10-04T06:27:15Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51776</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51776"/>
		<updated>2010-10-04T06:17:22Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Deriving to the area of the pentagon using squares */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Media:Example.ogg]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51775</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51775"/>
		<updated>2010-10-04T06:16:28Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.JPG]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51774</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51774"/>
		<updated>2010-10-04T06:15:27Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.png|200px|thumb|left|alt text]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51773</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51773"/>
		<updated>2010-10-04T06:14:56Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: /* Deriving to the area of the pentagon using squares */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:.jpg]]&lt;br /&gt;
[[File:P1120678.png|200px|thumb|left|alt text]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:P1120678.jpg&amp;diff=51772</id>
		<title>File:P1120678.jpg</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:P1120678.jpg&amp;diff=51772"/>
		<updated>2010-10-04T06:14:14Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: uploaded a new version of &amp;quot;File:P1120678.jpg&amp;quot;:&amp;amp;#32;alt text&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:P1120678.jpg&amp;diff=51771</id>
		<title>File:P1120678.jpg</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:P1120678.jpg&amp;diff=51771"/>
		<updated>2010-10-04T06:09:33Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51770</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 01</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_01&amp;diff=51770"/>
		<updated>2010-10-04T06:06:39Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Catherine Chen&lt;br /&gt;
* Curtis Doucette&lt;br /&gt;
* Tanya Jacob&lt;br /&gt;
* Albert König&lt;br /&gt;
* Shauna Maty&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
== Deriving to the area of the pentagon using squares ==&lt;br /&gt;
&lt;br /&gt;
[[File:P1120678.jpg]]&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=48375</id>
		<title>User:CatherinePuiLingChen</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=48375"/>
		<updated>2010-09-20T14:43:11Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: Parabola&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hello!&lt;br /&gt;
&lt;br /&gt;
This is Catherine Chen (Pui Ling)&lt;br /&gt;
&lt;br /&gt;
I am trying to love math!&lt;br /&gt;
&lt;br /&gt;
Parabola, some people read it like “Para-ba-la”, some say “Para-BO-la”. In either way, it sounds like a fun word to me and so I thought that the concept would be interesting by its name. &lt;br /&gt;
The parabola in physics was discovered by Galileo in the early 17th century, which is then proven mathematically by Issac Newton. The most well know example of a parabola in physics is “the trajectory of body in motion under the influence of a uniform gravitation field without air friction” (tripatlas.com) I’m not a physics student, nor am I a keen math person, and I never quite understand how they proved it right or wrong, mathematically or physics-wise. But if you tell me some real world examples of parabolae, I would probably get it. So there’s the example of cables on the suspension bridge. The free-hanging cables form curve, and appears in the shape of a parabola.It looks like a parabola that is horizontally stretched on a graph paper. In physics, there is also a principle called the “parabolic reflector” discovered in the 3rd century BC by Archimedes. It is a great idea that s a reflective device can concentrates light or other forms of electromagnetic radiation to a common focal point. We cannot prove whether or not the ancient Syracuse had used the principle to concentrate sunray to set fire on the Romans, but it is really amazing that such concept can be applied in the invention of microwaves and telescopes. After all, math is a bunch of great ideas!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Basic Parabola in equations&lt;br /&gt;
f(x)=x^2&lt;br /&gt;
f(x)=x^3  cubic function&lt;br /&gt;
f(x)=ax^2+bx+c quadratic function&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=48374</id>
		<title>User:CatherinePuiLingChen</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=48374"/>
		<updated>2010-09-20T14:40:57Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hello!&lt;br /&gt;
&lt;br /&gt;
This is Catherine Chen (Pui Ling)&lt;br /&gt;
&lt;br /&gt;
I am trying to love math!&lt;br /&gt;
&lt;br /&gt;
Parabola, some people read it like “Para-ba-la”, some say “Para-BO-la”. In either way, it sounds like a fun word to me and so I thought that the concept would be interesting by its name. &lt;br /&gt;
The parabola in physics was discovered by Galileo in the early 17th century, which is then proven mathematically by Issac Newton. The most well know example of a parabola in physics is “the trajectory of body in motion under the influence of a uniform gravitation field without air friction” (tripatlas.com) I’m not a physics student, nor am I a keen math person, and I never quite understand how they proved it right or wrong, mathematically or physics-wise. But if you tell me some real world examples of parabolae, I would probably get it. So there’s the example of cables on the suspension bridge. The free-hanging cables form curve, and appears in the shape of a parabola. In physics, there is also a principle called the “parabolic reflector” discovered in the 3rd century BC by Archimedes. It is a great idea that s a reflective device can concentrates light or other forms of electromagnetic radiation to a common focal point. We cannot prove whether or not the ancient Syracuse had used the principle to concentrate sunray to set fire on the Romans, but it is really amazing that such concept can be applied in the invention of microwaves and telescopes. After all, math is a bunch of great ideas!&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=47724</id>
		<title>User:CatherinePuiLingChen</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:CatherinePuiLingChen&amp;diff=47724"/>
		<updated>2010-09-17T21:04:10Z</updated>

		<summary type="html">&lt;p&gt;CatherinePuiLingChen: Created page with &amp;#039;Hello!  This is Catherine Chen (Pui Ling)  I am trying to love math!&amp;#039;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hello!&lt;br /&gt;
&lt;br /&gt;
This is Catherine Chen (Pui Ling)&lt;br /&gt;
&lt;br /&gt;
I am trying to love math!&lt;/div&gt;</summary>
		<author><name>CatherinePuiLingChen</name></author>
	</entry>
</feed>