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	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Aargau&amp;diff=71834</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Aargau</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Aargau&amp;diff=71834"/>
		<updated>2011-01-24T16:37:23Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Aargau&lt;br /&gt;
| member 1 = Daisy Gobina&lt;br /&gt;
| member 2 = Dominic Sunga&lt;br /&gt;
| member 3 = Shauna Maty&lt;br /&gt;
| member 4 = Yihong Chen&lt;br /&gt;
}}&lt;br /&gt;
In workshop G.&lt;br /&gt;
&lt;br /&gt;
   &amp;lt;br&amp;gt;SUBPAGES&amp;lt;/br&amp;gt;&lt;br /&gt;
 &amp;lt;br&amp;gt; 1.[[Documentation:Homework_11|Homework 11]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Jan 24, Monday - Keywords&lt;br /&gt;
&lt;br /&gt;
Shauna - Hockey&lt;br /&gt;
&lt;br /&gt;
Dominic - Aargau&lt;br /&gt;
&lt;br /&gt;
Daisy - Math&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Aargau&amp;diff=70895</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Aargau</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Aargau&amp;diff=70895"/>
		<updated>2011-01-19T08:00:30Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Aargau&lt;br /&gt;
| member 1 = Daisy Gobina&lt;br /&gt;
| member 2 = Dominic Sunga&lt;br /&gt;
| member 3 = Shauna Maty&lt;br /&gt;
| member 4 = Yihong Chen&lt;br /&gt;
}}&lt;br /&gt;
In workshop G.&lt;br /&gt;
&lt;br /&gt;
   &lt;br /&gt;
The marginal cost is the cost of producing one more unit.&lt;br /&gt;
 &lt;br /&gt;
The data that is given to us is:&lt;br /&gt;
 &lt;br /&gt;
It will cost 100 dollars to create 20 units, and it will cost 7 dollars per unit if you want to create more. &lt;br /&gt;
 &lt;br /&gt;
In this situation, we are given two pieces of information: one point on our graph (20,100) and the slope of the linear function, which is 7. We can find the equation. &lt;br /&gt;
 &lt;br /&gt;
y-y1=m(x-x1)&lt;br /&gt;
 &lt;br /&gt;
y-100=7(x-20)&lt;br /&gt;
 &lt;br /&gt;
y-100=7x-140&lt;br /&gt;
 &lt;br /&gt;
y=7x-40&lt;br /&gt;
 &lt;br /&gt;
We can find how much it would cost to produce 150 units using this equation. &lt;br /&gt;
 &lt;br /&gt;
Y=7(150)-40&lt;br /&gt;
 &lt;br /&gt;
Y=1050-40&lt;br /&gt;
 &lt;br /&gt;
Y=1010&lt;br /&gt;
 &lt;br /&gt;
It will cost 1010 dollars to produce 150 units. &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
Because this is a linear equation, the average cost of each unit as production increases remains constant. &lt;br /&gt;
 &lt;br /&gt;
We can find other models that show other things about the average cost as production increases.  We can do this by imagining that on the graphs, the y-axis is the average cost and the x-axis is the quantity produced.&lt;br /&gt;
 &lt;br /&gt;
In a situation where the average cost remains constant as production increases, the function would have to be linear and the slope 0. An example is y=5 or y=9&lt;br /&gt;
 &lt;br /&gt;
If the average cost diminishes as production increases, it means that the slope is negative, or that the marginal cost is negative. An example would be y=-7x-40&lt;br /&gt;
 &lt;br /&gt;
If the average cost is increasing, we need a positive slope. An example would be y=10x+5. It’s important to note that these functions are plotted where the y-axis is average cost and the x-axis is quantity produced. &lt;br /&gt;
 &lt;br /&gt;
In a situation when a company reaches economies of scale, the equation will have to  have a negative slope, similar to a problem that was explored earlier.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64827</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04/Basic Skills Project</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64827"/>
		<updated>2010-12-02T23:23:19Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion. &lt;br /&gt;
&lt;br /&gt;
What are functions?&lt;br /&gt;
&lt;br /&gt;
Functions in math are when two variables are considered and put into an equation, and how changes in these variables affect each other. The two variables are either the domain or the range.&lt;br /&gt;
&lt;br /&gt;
For example:&lt;br /&gt;
&lt;br /&gt;
Measuring the amount of rainfall (range, or the y-value) in a week (domain, or the x-value)&lt;br /&gt;
&lt;br /&gt;
Example of the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
It is important to note that there are many different types of functions, and they depend on what degree they are. &lt;br /&gt;
For example, if they are degree 1, they are linear functions. If they are degree 2, they are quadratic functions, degree 3 cubic functions and so on&lt;br /&gt;
&lt;br /&gt;
GRAPHING FUNCTIONS&lt;br /&gt;
There are many methods to graph functions. For quadratic functions, you can graph them by finding the vertex. For linear functions, by finding the slope and intercepts. &lt;br /&gt;
&lt;br /&gt;
One of the most common functions is a linear function, so we&#039;re gonna start off with how to graph a linear function. &lt;br /&gt;
&lt;br /&gt;
Graphing linear functions, for the most part, are simple. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
To graph this linear function, we need to find two points. But we can&#039;t just use any two points. We will draw the graph by finding the x-intercept and the y-intercept. &lt;br /&gt;
&lt;br /&gt;
To do this is fairly simple.&lt;br /&gt;
&lt;br /&gt;
To find the x-intercept, you have to substitute y for zero, and to find the y-intercept, you have to substitute x for zero&lt;br /&gt;
&lt;br /&gt;
FINDING THE X-INTERCEPT&lt;br /&gt;
&lt;br /&gt;
y=2x+1&lt;br /&gt;
0=2x+1&lt;br /&gt;
-1=2x&lt;br /&gt;
x=-1/2 - x-intercept&lt;br /&gt;
&lt;br /&gt;
FINDING THE Y-INTERCEPT&lt;br /&gt;
y=2x+1&lt;br /&gt;
y=2(0)+1&lt;br /&gt;
y=1 - y-intercept&lt;br /&gt;
&lt;br /&gt;
What we do now is simple. Plot the two points of the intercept.&lt;br /&gt;
Coordinates of x-intercept: (-1/2,0)&lt;br /&gt;
Coordinates of y0intercept: (0,1)&lt;br /&gt;
&lt;br /&gt;
Then simply draw a straight line through these two points, and that&#039;s the graph of the function!&lt;br /&gt;
&lt;br /&gt;
For a more visual guide, refer to the video:&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=mxBoni8N70Y&lt;br /&gt;
&lt;br /&gt;
To draw quadratic functions may be a more difficult task&lt;br /&gt;
&lt;br /&gt;
Although it&#039;s a little more complicated, there is a step by step process, and if you follow it, there should be no trouble at all. &lt;br /&gt;
&lt;br /&gt;
To illustrate how to draw a quadratic function, we have to understand what a quadratic function is.&lt;br /&gt;
&lt;br /&gt;
The standard format of a quadratic function is: y=ax^2+bx+c&lt;br /&gt;
&lt;br /&gt;
a, b and c are integers. A cannot equal zero. &lt;br /&gt;
&lt;br /&gt;
Step by step guide to how to graph a quadratic function:&lt;br /&gt;
&lt;br /&gt;
1) Determine concavity&lt;br /&gt;
2)Find vertex&lt;br /&gt;
3) Find x-intercepts (if any) and y-intercepts&lt;br /&gt;
&lt;br /&gt;
From here, we can use an example to demonstrate how to graph a quadratic function.&lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
So first, we have to find the concavity. In this case, we know that it is concave up because the a value is one, which is positive meaning that it is concave up. &lt;br /&gt;
&lt;br /&gt;
Second, we have to find the vertex. To do this we use the formula:&lt;br /&gt;
&lt;br /&gt;
-b/2a&lt;br /&gt;
&lt;br /&gt;
This formula gives us the x-value of the vertex, which means we can plug in that value to find the y-value&lt;br /&gt;
&lt;br /&gt;
So in this case, -b=-4 and 2a=2&lt;br /&gt;
&lt;br /&gt;
So it is -4/2, which is -2&lt;br /&gt;
&lt;br /&gt;
The x-value of the vertex is -2. &lt;br /&gt;
&lt;br /&gt;
To find the y-value, we simply plug in -2 for x&lt;br /&gt;
&lt;br /&gt;
So, y=(-2)^2+4(-2)+4&lt;br /&gt;
y=0&lt;br /&gt;
&lt;br /&gt;
So the coordinates of the vertex is (-2,0)&lt;br /&gt;
&lt;br /&gt;
Afterwards, we find the x and y intercepts. To do that, we can use the same method we used in determining the x and y intercepts for linear functions. We substitute x for zero if we want to find the y-intercept, and y for zero when we want to find the x-intercept.&lt;br /&gt;
&lt;br /&gt;
The x-intercept:&lt;br /&gt;
0=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
0=(x+2)^2&lt;br /&gt;
x=-2&lt;br /&gt;
&lt;br /&gt;
The y-intercept&lt;br /&gt;
y=(0)^2+4(0)+4&lt;br /&gt;
y=4&lt;br /&gt;
&lt;br /&gt;
With all the information we have now, which includes the vertex (-2,0), the x-intercept (-2,0) and the y-intercept (0,4) we can draw the graph. &lt;br /&gt;
&lt;br /&gt;
For a more visual demonstration, refer to the video:&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=mDwN1SqnMRU&lt;br /&gt;
&lt;br /&gt;
HOW TO USE THE GRAPH OF A FUNCTION&lt;br /&gt;
A graph of a function is simply a graphical representation of the said function. It can be used as a visual aid into discerning the behavior of the two variables the function is comparing.&lt;br /&gt;
&lt;br /&gt;
For example, let&#039;s say that a function is comparing the growth of a certain bacteria in Mars over time. This bacteria has a constant rate of growth, so it is a linear function. The x-axis of the graph will tell us the time, and the y-axis will tell us the growth of the bacteria. The graph will show us how much, visually, the bacteria has grown over a given period of time. &lt;br /&gt;
&lt;br /&gt;
In other instances, graphs of functions can show us the height of a ball after it was thrown, or the displacement of a swing on a playground. Essentially, they are visual aids that are used to help us better understand the things that we are trying to compare. &lt;br /&gt;
&lt;br /&gt;
WHAT IS THE DOMAIN OF A FUNCTION&lt;br /&gt;
&lt;br /&gt;
What is the domain? The domain is what x can be, including all real numbers, that will keep the function real. The best way to show this is through an example&lt;br /&gt;
&lt;br /&gt;
Consider the function: y=1/x&lt;br /&gt;
&lt;br /&gt;
The x-value in the function can be any number, and the function will still be real. However, if x was 0, there would be an error and it would no longer be a function. In fact, 1/0 is impossible. &lt;br /&gt;
&lt;br /&gt;
HOW TO FIND THE DOMAIN OF A FUNCTION&lt;br /&gt;
Finding the domain of a function is actually quite simple. All we need to remember are two rules: &lt;br /&gt;
&lt;br /&gt;
1) The denominator of a fraction in a function cannot be 0&lt;br /&gt;
&lt;br /&gt;
2) A negative number cannot be under a root in a function&lt;br /&gt;
&lt;br /&gt;
So basically, all x-values that make the denominator 0 is not part of the domain, and any x-value that makes a negative number under the root is not part of the domain&lt;br /&gt;
&lt;br /&gt;
For a more visual guide, refer to the video:&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=sbvSBH2Mo20&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64826</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04/Basic Skills Project</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64826"/>
		<updated>2010-12-02T23:22:34Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion. &lt;br /&gt;
&lt;br /&gt;
What are functions?&lt;br /&gt;
&lt;br /&gt;
Functions in math are when two variables are considered and put into an equation, and how changes in these variables affect each other. The two variables are either the domain or the range.&lt;br /&gt;
&lt;br /&gt;
For example:&lt;br /&gt;
&lt;br /&gt;
Measuring the amount of rainfall (range, or the y-value) in a week (domain, or the x-value)&lt;br /&gt;
&lt;br /&gt;
Example of the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
It is important to note that there are many different types of functions, and they depend on what degree they are. &lt;br /&gt;
For example, if they are degree 1, they are linear functions. If they are degree 2, they are quadratic functions, degree 3 cubic functions and so on&lt;br /&gt;
&lt;br /&gt;
GRAPHING FUNCTIONS&lt;br /&gt;
There are many methods to graph functions. For quadratic functions, you can graph them by finding the vertex. For linear functions, by finding the slope and intercepts. &lt;br /&gt;
&lt;br /&gt;
One of the most common functions is a linear function, so we&#039;re gonna start off with how to graph a linear function. &lt;br /&gt;
&lt;br /&gt;
Graphing linear functions, for the most part, are simple. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
To graph this linear function, we need to find two points. But we can&#039;t just use any two points. We will draw the graph by finding the x-intercept and the y-intercept. &lt;br /&gt;
&lt;br /&gt;
To do this is fairly simple.&lt;br /&gt;
&lt;br /&gt;
To find the x-intercept, you have to substitute y for zero, and to find the y-intercept, you have to substitute x for zero&lt;br /&gt;
&lt;br /&gt;
FINDING THE X-INTERCEPT&lt;br /&gt;
&lt;br /&gt;
y=2x+1&lt;br /&gt;
0=2x+1&lt;br /&gt;
-1=2x&lt;br /&gt;
x=-1/2 - x-intercept&lt;br /&gt;
&lt;br /&gt;
FINDING THE Y-INTERCEPT&lt;br /&gt;
y=2x+1&lt;br /&gt;
y=2(0)+1&lt;br /&gt;
y=1 - y-intercept&lt;br /&gt;
&lt;br /&gt;
What we do now is simple. Plot the two points of the intercept.&lt;br /&gt;
Coordinates of x-intercept: (-1/2,0)&lt;br /&gt;
Coordinates of y0intercept: (0,1)&lt;br /&gt;
&lt;br /&gt;
Then simply draw a straight line through these two points, and that&#039;s the graph of the function!&lt;br /&gt;
&lt;br /&gt;
For a more visual guide, refer to the video:&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=mxBoni8N70Y&lt;br /&gt;
&lt;br /&gt;
To draw quadratic functions may be a more difficult task&lt;br /&gt;
&lt;br /&gt;
Although it&#039;s a little more complicated, there is a step by step process, and if you follow it, there should be no trouble at all. &lt;br /&gt;
&lt;br /&gt;
To illustrate how to draw a quadratic function, we have to understand what a quadratic function is.&lt;br /&gt;
&lt;br /&gt;
The standard format of a quadratic function is: y=ax^2+bx+c&lt;br /&gt;
&lt;br /&gt;
a, b and c are integers. A cannot equal zero. &lt;br /&gt;
&lt;br /&gt;
Step by step guide to how to graph a quadratic function:&lt;br /&gt;
&lt;br /&gt;
1) Determine concavity&lt;br /&gt;
2)Find vertex&lt;br /&gt;
3) Find x-intercepts (if any) and y-intercepts&lt;br /&gt;
&lt;br /&gt;
From here, we can use an example to demonstrate how to graph a quadratic function.&lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
So first, we have to find the concavity. In this case, we know that it is concave up because the a value is one, which is positive meaning that it is concave up. &lt;br /&gt;
&lt;br /&gt;
Second, we have to find the vertex. To do this we use the formula:&lt;br /&gt;
&lt;br /&gt;
-b/2a&lt;br /&gt;
&lt;br /&gt;
This formula gives us the x-value of the vertex, which means we can plug in that value to find the y-value&lt;br /&gt;
&lt;br /&gt;
So in this case, -b=-4 and 2a=2&lt;br /&gt;
&lt;br /&gt;
So it is -4/2, which is -2&lt;br /&gt;
&lt;br /&gt;
The x-value of the vertex is -2. &lt;br /&gt;
&lt;br /&gt;
To find the y-value, we simply plug in -2 for x&lt;br /&gt;
&lt;br /&gt;
So, y=(-2)^2+4(-2)+4&lt;br /&gt;
y=0&lt;br /&gt;
&lt;br /&gt;
So the coordinates of the vertex is (-2,0)&lt;br /&gt;
&lt;br /&gt;
Afterwards, we find the x and y intercepts. To do that, we can use the same method we used in determining the x and y intercepts for linear functions. We substitute x for zero if we want to find the y-intercept, and y for zero when we want to find the x-intercept.&lt;br /&gt;
&lt;br /&gt;
The x-intercept:&lt;br /&gt;
0=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
0=(x+2)^2&lt;br /&gt;
x=-2&lt;br /&gt;
&lt;br /&gt;
The y-intercept&lt;br /&gt;
y=(0)^2+4(0)+4&lt;br /&gt;
y=4&lt;br /&gt;
&lt;br /&gt;
With all the information we have now, which includes the vertex (-2,0), the x-intercept (-2,0) and the y-intercept (0,4) we can draw the graph. &lt;br /&gt;
&lt;br /&gt;
For a more visual demonstration, refer to the video:&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=mDwN1SqnMRU&lt;br /&gt;
&lt;br /&gt;
HOW TO USE THE GRAPH OF A FUNCTION&lt;br /&gt;
A graph of a function is simply a graphical representation of the said function. It can be used as a visual aid into discerning the behavior of the two variables the function is comparing.&lt;br /&gt;
&lt;br /&gt;
For example, let&#039;s say that a function is comparing the growth of a certain bacteria in Mars over time. This bacteria has a constant rate of growth, so it is a linear function. The x-axis of the graph will tell us the time, and the y-axis will tell us the growth of the bacteria. The graph will show us how much, visually, the bacteria has grown over a given period of time. &lt;br /&gt;
&lt;br /&gt;
In other instances, graphs of functions can show us the height of a ball after it was thrown, or the displacement of a swing on a playground. Essentially, they are visual aids that are used to help us better understand the things that we are trying to compare. &lt;br /&gt;
&lt;br /&gt;
WHAT IS THE DOMAIN OF A FUNCTION&lt;br /&gt;
&lt;br /&gt;
What is the domain? The domain is what x can be, including all real numbers, that will keep the function real. The best way to show this is through an example&lt;br /&gt;
&lt;br /&gt;
Consider the function: y=1/x&lt;br /&gt;
&lt;br /&gt;
The x-value in the function can be any number, and the function will still be real. However, if x was 0, there would be an error and it would no longer be a function. In fact, 1/0 is impossible. &lt;br /&gt;
&lt;br /&gt;
HOW TO FIND THE DOMAIN OF A FUNCTION&lt;br /&gt;
Finding the domain of a function is actually quite simple. All we need to remember are two rules: &lt;br /&gt;
&lt;br /&gt;
1) The denominator of a fraction in a function cannot be 0&lt;br /&gt;
&lt;br /&gt;
2) A negative number cannot be under a root in a function&lt;br /&gt;
&lt;br /&gt;
So basically, all x-values that make the denominator 0 is not part of the domain, and any x-value that makes a negative number under the root is not part of the domain&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64818</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04/Basic Skills Project</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64818"/>
		<updated>2010-12-02T23:13:06Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion. &lt;br /&gt;
&lt;br /&gt;
What are functions?&lt;br /&gt;
&lt;br /&gt;
Functions in math are when two variables are considered and put into an equation, and how changes in these variables affect each other. The two variables are either the domain or the range.&lt;br /&gt;
&lt;br /&gt;
For example:&lt;br /&gt;
&lt;br /&gt;
Measuring the amount of rainfall (range, or the y-value) in a week (domain, or the x-value)&lt;br /&gt;
&lt;br /&gt;
Example of the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
It is important to note that there are many different types of functions, and they depend on what degree they are. &lt;br /&gt;
For example, if they are degree 1, they are linear functions. If they are degree 2, they are quadratic functions, degree 3 cubic functions and so on&lt;br /&gt;
&lt;br /&gt;
GRAPHING FUNCTIONS&lt;br /&gt;
There are many methods to graph functions. For quadratic functions, you can graph them by finding the vertex. For linear functions, by finding the slope and intercepts. &lt;br /&gt;
&lt;br /&gt;
One of the most common functions is a linear function, so we&#039;re gonna start off with how to graph a linear function. &lt;br /&gt;
&lt;br /&gt;
Graphing linear functions, for the most part, are simple. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
To graph this linear function, we need to find two points. But we can&#039;t just use any two points. We will draw the graph by finding the x-intercept and the y-intercept. &lt;br /&gt;
&lt;br /&gt;
To do this is fairly simple.&lt;br /&gt;
&lt;br /&gt;
To find the x-intercept, you have to substitute y for zero, and to find the y-intercept, you have to substitute x for zero&lt;br /&gt;
&lt;br /&gt;
FINDING THE X-INTERCEPT&lt;br /&gt;
&lt;br /&gt;
y=2x+1&lt;br /&gt;
0=2x+1&lt;br /&gt;
-1=2x&lt;br /&gt;
x=-1/2 - x-intercept&lt;br /&gt;
&lt;br /&gt;
FINDING THE Y-INTERCEPT&lt;br /&gt;
y=2x+1&lt;br /&gt;
y=2(0)+1&lt;br /&gt;
y=1 - y-intercept&lt;br /&gt;
&lt;br /&gt;
What we do now is simple. Plot the two points of the intercept.&lt;br /&gt;
Coordinates of x-intercept: (-1/2,0)&lt;br /&gt;
Coordinates of y0intercept: (0,1)&lt;br /&gt;
&lt;br /&gt;
Then simply draw a straight line through these two points, and that&#039;s the graph of the function!&lt;br /&gt;
&lt;br /&gt;
For a more visual guide, refer to the video:&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=mxBoni8N70Y&lt;br /&gt;
&lt;br /&gt;
To draw quadratic functions may be a more difficult task&lt;br /&gt;
&lt;br /&gt;
Although it&#039;s a little more complicated, there is a step by step process, and if you follow it, there should be no trouble at all. &lt;br /&gt;
&lt;br /&gt;
To illustrate how to draw a quadratic function, we have to understand what a quadratic function is.&lt;br /&gt;
&lt;br /&gt;
The standard format of a quadratic function is: y=ax^2+bx+c&lt;br /&gt;
&lt;br /&gt;
a, b and c are integers. A cannot equal zero. &lt;br /&gt;
&lt;br /&gt;
Step by step guide to how to graph a quadratic function:&lt;br /&gt;
&lt;br /&gt;
1) Determine concavity&lt;br /&gt;
2)Find vertex&lt;br /&gt;
3) Find x-intercepts (if any) and y-intercepts&lt;br /&gt;
&lt;br /&gt;
From here, we can use an example to demonstrate how to graph a quadratic function.&lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
So first, we have to find the concavity. In this case, we know that it is concave up because the a value is one, which is positive meaning that it is concave up. &lt;br /&gt;
&lt;br /&gt;
Second, we have to find the vertex. To do this we use the formula:&lt;br /&gt;
&lt;br /&gt;
-b/2a&lt;br /&gt;
&lt;br /&gt;
This formula gives us the x-value of the vertex, which means we can plug in that value to find the y-value&lt;br /&gt;
&lt;br /&gt;
So in this case, -b=-4 and 2a=2&lt;br /&gt;
&lt;br /&gt;
So it is -4/2, which is -2&lt;br /&gt;
&lt;br /&gt;
The x-value of the vertex is -2. &lt;br /&gt;
&lt;br /&gt;
To find the y-value, we simply plug in -2 for x&lt;br /&gt;
&lt;br /&gt;
So, y=(-2)^2+4(-2)+4&lt;br /&gt;
y=0&lt;br /&gt;
&lt;br /&gt;
So the coordinates of the vertex is (-2,0)&lt;br /&gt;
&lt;br /&gt;
Afterwards, we find the x and y intercepts. To do that, we can use the same method we used in determining the x and y intercepts for linear functions. We substitute x for zero if we want to find the y-intercept, and y for zero when we want to find the x-intercept.&lt;br /&gt;
&lt;br /&gt;
The x-intercept:&lt;br /&gt;
0=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
0=(x+2)^2&lt;br /&gt;
x=-2&lt;br /&gt;
&lt;br /&gt;
The y-intercept&lt;br /&gt;
y=(0)^2+4(0)+4&lt;br /&gt;
y=4&lt;br /&gt;
&lt;br /&gt;
With all the information we have now, which includes the vertex (-2,0), the x-intercept (-2,0) and the y-intercept (0,4) we can draw the graph. &lt;br /&gt;
&lt;br /&gt;
For a more visual demonstration, refer to the video:&lt;br /&gt;
&lt;br /&gt;
http://www.youtube.com/watch?v=mDwN1SqnMRU&lt;br /&gt;
&lt;br /&gt;
HOW TO USE THE GRAPH OF A FUNCTION&lt;br /&gt;
A graph of a function is simply a graphical representation of the said function. It can be used as a visual aid into discerning the behavior of the two variables the function is comparing.&lt;br /&gt;
&lt;br /&gt;
For example, let&#039;s say that a function is comparing the growth of a certain bacteria in Mars over time. This bacteria has a constant rate of growth, so it is a linear function. The x-axis of the graph will tell us the time, and the y-axis will tell us the growth of the bacteria. The graph will show us how much, visually, the bacteria has grown over a given period of time. &lt;br /&gt;
&lt;br /&gt;
In other instances, graphs of functions can show us the height of a ball after it was thrown, or the displacement of a swing on a playground. Essentially, they are visual aids that are used to help us better understand the things that we are trying to compare. &lt;br /&gt;
&lt;br /&gt;
WHAT IS THE DOMAIN OF A FUNCTION&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64811</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04/Basic Skills Project</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64811"/>
		<updated>2010-12-02T22:54:05Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion. &lt;br /&gt;
&lt;br /&gt;
What are functions?&lt;br /&gt;
&lt;br /&gt;
Functions in math are when two variables are considered and put into an equation, and how changes in these variables affect each other. The two variables are either the domain or the range.&lt;br /&gt;
&lt;br /&gt;
For example:&lt;br /&gt;
&lt;br /&gt;
Measuring the amount of rainfall (range, or the y-value) in a week (domain, or the x-value)&lt;br /&gt;
&lt;br /&gt;
Example of the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
It is important to note that there are many different types of functions, and they depend on what degree they are. &lt;br /&gt;
For example, if they are degree 1, they are linear functions. If they are degree 2, they are quadratic functions, degree 3 cubic functions and so on&lt;br /&gt;
&lt;br /&gt;
GRAPHING FUNCTIONS&lt;br /&gt;
There are many methods to graph functions. For quadratic functions, you can graph them by finding the vertex. For linear functions, by finding the slope and intercepts. &lt;br /&gt;
&lt;br /&gt;
One of the most common functions is a linear function, so we&#039;re gonna start off with how to graph a linear function. &lt;br /&gt;
&lt;br /&gt;
Graphing linear functions, for the most part, are simple. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
To graph this linear function, we need to find two points. But we can&#039;t just use any two points. We will draw the graph by finding the x-intercept and the y-intercept. &lt;br /&gt;
&lt;br /&gt;
To do this is fairly simple.&lt;br /&gt;
&lt;br /&gt;
To find the x-intercept, you have to substitute y for zero, and to find the y-intercept, you have to substitute x for zero&lt;br /&gt;
&lt;br /&gt;
FINDING THE X-INTERCEPT&lt;br /&gt;
&lt;br /&gt;
y=2x+1&lt;br /&gt;
0=2x+1&lt;br /&gt;
-1=2x&lt;br /&gt;
x=-1/2 - x-intercept&lt;br /&gt;
&lt;br /&gt;
FINDING THE Y-INTERCEPT&lt;br /&gt;
y=2x+1&lt;br /&gt;
y=2(0)+1&lt;br /&gt;
y=1 - y-intercept&lt;br /&gt;
&lt;br /&gt;
What we do now is simple. Plot the two points of the intercept.&lt;br /&gt;
Coordinates of x-intercept: (-1/2,0)&lt;br /&gt;
Coordinates of y0intercept: (0,1)&lt;br /&gt;
&lt;br /&gt;
Then simply draw a straight line through these two points, and that&#039;s the graph of the function!&lt;br /&gt;
&lt;br /&gt;
For a more visual guide, refer to the video:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;object width=&amp;quot;480&amp;quot; height=&amp;quot;385&amp;quot;&amp;gt;&amp;lt;param name=&amp;quot;movie&amp;quot; value=&amp;quot;http://www.youtube.com/v/mxBoni8N70Y?fs=1&amp;amp;amp;hl=en_US&amp;quot;&amp;gt;&amp;lt;/param&amp;gt;&amp;lt;param name=&amp;quot;allowFullScreen&amp;quot; value=&amp;quot;true&amp;quot;&amp;gt;&amp;lt;/param&amp;gt;&amp;lt;param name=&amp;quot;allowscriptaccess&amp;quot; value=&amp;quot;always&amp;quot;&amp;gt;&amp;lt;/param&amp;gt;&amp;lt;embed src=&amp;quot;http://www.youtube.com/v/mxBoni8N70Y?fs=1&amp;amp;amp;hl=en_US&amp;quot; type=&amp;quot;application/x-shockwave-flash&amp;quot; allowscriptaccess=&amp;quot;always&amp;quot; allowfullscreen=&amp;quot;true&amp;quot; width=&amp;quot;480&amp;quot; height=&amp;quot;385&amp;quot;&amp;gt;&amp;lt;/embed&amp;gt;&amp;lt;/object&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To draw quadratic functions may be a more difficult task&lt;br /&gt;
&lt;br /&gt;
Although it&#039;s a little more complicated, there is a step by step process, and if you follow it, there should be no trouble at all. &lt;br /&gt;
&lt;br /&gt;
To illustrate how to draw a quadratic function, we have to understand what a quadratic function is.&lt;br /&gt;
&lt;br /&gt;
The standard format of a quadratic function is: y=ax^2+bx+c&lt;br /&gt;
&lt;br /&gt;
a, b and c are integers. A cannot equal zero. &lt;br /&gt;
&lt;br /&gt;
Step by step guide to how to graph a quadratic function:&lt;br /&gt;
&lt;br /&gt;
1) Determine concavity&lt;br /&gt;
2)Find vertex&lt;br /&gt;
3) Find x-intercepts (if any) and y-intercepts&lt;br /&gt;
&lt;br /&gt;
From here, we can use an example to demonstrate how to graph a quadratic function.&lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
So first, we have to find the concavity. In this case, we know that it is concave up because the a value is one, which is positive meaning that it is concave up. &lt;br /&gt;
&lt;br /&gt;
Second, we have to find the vertex. To do this we use the formula:&lt;br /&gt;
&lt;br /&gt;
-b/2a&lt;br /&gt;
&lt;br /&gt;
This formula gives us the x-value of the vertex, which means we can plug in that value to find the y-value&lt;br /&gt;
&lt;br /&gt;
So in this case, -b=-4 and 2a=2&lt;br /&gt;
&lt;br /&gt;
So it is -4/2, which is -2&lt;br /&gt;
&lt;br /&gt;
The x-value of the vertex is -2. &lt;br /&gt;
&lt;br /&gt;
To find the y-value, we simply plug in -2 for x&lt;br /&gt;
&lt;br /&gt;
So, y=(-2)^2+4(-2)+8&lt;br /&gt;
y=4&lt;br /&gt;
&lt;br /&gt;
So the coordinates of the vertex is (-2,4)&lt;br /&gt;
&lt;br /&gt;
Afterwards, we find the x and y intercepts. To do that, we can use the same method we used in determining the x and y intercepts for linear functions. We substitute x for zero if we want to find the y-intercept, and y for zero when we want to find the x-intercept.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64809</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04/Basic Skills Project</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=64809"/>
		<updated>2010-12-02T22:53:42Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion. &lt;br /&gt;
&lt;br /&gt;
What are functions?&lt;br /&gt;
&lt;br /&gt;
Functions in math are when two variables are considered and put into an equation, and how changes in these variables affect each other. The two variables are either the domain or the range.&lt;br /&gt;
&lt;br /&gt;
For example:&lt;br /&gt;
&lt;br /&gt;
Measuring the amount of rainfall (range, or the y-value) in a week (domain, or the x-value)&lt;br /&gt;
&lt;br /&gt;
Example of the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
It is important to note that there are many different types of functions, and they depend on what degree they are. &lt;br /&gt;
For example, if they are degree 1, they are linear functions. If they are degree 2, they are quadratic functions, degree 3 cubic functions and so on&lt;br /&gt;
&lt;br /&gt;
GRAPHING FUNCTIONS&lt;br /&gt;
There are many methods to graph functions. For quadratic functions, you can graph them by finding the vertex. For linear functions, by finding the slope and intercepts. &lt;br /&gt;
&lt;br /&gt;
One of the most common functions is a linear function, so we&#039;re gonna start off with how to graph a linear function. &lt;br /&gt;
&lt;br /&gt;
Graphing linear functions, for the most part, are simple. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=2x+1&lt;br /&gt;
&lt;br /&gt;
To graph this linear function, we need to find two points. But we can&#039;t just use any two points. We will draw the graph by finding the x-intercept and the y-intercept. &lt;br /&gt;
&lt;br /&gt;
To do this is fairly simple.&lt;br /&gt;
&lt;br /&gt;
To find the x-intercept, you have to substitute y for zero, and to find the y-intercept, you have to substitute x for zero&lt;br /&gt;
&lt;br /&gt;
FINDING THE X-INTERCEPT&lt;br /&gt;
&lt;br /&gt;
y=2x+1&lt;br /&gt;
0=2x+1&lt;br /&gt;
-1=2x&lt;br /&gt;
x=-1/2 - x-intercept&lt;br /&gt;
&lt;br /&gt;
FINDING THE Y-INTERCEPT&lt;br /&gt;
y=2x+1&lt;br /&gt;
y=2(0)+1&lt;br /&gt;
y=1 - y-intercept&lt;br /&gt;
&lt;br /&gt;
What we do now is simple. Plot the two points of the intercept.&lt;br /&gt;
Coordinates of x-intercept: (-1/2,0)&lt;br /&gt;
Coordinates of y0intercept: (0,1)&lt;br /&gt;
&lt;br /&gt;
Then simply draw a straight line through these two points, and that&#039;s the graph of the function!&lt;br /&gt;
&lt;br /&gt;
For a more visual guide, refer to the video:&lt;br /&gt;
&amp;lt;object width=&amp;quot;480&amp;quot; height=&amp;quot;385&amp;quot;&amp;gt;&amp;lt;param name=&amp;quot;movie&amp;quot; value=&amp;quot;http://www.youtube.com/v/mxBoni8N70Y?fs=1&amp;amp;amp;hl=en_US&amp;quot;&amp;gt;&amp;lt;/param&amp;gt;&amp;lt;param name=&amp;quot;allowFullScreen&amp;quot; value=&amp;quot;true&amp;quot;&amp;gt;&amp;lt;/param&amp;gt;&amp;lt;param name=&amp;quot;allowscriptaccess&amp;quot; value=&amp;quot;always&amp;quot;&amp;gt;&amp;lt;/param&amp;gt;&amp;lt;embed src=&amp;quot;http://www.youtube.com/v/mxBoni8N70Y?fs=1&amp;amp;amp;hl=en_US&amp;quot; type=&amp;quot;application/x-shockwave-flash&amp;quot; allowscriptaccess=&amp;quot;always&amp;quot; allowfullscreen=&amp;quot;true&amp;quot; width=&amp;quot;480&amp;quot; height=&amp;quot;385&amp;quot;&amp;gt;&amp;lt;/embed&amp;gt;&amp;lt;/object&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To draw quadratic functions may be a more difficult task&lt;br /&gt;
&lt;br /&gt;
Although it&#039;s a little more complicated, there is a step by step process, and if you follow it, there should be no trouble at all. &lt;br /&gt;
&lt;br /&gt;
To illustrate how to draw a quadratic function, we have to understand what a quadratic function is.&lt;br /&gt;
&lt;br /&gt;
The standard format of a quadratic function is: y=ax^2+bx+c&lt;br /&gt;
&lt;br /&gt;
a, b and c are integers. A cannot equal zero. &lt;br /&gt;
&lt;br /&gt;
Step by step guide to how to graph a quadratic function:&lt;br /&gt;
&lt;br /&gt;
1) Determine concavity&lt;br /&gt;
2)Find vertex&lt;br /&gt;
3) Find x-intercepts (if any) and y-intercepts&lt;br /&gt;
&lt;br /&gt;
From here, we can use an example to demonstrate how to graph a quadratic function.&lt;br /&gt;
&lt;br /&gt;
Let&#039;s use the function: y=x^2+4x+4&lt;br /&gt;
&lt;br /&gt;
So first, we have to find the concavity. In this case, we know that it is concave up because the a value is one, which is positive meaning that it is concave up. &lt;br /&gt;
&lt;br /&gt;
Second, we have to find the vertex. To do this we use the formula:&lt;br /&gt;
&lt;br /&gt;
-b/2a&lt;br /&gt;
&lt;br /&gt;
This formula gives us the x-value of the vertex, which means we can plug in that value to find the y-value&lt;br /&gt;
&lt;br /&gt;
So in this case, -b=-4 and 2a=2&lt;br /&gt;
&lt;br /&gt;
So it is -4/2, which is -2&lt;br /&gt;
&lt;br /&gt;
The x-value of the vertex is -2. &lt;br /&gt;
&lt;br /&gt;
To find the y-value, we simply plug in -2 for x&lt;br /&gt;
&lt;br /&gt;
So, y=(-2)^2+4(-2)+8&lt;br /&gt;
y=4&lt;br /&gt;
&lt;br /&gt;
So the coordinates of the vertex is (-2,4)&lt;br /&gt;
&lt;br /&gt;
Afterwards, we find the x and y intercepts. To do that, we can use the same method we used in determining the x and y intercepts for linear functions. We substitute x for zero if we want to find the y-intercept, and y for zero when we want to find the x-intercept.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=63524</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04/Basic Skills Project</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=63524"/>
		<updated>2010-11-28T06:36:13Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion. &lt;br /&gt;
&lt;br /&gt;
What are functions?&lt;br /&gt;
&lt;br /&gt;
Functions in math are when two variables are considered and put into an equation, and how changes in these variables affect each other. The two variables are either the domain or the range.&lt;br /&gt;
&lt;br /&gt;
For example:&lt;br /&gt;
&lt;br /&gt;
Measuring the amount of rainfall (range, or the y-value) in a week (domain, or the x-value)&lt;br /&gt;
&lt;br /&gt;
Example of the function: y=2x+1&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=63523</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=63523"/>
		<updated>2010-11-28T06:27:33Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 4&lt;br /&gt;
| member 1 = Owen Johnston&lt;br /&gt;
| member 2 = Charlene Ng&lt;br /&gt;
| member 3 = Ling Sun&lt;br /&gt;
| member 4 = [[User:Dominicjansunga|Dominic Sunga]]&lt;br /&gt;
| member 5 = &lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
SUBPAGES&lt;br /&gt;
Basic Skills Project: http://wiki.ubc.ca/Course:MATH110/003/Groups/Group_04/Basic_Skills_Project&lt;br /&gt;
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Homework - Due October 20&lt;br /&gt;
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THOSE THAT POSE NO PROBLEM TO THE GROUP&lt;br /&gt;
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Basic Functions&lt;br /&gt;
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Equations&lt;br /&gt;
Composition of functions&lt;br /&gt;
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Intersection of functions&lt;br /&gt;
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Distances of lines&lt;br /&gt;
Trigonometry and Pythagorean Theorem&lt;br /&gt;
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Areas and Volume&lt;br /&gt;
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Mathematical writing&lt;br /&gt;
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polynomial Long Divison&lt;br /&gt;
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Construction of graphs&lt;br /&gt;
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inequalities&lt;br /&gt;
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properties of functions&lt;br /&gt;
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THOSE THAT POSE PROBLEMS FOR SOME, BUT NOT ALL&lt;br /&gt;
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Graphs of functions&lt;br /&gt;
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Operations of graphs of functions&lt;br /&gt;
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THOSE THAT NOONE KNOWS HOW TO HANDLE&lt;br /&gt;
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=Homework - Due October 20=&lt;br /&gt;
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1)&lt;br /&gt;
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We know that the turtle belongs to Suzan because every other pet was given a owner, except for the Turtle. &lt;br /&gt;
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We are given two names of the animals – Suzan the frog and Jun the Cat. From this, we can deduce the names of the other animals – Jaela the turtle, Bianca the Snake and Tosh the Parrot&lt;br /&gt;
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With this information, we can create a table:&lt;br /&gt;
&lt;br /&gt;
Owner	Gender	Pet	Pet name&lt;br /&gt;
Tosh	Male	Cat	Jun&lt;br /&gt;
Bianca	Female	Frog 	Suzan&lt;br /&gt;
Suzan’s Mom	Female	Turtle	Jaela&lt;br /&gt;
Jun	Male	Snake	Bianca&lt;br /&gt;
Jaela	Female	Parrot 	Tosh&lt;br /&gt;
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The question is asking us which pet belongs to Suzan’s mother. The answer is Jaela the Turtle. &lt;br /&gt;
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3)&lt;br /&gt;
Use a graph where the column headings are for the names of the people and the row headings are for the different positions. Then use the given information to eliminate and deduce what positions each person will play.&lt;br /&gt;
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Ed- pitcher &lt;br /&gt;
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Pascal - left outfielder &lt;br /&gt;
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Mathieu - right outfielder &lt;br /&gt;
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Sung - first base &lt;br /&gt;
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Adam - centre outfielder &lt;br /&gt;
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Hasan - second base &lt;br /&gt;
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Bobo - catcher &lt;br /&gt;
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Jason - third base &lt;br /&gt;
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Charles - shortstop &lt;br /&gt;
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4) &lt;br /&gt;
Understand the problem&lt;br /&gt;
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The question is asking me who will be playing with Fernanda on the fifth day.  The question state that 3 matches are played simultaneously during each day and each player plays each of the others once. &lt;br /&gt;
1st	Carla vs. Petra	&lt;br /&gt;
2nd	Carla vs. Janet&lt;br /&gt;
3rd	Janet vs. Li&lt;br /&gt;
4th	Petra vs. Sandra&lt;br /&gt;
5th 	Fernanda vs. ?&lt;br /&gt;
&lt;br /&gt;
Plan a strategy for solving the problem &lt;br /&gt;
Since the question already gives us some of the teams during these five days and we already know that each player plays each of the other once, 3 games per day, and played simultaneously which means that one player will only play one game per day. By knowing all these circumstances, we can create a chart to list out all the possible teams that can play on each day by elimination. &lt;br /&gt;
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Execute the strategy&lt;br /&gt;
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1st	Carla vs. Petra 	J vs. S	 F vs. L	or	J vs. F	S vs. L	&lt;br /&gt;
Janet can’t play with Li → they will be playing on the 3rd day, we eliminate the second choice since on the 4th day F vs. J will be playing.&lt;br /&gt;
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2nd	Carla vs. Janet 	P vs. L	 S vs. F	or	P vs. F	S vs. L&lt;br /&gt;
Petra can’t play with Sandra → they will be playing on the 4th day; we eliminate the second choice since on the 3rd day will be P vs. F.&lt;br /&gt;
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3rd	Janet vs. Li	        C vs. S  P vs. F	or	C vs. F	S vs. P&lt;br /&gt;
Carla can’t play with Petra → they played on the 1st day; we eliminate the second choice since we can tell from the table they will be playing on the 4th day. &lt;br /&gt;
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4th	Petra vs. Sandra	C vs. Li F vs. J	or	C vs. F	J vs. Li&lt;br /&gt;
Carl can’t play with Janet → they played on the 2nd day, and we eliminate the second choice since we can tell from the table Janet vs. Li played on the 3rd day. &lt;br /&gt;
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→ from the table above we can list out the teams will be playing on the 5th day&lt;br /&gt;
5th 	Fernanda vs. Carla	Petra vs. Janet		Li vs. Sandra&lt;br /&gt;
The table above list out all the possible arrangement of each teams, and we eliminate all the impossible guesses.  Each of the players only played with each other once during 5 days. &lt;br /&gt;
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[[on the 5th day, Fernanda is playing with Carla]]&lt;br /&gt;
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=Homework - Due October 13=&lt;br /&gt;
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Question 1:&lt;br /&gt;
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Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
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Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
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Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
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Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
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Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
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Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
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Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
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Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
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Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
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Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
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Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
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Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
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If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
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Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
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Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
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Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
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Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
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Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
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Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
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Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
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Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
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Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
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Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
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Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
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Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
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Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
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Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
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Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
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Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
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Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
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Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
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Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
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Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
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Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
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Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
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a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
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Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
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Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
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Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
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Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
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Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
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Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
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Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
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Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
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Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
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Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
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Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
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Step 4:&lt;br /&gt;
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Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
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Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
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Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
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Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
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Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
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Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
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Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
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Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
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Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
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Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;br /&gt;
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16. 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;
Let x represent the number of daughters&lt;br /&gt;
Let y represent the number of brothers&lt;br /&gt;
&lt;br /&gt;
From the information given in the question, we can deduce the following equations.&lt;br /&gt;
1)	y = x-1&lt;br /&gt;
&lt;br /&gt;
2)	x = 2(y-1)&lt;br /&gt;
   = 2y - 2&lt;br /&gt;
&lt;br /&gt;
Now let us substitute equation 1 into equation 2 to solve for x.&lt;br /&gt;
x = 2y – 2&lt;br /&gt;
x = 2(x-1) – 2&lt;br /&gt;
x = 2x – 2 – 2&lt;br /&gt;
x = 2x – 4&lt;br /&gt;
x = 4&lt;br /&gt;
&lt;br /&gt;
Substitute x = 4 into equation 1.&lt;br /&gt;
	y = x – 1&lt;br /&gt;
	y = 4 – 1&lt;br /&gt;
	y = 3&lt;br /&gt;
&lt;br /&gt;
	Therefore, the family has 4 girls and 3 boys&lt;br /&gt;
&lt;br /&gt;
17. 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;
Let x represent Dan’s actual weight.&lt;br /&gt;
Let y represent Sarah’s actual weight.&lt;br /&gt;
Let z represent the error on the scale.&lt;br /&gt;
&lt;br /&gt;
Then using the information given in the question we can come up with the following 3 equations.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
3)	105kg = (x + y) + z&lt;br /&gt;
Now we will reorganize the variables in equations 1 and 2.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
            x = 60 – z&lt;br /&gt;
&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
      y = 50kg – z&lt;br /&gt;
&lt;br /&gt;
Now substitute x = 60 – z and y = 50kg – z into equation 3&lt;br /&gt;
105kg = (x + y) + z&lt;br /&gt;
105kg = [(60-z) + (50kg – z )] + z&lt;br /&gt;
105kg = (60 – z + 50kg – z) + z&lt;br /&gt;
105kg = 110kg – z&lt;br /&gt;
	z = 5kg&lt;br /&gt;
&lt;br /&gt;
	Therefore, the scale reads too high by 5kg.&lt;br /&gt;
&lt;br /&gt;
18. 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;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar. &lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following 4 equations.&lt;br /&gt;
1)	y = x/3 &lt;br /&gt;
2)	z = (1/3)(x - x/3)&lt;br /&gt;
3)	w = (1/3)[x – (1/3)(x - x/3) – x/3]&lt;br /&gt;
4)	x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Substitute the values of y, z and w into equation 4.&lt;br /&gt;
x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40&lt;br /&gt;
	      x = (x/3) + (2/9)x + (4/27)x + 40&lt;br /&gt;
	      x = (19/27)x + 40&lt;br /&gt;
        (8/27)x = 40	      &lt;br /&gt;
      	      x = 135&lt;br /&gt;
	&lt;br /&gt;
	Therefore, there are 135 pennies in total at the start.&lt;br /&gt;
&lt;br /&gt;
19. 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;
Let b represent the total amount of milk.&lt;br /&gt;
Let a represent the total amount of coffee.&lt;br /&gt;
Let n represent the number of people in Angela’s family.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following equation.&lt;br /&gt;
&lt;br /&gt;
(a/6 + b/4)n = a + b&lt;br /&gt;
2a(6-n) = 3b(n-4)&lt;br /&gt;
Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.&lt;br /&gt;
&lt;br /&gt;
Therefore, there are at least 5 people in the family.&lt;br /&gt;
	&lt;br /&gt;
20. 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;
If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour&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;
Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.&lt;br /&gt;
&lt;br /&gt;
22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&lt;br /&gt;
&lt;br /&gt;
Let R represent a rainy day.&lt;br /&gt;
Let S represent a sunny day.&lt;br /&gt;
It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons.&lt;br /&gt;
Morning - R S S S S R R S S S S S R R S S R R &lt;br /&gt;
Noon - S S R R S S S R R S R S S S R S S S&lt;br /&gt;
There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.&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;
Let x represent the oldest child.&lt;br /&gt;
Let y represent the second child.&lt;br /&gt;
Let z represent the third child.&lt;br /&gt;
&lt;br /&gt;
Given that the produce of their ages is 36. We can deduce equation 1.&lt;br /&gt;
1)	36 = xyz&lt;br /&gt;
&lt;br /&gt;
Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.&lt;br /&gt;
&lt;br /&gt;
Ages		Product	Sum&lt;br /&gt;
	1 x 1 x 36	36		38&lt;br /&gt;
2 x 3 x 6	36		11&lt;br /&gt;
	1 x 3 x 12	36		16	&lt;br /&gt;
2 x 2 x 9	36		13&lt;br /&gt;
	1 x 4 x 9	36		14&lt;br /&gt;
3 x 3 x 4	36		10&lt;br /&gt;
	1 x 2 x 18	36		21&lt;br /&gt;
	1 x 6 x 6	36		13&lt;br /&gt;
&lt;br /&gt;
	The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.&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 A and B represent the 2 candles.&lt;br /&gt;
Let A represent the candle that takes 6 hours to burn out.&lt;br /&gt;
Let B represent the candle that takes 3 hours to burn out.&lt;br /&gt;
&lt;br /&gt;
Candle A takes twice as long to burn out than Candle B. We can form equation 1.&lt;br /&gt;
1)	A = 2B&lt;br /&gt;
&lt;br /&gt;
Assume that time starts at 0 and that the burning of the candle is constant.&lt;br /&gt;
Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour.&lt;br /&gt;
Let x represent the time in hours. We can form equation 2 and 3.&lt;br /&gt;
2)	A = (6-x)/6&lt;br /&gt;
3)	B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
Substitute equation 2 and 3 into equation 1.&lt;br /&gt;
A = 2B&lt;br /&gt;
(6-x)/6 = 2[(3-x)/6]&lt;br /&gt;
x = 2&lt;br /&gt;
&lt;br /&gt;
Therefore, it takes 2 hours for one candle to be half of the other.&lt;br /&gt;
&lt;br /&gt;
25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L. &lt;br /&gt;
&lt;br /&gt;
Given:&lt;br /&gt;
Candle X has length L and was lit at 6 hour.&lt;br /&gt;
Candle Y has length L + 1 and was lit at 4.5 hour.&lt;br /&gt;
At 8.5 hour, the length of Candle A is equal to the length of Candle B.&lt;br /&gt;
Candle A burns out at 10 hours.&lt;br /&gt;
Candle B burns out at 10.5 hours.&lt;br /&gt;
&lt;br /&gt;
Then the total burning time of Candle A is: 10 – 6 = 4 hours&lt;br /&gt;
Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours&lt;br /&gt;
&lt;br /&gt;
Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours.&lt;br /&gt;
Burning time till 8.5h for Candle B  is: 8.5 - 4.5 = 4 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Let x represent the length of Candle A&lt;br /&gt;
Let y represent the length of Candle B&lt;br /&gt;
&lt;br /&gt;
y = x + 1&lt;br /&gt;
&lt;br /&gt;
Let B1 represent the burning constant for x&lt;br /&gt;
Let B2 represent the burning constant for y&lt;br /&gt;
We know that x=y at 8.5 hours.&lt;br /&gt;
&lt;br /&gt;
 (x-4) B1 = x + 1 – 2.5B2&lt;br /&gt;
&lt;br /&gt;
y = 6 B2 at end time&lt;br /&gt;
x = 4B1 at end time&lt;br /&gt;
6B2 = 4B1 + 1 &lt;br /&gt;
B1 = (6B1 -1)/4&lt;br /&gt;
&lt;br /&gt;
Using the above equations and substituting back into the original.&lt;br /&gt;
4B2 = 1 – 2.5B1&lt;br /&gt;
4B2 = 1 – 2.5((6B1 -1)/4)&lt;br /&gt;
B2 = 1.5&lt;br /&gt;
&lt;br /&gt;
Substitute B2 into y = 6B2&lt;br /&gt;
4((4B1 + 1)/6) = 1 – 2.5B1&lt;br /&gt;
B1 = 2&lt;br /&gt;
Substitute B1 into x = 4B1&lt;br /&gt;
&lt;br /&gt;
y = 6B2&lt;br /&gt;
y = 6(1.5) = 9&lt;br /&gt;
&lt;br /&gt;
x = 4B1&lt;br /&gt;
x = 4(2) = 8&lt;br /&gt;
&lt;br /&gt;
Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=63522</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04/Basic Skills Project</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04/Basic_Skills_Project&amp;diff=63522"/>
		<updated>2010-11-28T06:25:59Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: Created page with &amp;#039;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion.   What are functions?&amp;#039;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;For the basic skills project, group 4 plans to outline the properties of functions in a detailed fashion. &lt;br /&gt;
&lt;br /&gt;
What are functions?&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course_talk:MATH110/003/Groups/Group_04&amp;diff=61001</id>
		<title>Course talk:MATH110/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course_talk:MATH110/003/Groups/Group_04&amp;diff=61001"/>
		<updated>2010-11-12T07:34:12Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: /* Basic Skills Project */ new section&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Comments for Homework 4=&lt;br /&gt;
What happened to question 2 and 5? Please communicate with your group members to make sure at least all the questions are done! &lt;br /&gt;
&lt;br /&gt;
Also, please provide more thought process. Just an answer or a brief word about process of elimination is not going to get you a lot of marks. More over, if you say you are using a graph or chart, SHOW the graph or chart!&lt;br /&gt;
&lt;br /&gt;
[[User:JingFeiYu|JingFeiYu]]&lt;br /&gt;
&lt;br /&gt;
=Comments for Homework 3=&lt;br /&gt;
Good job in getting most of the questions! Below is a few pointers for improvements&lt;br /&gt;
&lt;br /&gt;
1) Please format your work in a more readable manner. Right now it&#039;s difficult to distinguish between questions &lt;br /&gt;
&lt;br /&gt;
2) Please also stick to one way of presenting. In the beginning each questions was broken down to steps accounting to the problem &lt;br /&gt;
solving steps given in the question page, but latter you stopped doing that. Just stick to one uniform way throughout!&lt;br /&gt;
&lt;br /&gt;
3) Please write down more of your thought processes! It is hard to tell how you solved the problems, after all, that&#039;s the most important exercise in this homework. &lt;br /&gt;
&lt;br /&gt;
4) Read the questions more carefully! Do not make groundless assumptions or jumping to conclusions. Check your answers after you&#039;ve arrived at them!&lt;br /&gt;
&lt;br /&gt;
[[User:JingFeiYu|JingFeiYu]]&lt;br /&gt;
&lt;br /&gt;
== Basic Skills Project ==&lt;br /&gt;
&lt;br /&gt;
Our group plans to contribute to his project by doing the following:&lt;br /&gt;
&lt;br /&gt;
Providing examples of how to solve particular problems, which includes a step by step procedure of how to go about it as well as descriptions at each step.&lt;br /&gt;
&lt;br /&gt;
When it comes to theory, providing real world examples so as to remember the theories more clearly and be able to apply them&lt;br /&gt;
The real world examples will also come with example problems and methods.&lt;br /&gt;
&lt;br /&gt;
Creating a program of some sort to create a study group when people want them. Basically, if people are having a hard time with a section of the course, they can go to the project, request a study group and then they can meet up and work out the problems together. This way, students learn from each other, times are flexible because everyone is involved and it will be very dynamic as it won&#039;t be the same exact study group every single time. &lt;br /&gt;
&lt;br /&gt;
Find resources on the internet that give concise instructions as to how to solve problems that students can refer to at their leisure.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58365</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58365"/>
		<updated>2010-10-29T03:33:47Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 4&lt;br /&gt;
| member 1 = Anoud Al-Moqbil&lt;br /&gt;
| member 2 = Owen Johnston&lt;br /&gt;
| member 3 = Charlene Ng&lt;br /&gt;
| member 4 = Ling Sun&lt;br /&gt;
| member 5 = [[User:Dominicjansunga|Dominic Sunga]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
Homework - Due October 20&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE NO PROBLEM TO THE GROUP&lt;br /&gt;
&lt;br /&gt;
Basic Functions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Equations&lt;br /&gt;
Composition of functions&lt;br /&gt;
&lt;br /&gt;
Intersection of functions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Distances of lines&lt;br /&gt;
Trigonometry and Pythagorean Theorem&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Areas and Volume&lt;br /&gt;
&lt;br /&gt;
Mathematical writing&lt;br /&gt;
&lt;br /&gt;
polynomial Long Divison&lt;br /&gt;
&lt;br /&gt;
Construction of graphs&lt;br /&gt;
&lt;br /&gt;
inequalities&lt;br /&gt;
&lt;br /&gt;
properties of functions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE PROBLEMS FOR SOME, BUT NOT ALL&lt;br /&gt;
&lt;br /&gt;
Graphs of functions&lt;br /&gt;
&lt;br /&gt;
Operations of graphs of functions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT NOONE KNOWS HOW TO HANDLE&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 20=&lt;br /&gt;
&lt;br /&gt;
1)&lt;br /&gt;
&lt;br /&gt;
We know that the turtle belongs to Suzan because every other pet was given a owner, except for the Turtle. &lt;br /&gt;
&lt;br /&gt;
We are given two names of the animals – Suzan the frog and Jun the Cat. From this, we can deduce the names of the other animals – Jaela the turtle, Bianca the Snake and Tosh the Parrot&lt;br /&gt;
&lt;br /&gt;
With this information, we can create a table:&lt;br /&gt;
&lt;br /&gt;
Owner	Gender	Pet	Pet name&lt;br /&gt;
Tosh	Male	Cat	Jun&lt;br /&gt;
Bianca	Female	Frog 	Suzan&lt;br /&gt;
Suzan’s Mom	Female	Turtle	Jaela&lt;br /&gt;
Jun	Male	Snake	Bianca&lt;br /&gt;
Jaela	Female	Parrot 	Tosh&lt;br /&gt;
&lt;br /&gt;
The question is asking us which pet belongs to Suzan’s mother. The answer is Jaela the Turtle. &lt;br /&gt;
&lt;br /&gt;
3)&lt;br /&gt;
Use a graph where the column headings are for the names of the people and the row headings are for the different positions. Then use the given information to eliminate and deduce what positions each person will play.&lt;br /&gt;
&lt;br /&gt;
Ed- pitcher &lt;br /&gt;
&lt;br /&gt;
Pascal - left outfielder &lt;br /&gt;
&lt;br /&gt;
Mathieu - right outfielder &lt;br /&gt;
&lt;br /&gt;
Sung - first base &lt;br /&gt;
&lt;br /&gt;
Adam - centre outfielder &lt;br /&gt;
&lt;br /&gt;
Hasan - second base &lt;br /&gt;
&lt;br /&gt;
Bobo - catcher &lt;br /&gt;
&lt;br /&gt;
Jason - third base &lt;br /&gt;
&lt;br /&gt;
Charles - shortstop &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4) &lt;br /&gt;
Understand the problem&lt;br /&gt;
&lt;br /&gt;
The question is asking me who will be playing with Fernanda on the fifth day.  The question state that 3 matches are played simultaneously during each day and each player plays each of the others once. &lt;br /&gt;
1st	Carla vs. Petra	&lt;br /&gt;
2nd	Carla vs. Janet&lt;br /&gt;
3rd	Janet vs. Li&lt;br /&gt;
4th	Petra vs. Sandra&lt;br /&gt;
5th 	Fernanda vs. ?&lt;br /&gt;
&lt;br /&gt;
Plan a strategy for solving the problem &lt;br /&gt;
Since the question already gives us some of the teams during these five days and we already know that each player plays each of the other once, 3 games per day, and played simultaneously which means that one player will only play one game per day. By knowing all these circumstances, we can create a chart to list out all the possible teams that can play on each day by elimination. &lt;br /&gt;
&lt;br /&gt;
Execute the strategy&lt;br /&gt;
&lt;br /&gt;
1st	Carla vs. Petra 	J vs. S	 F vs. L	or	J vs. F	S vs. L	&lt;br /&gt;
Janet can’t play with Li → they will be playing on the 3rd day, we eliminate the second choice since on the 4th day F vs. J will be playing.&lt;br /&gt;
&lt;br /&gt;
2nd	Carla vs. Janet 	P vs. L	 S vs. F	or	P vs. F	S vs. L&lt;br /&gt;
Petra can’t play with Sandra → they will be playing on the 4th day; we eliminate the second choice since on the 3rd day will be P vs. F.&lt;br /&gt;
&lt;br /&gt;
3rd	Janet vs. Li	        C vs. S  P vs. F	or	C vs. F	S vs. P&lt;br /&gt;
Carla can’t play with Petra → they played on the 1st day; we eliminate the second choice since we can tell from the table they will be playing on the 4th day. &lt;br /&gt;
&lt;br /&gt;
4th	Petra vs. Sandra	C vs. Li F vs. J	or	C vs. F	J vs. Li&lt;br /&gt;
Carl can’t play with Janet → they played on the 2nd day, and we eliminate the second choice since we can tell from the table Janet vs. Li played on the 3rd day. &lt;br /&gt;
&lt;br /&gt;
→ from the table above we can list out the teams will be playing on the 5th day&lt;br /&gt;
5th 	Fernanda vs. Carla	Petra vs. Janet		Li vs. Sandra&lt;br /&gt;
The table above list out all the possible arrangement of each teams, and we eliminate all the impossible guesses.  Each of the players only played with each other once during 5 days. &lt;br /&gt;
&lt;br /&gt;
[[on the 5th day, Fernanda is playing with Carla]]&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 13=&lt;br /&gt;
&lt;br /&gt;
Question 1:&lt;br /&gt;
&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
&lt;br /&gt;
Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
&lt;br /&gt;
Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
&lt;br /&gt;
Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
&lt;br /&gt;
Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
&lt;br /&gt;
Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
&lt;br /&gt;
Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
&lt;br /&gt;
a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
&lt;br /&gt;
Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
&lt;br /&gt;
Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
&lt;br /&gt;
Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
&lt;br /&gt;
Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
&lt;br /&gt;
Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;br /&gt;
&lt;br /&gt;
16. 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;
Let x represent the number of daughters&lt;br /&gt;
Let y represent the number of brothers&lt;br /&gt;
&lt;br /&gt;
From the information given in the question, we can deduce the following equations.&lt;br /&gt;
1)	y = x-1&lt;br /&gt;
&lt;br /&gt;
2)	x = 2(y-1)&lt;br /&gt;
   = 2y - 2&lt;br /&gt;
&lt;br /&gt;
Now let us substitute equation 1 into equation 2 to solve for x.&lt;br /&gt;
x = 2y – 2&lt;br /&gt;
x = 2(x-1) – 2&lt;br /&gt;
x = 2x – 2 – 2&lt;br /&gt;
x = 2x – 4&lt;br /&gt;
x = 4&lt;br /&gt;
&lt;br /&gt;
Substitute x = 4 into equation 1.&lt;br /&gt;
	y = x – 1&lt;br /&gt;
	y = 4 – 1&lt;br /&gt;
	y = 3&lt;br /&gt;
&lt;br /&gt;
	Therefore, the family has 4 girls and 3 boys&lt;br /&gt;
&lt;br /&gt;
17. 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;
Let x represent Dan’s actual weight.&lt;br /&gt;
Let y represent Sarah’s actual weight.&lt;br /&gt;
Let z represent the error on the scale.&lt;br /&gt;
&lt;br /&gt;
Then using the information given in the question we can come up with the following 3 equations.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
3)	105kg = (x + y) + z&lt;br /&gt;
Now we will reorganize the variables in equations 1 and 2.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
            x = 60 – z&lt;br /&gt;
&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
      y = 50kg – z&lt;br /&gt;
&lt;br /&gt;
Now substitute x = 60 – z and y = 50kg – z into equation 3&lt;br /&gt;
105kg = (x + y) + z&lt;br /&gt;
105kg = [(60-z) + (50kg – z )] + z&lt;br /&gt;
105kg = (60 – z + 50kg – z) + z&lt;br /&gt;
105kg = 110kg – z&lt;br /&gt;
	z = 5kg&lt;br /&gt;
&lt;br /&gt;
	Therefore, the scale reads too high by 5kg.&lt;br /&gt;
&lt;br /&gt;
18. 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;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar. &lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following 4 equations.&lt;br /&gt;
1)	y = x/3 &lt;br /&gt;
2)	z = (1/3)(x - x/3)&lt;br /&gt;
3)	w = (1/3)[x – (1/3)(x - x/3) – x/3]&lt;br /&gt;
4)	x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Substitute the values of y, z and w into equation 4.&lt;br /&gt;
x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40&lt;br /&gt;
	      x = (x/3) + (2/9)x + (4/27)x + 40&lt;br /&gt;
	      x = (19/27)x + 40&lt;br /&gt;
        (8/27)x = 40	      &lt;br /&gt;
      	      x = 135&lt;br /&gt;
	&lt;br /&gt;
	Therefore, there are 135 pennies in total at the start.&lt;br /&gt;
&lt;br /&gt;
19. 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;
Let b represent the total amount of milk.&lt;br /&gt;
Let a represent the total amount of coffee.&lt;br /&gt;
Let n represent the number of people in Angela’s family.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following equation.&lt;br /&gt;
&lt;br /&gt;
(a/6 + b/4)n = a + b&lt;br /&gt;
2a(6-n) = 3b(n-4)&lt;br /&gt;
Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.&lt;br /&gt;
&lt;br /&gt;
Therefore, there are at least 5 people in the family.&lt;br /&gt;
	&lt;br /&gt;
20. 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;
If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour&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;
Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.&lt;br /&gt;
&lt;br /&gt;
22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&lt;br /&gt;
&lt;br /&gt;
Let R represent a rainy day.&lt;br /&gt;
Let S represent a sunny day.&lt;br /&gt;
It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons.&lt;br /&gt;
Morning - R S S S S R R S S S S S R R S S R R &lt;br /&gt;
Noon - S S R R S S S R R S R S S S R S S S&lt;br /&gt;
There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.&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;
Let x represent the oldest child.&lt;br /&gt;
Let y represent the second child.&lt;br /&gt;
Let z represent the third child.&lt;br /&gt;
&lt;br /&gt;
Given that the produce of their ages is 36. We can deduce equation 1.&lt;br /&gt;
1)	36 = xyz&lt;br /&gt;
&lt;br /&gt;
Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.&lt;br /&gt;
&lt;br /&gt;
Ages		Product	Sum&lt;br /&gt;
	1 x 1 x 36	36		38&lt;br /&gt;
2 x 3 x 6	36		11&lt;br /&gt;
	1 x 3 x 12	36		16	&lt;br /&gt;
2 x 2 x 9	36		13&lt;br /&gt;
	1 x 4 x 9	36		14&lt;br /&gt;
3 x 3 x 4	36		10&lt;br /&gt;
	1 x 2 x 18	36		21&lt;br /&gt;
	1 x 6 x 6	36		13&lt;br /&gt;
&lt;br /&gt;
	The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.&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 A and B represent the 2 candles.&lt;br /&gt;
Let A represent the candle that takes 6 hours to burn out.&lt;br /&gt;
Let B represent the candle that takes 3 hours to burn out.&lt;br /&gt;
&lt;br /&gt;
Candle A takes twice as long to burn out than Candle B. We can form equation 1.&lt;br /&gt;
1)	A = 2B&lt;br /&gt;
&lt;br /&gt;
Assume that time starts at 0 and that the burning of the candle is constant.&lt;br /&gt;
Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour.&lt;br /&gt;
Let x represent the time in hours. We can form equation 2 and 3.&lt;br /&gt;
2)	A = (6-x)/6&lt;br /&gt;
3)	B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
Substitute equation 2 and 3 into equation 1.&lt;br /&gt;
A = 2B&lt;br /&gt;
(6-x)/6 = 2[(3-x)/6]&lt;br /&gt;
x = 2&lt;br /&gt;
&lt;br /&gt;
Therefore, it takes 2 hours for one candle to be half of the other.&lt;br /&gt;
&lt;br /&gt;
25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L. &lt;br /&gt;
&lt;br /&gt;
Given:&lt;br /&gt;
Candle X has length L and was lit at 6 hour.&lt;br /&gt;
Candle Y has length L + 1 and was lit at 4.5 hour.&lt;br /&gt;
At 8.5 hour, the length of Candle A is equal to the length of Candle B.&lt;br /&gt;
Candle A burns out at 10 hours.&lt;br /&gt;
Candle B burns out at 10.5 hours.&lt;br /&gt;
&lt;br /&gt;
Then the total burning time of Candle A is: 10 – 6 = 4 hours&lt;br /&gt;
Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours&lt;br /&gt;
&lt;br /&gt;
Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours.&lt;br /&gt;
Burning time till 8.5h for Candle B  is: 8.5 - 4.5 = 4 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Let x represent the length of Candle A&lt;br /&gt;
Let y represent the length of Candle B&lt;br /&gt;
&lt;br /&gt;
y = x + 1&lt;br /&gt;
&lt;br /&gt;
Let B1 represent the burning constant for x&lt;br /&gt;
Let B2 represent the burning constant for y&lt;br /&gt;
We know that x=y at 8.5 hours.&lt;br /&gt;
&lt;br /&gt;
 (x-4) B1 = x + 1 – 2.5B2&lt;br /&gt;
&lt;br /&gt;
y = 6 B2 at end time&lt;br /&gt;
x = 4B1 at end time&lt;br /&gt;
6B2 = 4B1 + 1 &lt;br /&gt;
B1 = (6B1 -1)/4&lt;br /&gt;
&lt;br /&gt;
Using the above equations and substituting back into the original.&lt;br /&gt;
4B2 = 1 – 2.5B1&lt;br /&gt;
4B2 = 1 – 2.5((6B1 -1)/4)&lt;br /&gt;
B2 = 1.5&lt;br /&gt;
&lt;br /&gt;
Substitute B2 into y = 6B2&lt;br /&gt;
4((4B1 + 1)/6) = 1 – 2.5B1&lt;br /&gt;
B1 = 2&lt;br /&gt;
Substitute B1 into x = 4B1&lt;br /&gt;
&lt;br /&gt;
y = 6B2&lt;br /&gt;
y = 6(1.5) = 9&lt;br /&gt;
&lt;br /&gt;
x = 4B1&lt;br /&gt;
x = 4(2) = 8&lt;br /&gt;
&lt;br /&gt;
Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58364</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58364"/>
		<updated>2010-10-29T03:25:13Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 4&lt;br /&gt;
| member 1 = Anoud Al-Moqbil&lt;br /&gt;
| member 2 = Owen Johnston&lt;br /&gt;
| member 3 = Charlene Ng&lt;br /&gt;
| member 4 = Ling Sun&lt;br /&gt;
| member 5 = [[User:Dominicjansunga|Dominic Sunga]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
Homework - Due October 20&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE NO PROBLEM TO THE GROUP&lt;br /&gt;
&lt;br /&gt;
Basic Functions&lt;br /&gt;
Equations&lt;br /&gt;
Composition of functions&lt;br /&gt;
Intersection of functions&lt;br /&gt;
Distances of lines&lt;br /&gt;
Trigonometry and Pythagorean Theorem&lt;br /&gt;
Areas and Volume&lt;br /&gt;
Mathematical writing&lt;br /&gt;
polynomial Long Divison&lt;br /&gt;
Construction of graphs&lt;br /&gt;
inequalities&lt;br /&gt;
properties of functions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE PROBLEMS FOR SOME, BUT NOT ALL&lt;br /&gt;
&lt;br /&gt;
Graphs of functions&lt;br /&gt;
&lt;br /&gt;
Operations of graphs of functions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT NOONE KNOWS HOW TO HANDLE&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 20=&lt;br /&gt;
&lt;br /&gt;
1)&lt;br /&gt;
&lt;br /&gt;
We know that the turtle belongs to Suzan because every other pet was given a owner, except for the Turtle. &lt;br /&gt;
&lt;br /&gt;
We are given two names of the animals – Suzan the frog and Jun the Cat. From this, we can deduce the names of the other animals – Jaela the turtle, Bianca the Snake and Tosh the Parrot&lt;br /&gt;
&lt;br /&gt;
With this information, we can create a table:&lt;br /&gt;
&lt;br /&gt;
Owner	Gender	Pet	Pet name&lt;br /&gt;
Tosh	Male	Cat	Jun&lt;br /&gt;
Bianca	Female	Frog 	Suzan&lt;br /&gt;
Suzan’s Mom	Female	Turtle	Jaela&lt;br /&gt;
Jun	Male	Snake	Bianca&lt;br /&gt;
Jaela	Female	Parrot 	Tosh&lt;br /&gt;
&lt;br /&gt;
The question is asking us which pet belongs to Suzan’s mother. The answer is Jaela the Turtle. &lt;br /&gt;
&lt;br /&gt;
3)&lt;br /&gt;
Use a graph where the column headings are for the names of the people and the row headings are for the different positions. Then use the given information to eliminate and deduce what positions each person will play.&lt;br /&gt;
&lt;br /&gt;
Ed- pitcher &lt;br /&gt;
&lt;br /&gt;
Pascal - left outfielder &lt;br /&gt;
&lt;br /&gt;
Mathieu - right outfielder &lt;br /&gt;
&lt;br /&gt;
Sung - first base &lt;br /&gt;
&lt;br /&gt;
Adam - centre outfielder &lt;br /&gt;
&lt;br /&gt;
Hasan - second base &lt;br /&gt;
&lt;br /&gt;
Bobo - catcher &lt;br /&gt;
&lt;br /&gt;
Jason - third base &lt;br /&gt;
&lt;br /&gt;
Charles - shortstop &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4) &lt;br /&gt;
Understand the problem&lt;br /&gt;
&lt;br /&gt;
The question is asking me who will be playing with Fernanda on the fifth day.  The question state that 3 matches are played simultaneously during each day and each player plays each of the others once. &lt;br /&gt;
1st	Carla vs. Petra	&lt;br /&gt;
2nd	Carla vs. Janet&lt;br /&gt;
3rd	Janet vs. Li&lt;br /&gt;
4th	Petra vs. Sandra&lt;br /&gt;
5th 	Fernanda vs. ?&lt;br /&gt;
&lt;br /&gt;
Plan a strategy for solving the problem &lt;br /&gt;
Since the question already gives us some of the teams during these five days and we already know that each player plays each of the other once, 3 games per day, and played simultaneously which means that one player will only play one game per day. By knowing all these circumstances, we can create a chart to list out all the possible teams that can play on each day by elimination. &lt;br /&gt;
&lt;br /&gt;
Execute the strategy&lt;br /&gt;
&lt;br /&gt;
1st	Carla vs. Petra 	J vs. S	 F vs. L	or	J vs. F	S vs. L	&lt;br /&gt;
Janet can’t play with Li → they will be playing on the 3rd day, we eliminate the second choice since on the 4th day F vs. J will be playing.&lt;br /&gt;
&lt;br /&gt;
2nd	Carla vs. Janet 	P vs. L	 S vs. F	or	P vs. F	S vs. L&lt;br /&gt;
Petra can’t play with Sandra → they will be playing on the 4th day; we eliminate the second choice since on the 3rd day will be P vs. F.&lt;br /&gt;
&lt;br /&gt;
3rd	Janet vs. Li	        C vs. S  P vs. F	or	C vs. F	S vs. P&lt;br /&gt;
Carla can’t play with Petra → they played on the 1st day; we eliminate the second choice since we can tell from the table they will be playing on the 4th day. &lt;br /&gt;
&lt;br /&gt;
4th	Petra vs. Sandra	C vs. Li F vs. J	or	C vs. F	J vs. Li&lt;br /&gt;
Carl can’t play with Janet → they played on the 2nd day, and we eliminate the second choice since we can tell from the table Janet vs. Li played on the 3rd day. &lt;br /&gt;
&lt;br /&gt;
→ from the table above we can list out the teams will be playing on the 5th day&lt;br /&gt;
5th 	Fernanda vs. Carla	Petra vs. Janet		Li vs. Sandra&lt;br /&gt;
The table above list out all the possible arrangement of each teams, and we eliminate all the impossible guesses.  Each of the players only played with each other once during 5 days. &lt;br /&gt;
&lt;br /&gt;
[[on the 5th day, Fernanda is playing with Carla]]&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 13=&lt;br /&gt;
&lt;br /&gt;
Question 1:&lt;br /&gt;
&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
&lt;br /&gt;
Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
&lt;br /&gt;
Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
&lt;br /&gt;
Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
&lt;br /&gt;
Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
&lt;br /&gt;
Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
&lt;br /&gt;
Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
&lt;br /&gt;
a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
&lt;br /&gt;
Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
&lt;br /&gt;
Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
&lt;br /&gt;
Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
&lt;br /&gt;
Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
&lt;br /&gt;
Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;br /&gt;
&lt;br /&gt;
16. 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;
Let x represent the number of daughters&lt;br /&gt;
Let y represent the number of brothers&lt;br /&gt;
&lt;br /&gt;
From the information given in the question, we can deduce the following equations.&lt;br /&gt;
1)	y = x-1&lt;br /&gt;
&lt;br /&gt;
2)	x = 2(y-1)&lt;br /&gt;
   = 2y - 2&lt;br /&gt;
&lt;br /&gt;
Now let us substitute equation 1 into equation 2 to solve for x.&lt;br /&gt;
x = 2y – 2&lt;br /&gt;
x = 2(x-1) – 2&lt;br /&gt;
x = 2x – 2 – 2&lt;br /&gt;
x = 2x – 4&lt;br /&gt;
x = 4&lt;br /&gt;
&lt;br /&gt;
Substitute x = 4 into equation 1.&lt;br /&gt;
	y = x – 1&lt;br /&gt;
	y = 4 – 1&lt;br /&gt;
	y = 3&lt;br /&gt;
&lt;br /&gt;
	Therefore, the family has 4 girls and 3 boys&lt;br /&gt;
&lt;br /&gt;
17. 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;
Let x represent Dan’s actual weight.&lt;br /&gt;
Let y represent Sarah’s actual weight.&lt;br /&gt;
Let z represent the error on the scale.&lt;br /&gt;
&lt;br /&gt;
Then using the information given in the question we can come up with the following 3 equations.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
3)	105kg = (x + y) + z&lt;br /&gt;
Now we will reorganize the variables in equations 1 and 2.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
            x = 60 – z&lt;br /&gt;
&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
      y = 50kg – z&lt;br /&gt;
&lt;br /&gt;
Now substitute x = 60 – z and y = 50kg – z into equation 3&lt;br /&gt;
105kg = (x + y) + z&lt;br /&gt;
105kg = [(60-z) + (50kg – z )] + z&lt;br /&gt;
105kg = (60 – z + 50kg – z) + z&lt;br /&gt;
105kg = 110kg – z&lt;br /&gt;
	z = 5kg&lt;br /&gt;
&lt;br /&gt;
	Therefore, the scale reads too high by 5kg.&lt;br /&gt;
&lt;br /&gt;
18. 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;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar. &lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following 4 equations.&lt;br /&gt;
1)	y = x/3 &lt;br /&gt;
2)	z = (1/3)(x - x/3)&lt;br /&gt;
3)	w = (1/3)[x – (1/3)(x - x/3) – x/3]&lt;br /&gt;
4)	x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Substitute the values of y, z and w into equation 4.&lt;br /&gt;
x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40&lt;br /&gt;
	      x = (x/3) + (2/9)x + (4/27)x + 40&lt;br /&gt;
	      x = (19/27)x + 40&lt;br /&gt;
        (8/27)x = 40	      &lt;br /&gt;
      	      x = 135&lt;br /&gt;
	&lt;br /&gt;
	Therefore, there are 135 pennies in total at the start.&lt;br /&gt;
&lt;br /&gt;
19. 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;
Let b represent the total amount of milk.&lt;br /&gt;
Let a represent the total amount of coffee.&lt;br /&gt;
Let n represent the number of people in Angela’s family.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following equation.&lt;br /&gt;
&lt;br /&gt;
(a/6 + b/4)n = a + b&lt;br /&gt;
2a(6-n) = 3b(n-4)&lt;br /&gt;
Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.&lt;br /&gt;
&lt;br /&gt;
Therefore, there are at least 5 people in the family.&lt;br /&gt;
	&lt;br /&gt;
20. 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;
If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour&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;
Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.&lt;br /&gt;
&lt;br /&gt;
22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&lt;br /&gt;
&lt;br /&gt;
Let R represent a rainy day.&lt;br /&gt;
Let S represent a sunny day.&lt;br /&gt;
It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons.&lt;br /&gt;
Morning - R S S S S R R S S S S S R R S S R R &lt;br /&gt;
Noon - S S R R S S S R R S R S S S R S S S&lt;br /&gt;
There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.&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;
Let x represent the oldest child.&lt;br /&gt;
Let y represent the second child.&lt;br /&gt;
Let z represent the third child.&lt;br /&gt;
&lt;br /&gt;
Given that the produce of their ages is 36. We can deduce equation 1.&lt;br /&gt;
1)	36 = xyz&lt;br /&gt;
&lt;br /&gt;
Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.&lt;br /&gt;
&lt;br /&gt;
Ages		Product	Sum&lt;br /&gt;
	1 x 1 x 36	36		38&lt;br /&gt;
2 x 3 x 6	36		11&lt;br /&gt;
	1 x 3 x 12	36		16	&lt;br /&gt;
2 x 2 x 9	36		13&lt;br /&gt;
	1 x 4 x 9	36		14&lt;br /&gt;
3 x 3 x 4	36		10&lt;br /&gt;
	1 x 2 x 18	36		21&lt;br /&gt;
	1 x 6 x 6	36		13&lt;br /&gt;
&lt;br /&gt;
	The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.&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 A and B represent the 2 candles.&lt;br /&gt;
Let A represent the candle that takes 6 hours to burn out.&lt;br /&gt;
Let B represent the candle that takes 3 hours to burn out.&lt;br /&gt;
&lt;br /&gt;
Candle A takes twice as long to burn out than Candle B. We can form equation 1.&lt;br /&gt;
1)	A = 2B&lt;br /&gt;
&lt;br /&gt;
Assume that time starts at 0 and that the burning of the candle is constant.&lt;br /&gt;
Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour.&lt;br /&gt;
Let x represent the time in hours. We can form equation 2 and 3.&lt;br /&gt;
2)	A = (6-x)/6&lt;br /&gt;
3)	B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
Substitute equation 2 and 3 into equation 1.&lt;br /&gt;
A = 2B&lt;br /&gt;
(6-x)/6 = 2[(3-x)/6]&lt;br /&gt;
x = 2&lt;br /&gt;
&lt;br /&gt;
Therefore, it takes 2 hours for one candle to be half of the other.&lt;br /&gt;
&lt;br /&gt;
25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L. &lt;br /&gt;
&lt;br /&gt;
Given:&lt;br /&gt;
Candle X has length L and was lit at 6 hour.&lt;br /&gt;
Candle Y has length L + 1 and was lit at 4.5 hour.&lt;br /&gt;
At 8.5 hour, the length of Candle A is equal to the length of Candle B.&lt;br /&gt;
Candle A burns out at 10 hours.&lt;br /&gt;
Candle B burns out at 10.5 hours.&lt;br /&gt;
&lt;br /&gt;
Then the total burning time of Candle A is: 10 – 6 = 4 hours&lt;br /&gt;
Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours&lt;br /&gt;
&lt;br /&gt;
Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours.&lt;br /&gt;
Burning time till 8.5h for Candle B  is: 8.5 - 4.5 = 4 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Let x represent the length of Candle A&lt;br /&gt;
Let y represent the length of Candle B&lt;br /&gt;
&lt;br /&gt;
y = x + 1&lt;br /&gt;
&lt;br /&gt;
Let B1 represent the burning constant for x&lt;br /&gt;
Let B2 represent the burning constant for y&lt;br /&gt;
We know that x=y at 8.5 hours.&lt;br /&gt;
&lt;br /&gt;
 (x-4) B1 = x + 1 – 2.5B2&lt;br /&gt;
&lt;br /&gt;
y = 6 B2 at end time&lt;br /&gt;
x = 4B1 at end time&lt;br /&gt;
6B2 = 4B1 + 1 &lt;br /&gt;
B1 = (6B1 -1)/4&lt;br /&gt;
&lt;br /&gt;
Using the above equations and substituting back into the original.&lt;br /&gt;
4B2 = 1 – 2.5B1&lt;br /&gt;
4B2 = 1 – 2.5((6B1 -1)/4)&lt;br /&gt;
B2 = 1.5&lt;br /&gt;
&lt;br /&gt;
Substitute B2 into y = 6B2&lt;br /&gt;
4((4B1 + 1)/6) = 1 – 2.5B1&lt;br /&gt;
B1 = 2&lt;br /&gt;
Substitute B1 into x = 4B1&lt;br /&gt;
&lt;br /&gt;
y = 6B2&lt;br /&gt;
y = 6(1.5) = 9&lt;br /&gt;
&lt;br /&gt;
x = 4B1&lt;br /&gt;
x = 4(2) = 8&lt;br /&gt;
&lt;br /&gt;
Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58361</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58361"/>
		<updated>2010-10-29T03:15:33Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 4&lt;br /&gt;
| member 1 = Anoud Al-Moqbil&lt;br /&gt;
| member 2 = Owen Johnston&lt;br /&gt;
| member 3 = Charlene Ng&lt;br /&gt;
| member 4 = Ling Sun&lt;br /&gt;
| member 5 = [[User:Dominicjansunga|Dominic Sunga]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
Homework - Due October 20&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE NO PROBLEM TO THE GROUP&lt;br /&gt;
Basic Functions&lt;br /&gt;
Equations&lt;br /&gt;
Composition of functions&lt;br /&gt;
Intersection of functions&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE PROBLEMS FOR SOME, BUT NOT ALL&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT NOONE KNOWS HOW TO HANDLE&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 20=&lt;br /&gt;
&lt;br /&gt;
1)&lt;br /&gt;
&lt;br /&gt;
We know that the turtle belongs to Suzan because every other pet was given a owner, except for the Turtle. &lt;br /&gt;
&lt;br /&gt;
We are given two names of the animals – Suzan the frog and Jun the Cat. From this, we can deduce the names of the other animals – Jaela the turtle, Bianca the Snake and Tosh the Parrot&lt;br /&gt;
&lt;br /&gt;
With this information, we can create a table:&lt;br /&gt;
&lt;br /&gt;
Owner	Gender	Pet	Pet name&lt;br /&gt;
Tosh	Male	Cat	Jun&lt;br /&gt;
Bianca	Female	Frog 	Suzan&lt;br /&gt;
Suzan’s Mom	Female	Turtle	Jaela&lt;br /&gt;
Jun	Male	Snake	Bianca&lt;br /&gt;
Jaela	Female	Parrot 	Tosh&lt;br /&gt;
&lt;br /&gt;
The question is asking us which pet belongs to Suzan’s mother. The answer is Jaela the Turtle. &lt;br /&gt;
&lt;br /&gt;
3)&lt;br /&gt;
Use a graph where the column headings are for the names of the people and the row headings are for the different positions. Then use the given information to eliminate and deduce what positions each person will play.&lt;br /&gt;
&lt;br /&gt;
Ed- pitcher &lt;br /&gt;
&lt;br /&gt;
Pascal - left outfielder &lt;br /&gt;
&lt;br /&gt;
Mathieu - right outfielder &lt;br /&gt;
&lt;br /&gt;
Sung - first base &lt;br /&gt;
&lt;br /&gt;
Adam - centre outfielder &lt;br /&gt;
&lt;br /&gt;
Hasan - second base &lt;br /&gt;
&lt;br /&gt;
Bobo - catcher &lt;br /&gt;
&lt;br /&gt;
Jason - third base &lt;br /&gt;
&lt;br /&gt;
Charles - shortstop &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4) &lt;br /&gt;
Understand the problem&lt;br /&gt;
&lt;br /&gt;
The question is asking me who will be playing with Fernanda on the fifth day.  The question state that 3 matches are played simultaneously during each day and each player plays each of the others once. &lt;br /&gt;
1st	Carla vs. Petra	&lt;br /&gt;
2nd	Carla vs. Janet&lt;br /&gt;
3rd	Janet vs. Li&lt;br /&gt;
4th	Petra vs. Sandra&lt;br /&gt;
5th 	Fernanda vs. ?&lt;br /&gt;
&lt;br /&gt;
Plan a strategy for solving the problem &lt;br /&gt;
Since the question already gives us some of the teams during these five days and we already know that each player plays each of the other once, 3 games per day, and played simultaneously which means that one player will only play one game per day. By knowing all these circumstances, we can create a chart to list out all the possible teams that can play on each day by elimination. &lt;br /&gt;
&lt;br /&gt;
Execute the strategy&lt;br /&gt;
&lt;br /&gt;
1st	Carla vs. Petra 	J vs. S	 F vs. L	or	J vs. F	S vs. L	&lt;br /&gt;
Janet can’t play with Li → they will be playing on the 3rd day, we eliminate the second choice since on the 4th day F vs. J will be playing.&lt;br /&gt;
&lt;br /&gt;
2nd	Carla vs. Janet 	P vs. L	 S vs. F	or	P vs. F	S vs. L&lt;br /&gt;
Petra can’t play with Sandra → they will be playing on the 4th day; we eliminate the second choice since on the 3rd day will be P vs. F.&lt;br /&gt;
&lt;br /&gt;
3rd	Janet vs. Li	        C vs. S  P vs. F	or	C vs. F	S vs. P&lt;br /&gt;
Carla can’t play with Petra → they played on the 1st day; we eliminate the second choice since we can tell from the table they will be playing on the 4th day. &lt;br /&gt;
&lt;br /&gt;
4th	Petra vs. Sandra	C vs. Li F vs. J	or	C vs. F	J vs. Li&lt;br /&gt;
Carl can’t play with Janet → they played on the 2nd day, and we eliminate the second choice since we can tell from the table Janet vs. Li played on the 3rd day. &lt;br /&gt;
&lt;br /&gt;
→ from the table above we can list out the teams will be playing on the 5th day&lt;br /&gt;
5th 	Fernanda vs. Carla	Petra vs. Janet		Li vs. Sandra&lt;br /&gt;
The table above list out all the possible arrangement of each teams, and we eliminate all the impossible guesses.  Each of the players only played with each other once during 5 days. &lt;br /&gt;
&lt;br /&gt;
[[on the 5th day, Fernanda is playing with Carla]]&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 13=&lt;br /&gt;
&lt;br /&gt;
Question 1:&lt;br /&gt;
&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
&lt;br /&gt;
Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
&lt;br /&gt;
Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
&lt;br /&gt;
Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
&lt;br /&gt;
Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
&lt;br /&gt;
Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
&lt;br /&gt;
Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
&lt;br /&gt;
a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
&lt;br /&gt;
Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
&lt;br /&gt;
Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
&lt;br /&gt;
Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
&lt;br /&gt;
Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
&lt;br /&gt;
Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;br /&gt;
&lt;br /&gt;
16. 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;
Let x represent the number of daughters&lt;br /&gt;
Let y represent the number of brothers&lt;br /&gt;
&lt;br /&gt;
From the information given in the question, we can deduce the following equations.&lt;br /&gt;
1)	y = x-1&lt;br /&gt;
&lt;br /&gt;
2)	x = 2(y-1)&lt;br /&gt;
   = 2y - 2&lt;br /&gt;
&lt;br /&gt;
Now let us substitute equation 1 into equation 2 to solve for x.&lt;br /&gt;
x = 2y – 2&lt;br /&gt;
x = 2(x-1) – 2&lt;br /&gt;
x = 2x – 2 – 2&lt;br /&gt;
x = 2x – 4&lt;br /&gt;
x = 4&lt;br /&gt;
&lt;br /&gt;
Substitute x = 4 into equation 1.&lt;br /&gt;
	y = x – 1&lt;br /&gt;
	y = 4 – 1&lt;br /&gt;
	y = 3&lt;br /&gt;
&lt;br /&gt;
	Therefore, the family has 4 girls and 3 boys&lt;br /&gt;
&lt;br /&gt;
17. 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;
Let x represent Dan’s actual weight.&lt;br /&gt;
Let y represent Sarah’s actual weight.&lt;br /&gt;
Let z represent the error on the scale.&lt;br /&gt;
&lt;br /&gt;
Then using the information given in the question we can come up with the following 3 equations.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
3)	105kg = (x + y) + z&lt;br /&gt;
Now we will reorganize the variables in equations 1 and 2.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
            x = 60 – z&lt;br /&gt;
&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
      y = 50kg – z&lt;br /&gt;
&lt;br /&gt;
Now substitute x = 60 – z and y = 50kg – z into equation 3&lt;br /&gt;
105kg = (x + y) + z&lt;br /&gt;
105kg = [(60-z) + (50kg – z )] + z&lt;br /&gt;
105kg = (60 – z + 50kg – z) + z&lt;br /&gt;
105kg = 110kg – z&lt;br /&gt;
	z = 5kg&lt;br /&gt;
&lt;br /&gt;
	Therefore, the scale reads too high by 5kg.&lt;br /&gt;
&lt;br /&gt;
18. 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;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar. &lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following 4 equations.&lt;br /&gt;
1)	y = x/3 &lt;br /&gt;
2)	z = (1/3)(x - x/3)&lt;br /&gt;
3)	w = (1/3)[x – (1/3)(x - x/3) – x/3]&lt;br /&gt;
4)	x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Substitute the values of y, z and w into equation 4.&lt;br /&gt;
x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40&lt;br /&gt;
	      x = (x/3) + (2/9)x + (4/27)x + 40&lt;br /&gt;
	      x = (19/27)x + 40&lt;br /&gt;
        (8/27)x = 40	      &lt;br /&gt;
      	      x = 135&lt;br /&gt;
	&lt;br /&gt;
	Therefore, there are 135 pennies in total at the start.&lt;br /&gt;
&lt;br /&gt;
19. 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;
Let b represent the total amount of milk.&lt;br /&gt;
Let a represent the total amount of coffee.&lt;br /&gt;
Let n represent the number of people in Angela’s family.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following equation.&lt;br /&gt;
&lt;br /&gt;
(a/6 + b/4)n = a + b&lt;br /&gt;
2a(6-n) = 3b(n-4)&lt;br /&gt;
Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.&lt;br /&gt;
&lt;br /&gt;
Therefore, there are at least 5 people in the family.&lt;br /&gt;
	&lt;br /&gt;
20. 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;
If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour&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;
Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.&lt;br /&gt;
&lt;br /&gt;
22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&lt;br /&gt;
&lt;br /&gt;
Let R represent a rainy day.&lt;br /&gt;
Let S represent a sunny day.&lt;br /&gt;
It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons.&lt;br /&gt;
Morning - R S S S S R R S S S S S R R S S R R &lt;br /&gt;
Noon - S S R R S S S R R S R S S S R S S S&lt;br /&gt;
There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.&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;
Let x represent the oldest child.&lt;br /&gt;
Let y represent the second child.&lt;br /&gt;
Let z represent the third child.&lt;br /&gt;
&lt;br /&gt;
Given that the produce of their ages is 36. We can deduce equation 1.&lt;br /&gt;
1)	36 = xyz&lt;br /&gt;
&lt;br /&gt;
Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.&lt;br /&gt;
&lt;br /&gt;
Ages		Product	Sum&lt;br /&gt;
	1 x 1 x 36	36		38&lt;br /&gt;
2 x 3 x 6	36		11&lt;br /&gt;
	1 x 3 x 12	36		16	&lt;br /&gt;
2 x 2 x 9	36		13&lt;br /&gt;
	1 x 4 x 9	36		14&lt;br /&gt;
3 x 3 x 4	36		10&lt;br /&gt;
	1 x 2 x 18	36		21&lt;br /&gt;
	1 x 6 x 6	36		13&lt;br /&gt;
&lt;br /&gt;
	The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.&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 A and B represent the 2 candles.&lt;br /&gt;
Let A represent the candle that takes 6 hours to burn out.&lt;br /&gt;
Let B represent the candle that takes 3 hours to burn out.&lt;br /&gt;
&lt;br /&gt;
Candle A takes twice as long to burn out than Candle B. We can form equation 1.&lt;br /&gt;
1)	A = 2B&lt;br /&gt;
&lt;br /&gt;
Assume that time starts at 0 and that the burning of the candle is constant.&lt;br /&gt;
Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour.&lt;br /&gt;
Let x represent the time in hours. We can form equation 2 and 3.&lt;br /&gt;
2)	A = (6-x)/6&lt;br /&gt;
3)	B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
Substitute equation 2 and 3 into equation 1.&lt;br /&gt;
A = 2B&lt;br /&gt;
(6-x)/6 = 2[(3-x)/6]&lt;br /&gt;
x = 2&lt;br /&gt;
&lt;br /&gt;
Therefore, it takes 2 hours for one candle to be half of the other.&lt;br /&gt;
&lt;br /&gt;
25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L. &lt;br /&gt;
&lt;br /&gt;
Given:&lt;br /&gt;
Candle X has length L and was lit at 6 hour.&lt;br /&gt;
Candle Y has length L + 1 and was lit at 4.5 hour.&lt;br /&gt;
At 8.5 hour, the length of Candle A is equal to the length of Candle B.&lt;br /&gt;
Candle A burns out at 10 hours.&lt;br /&gt;
Candle B burns out at 10.5 hours.&lt;br /&gt;
&lt;br /&gt;
Then the total burning time of Candle A is: 10 – 6 = 4 hours&lt;br /&gt;
Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours&lt;br /&gt;
&lt;br /&gt;
Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours.&lt;br /&gt;
Burning time till 8.5h for Candle B  is: 8.5 - 4.5 = 4 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Let x represent the length of Candle A&lt;br /&gt;
Let y represent the length of Candle B&lt;br /&gt;
&lt;br /&gt;
y = x + 1&lt;br /&gt;
&lt;br /&gt;
Let B1 represent the burning constant for x&lt;br /&gt;
Let B2 represent the burning constant for y&lt;br /&gt;
We know that x=y at 8.5 hours.&lt;br /&gt;
&lt;br /&gt;
 (x-4) B1 = x + 1 – 2.5B2&lt;br /&gt;
&lt;br /&gt;
y = 6 B2 at end time&lt;br /&gt;
x = 4B1 at end time&lt;br /&gt;
6B2 = 4B1 + 1 &lt;br /&gt;
B1 = (6B1 -1)/4&lt;br /&gt;
&lt;br /&gt;
Using the above equations and substituting back into the original.&lt;br /&gt;
4B2 = 1 – 2.5B1&lt;br /&gt;
4B2 = 1 – 2.5((6B1 -1)/4)&lt;br /&gt;
B2 = 1.5&lt;br /&gt;
&lt;br /&gt;
Substitute B2 into y = 6B2&lt;br /&gt;
4((4B1 + 1)/6) = 1 – 2.5B1&lt;br /&gt;
B1 = 2&lt;br /&gt;
Substitute B1 into x = 4B1&lt;br /&gt;
&lt;br /&gt;
y = 6B2&lt;br /&gt;
y = 6(1.5) = 9&lt;br /&gt;
&lt;br /&gt;
x = 4B1&lt;br /&gt;
x = 4(2) = 8&lt;br /&gt;
&lt;br /&gt;
Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58360</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=58360"/>
		<updated>2010-10-29T03:13:33Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 4&lt;br /&gt;
| member 1 = Anoud Al-Moqbil&lt;br /&gt;
| member 2 = Owen Johnston&lt;br /&gt;
| member 3 = Charlene Ng&lt;br /&gt;
| member 4 = Ling Sun&lt;br /&gt;
| member 5 = [[User:Dominicjansunga|Dominic Sunga]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
Homework - Due October 20&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE NO PROBLEM TO THE GROUP&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT POSE PROBLEMS FOR SOME, BUT NOT ALL&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
THOSE THAT NOONE KNOWS HOW TO HANDLE&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 20=&lt;br /&gt;
&lt;br /&gt;
1)&lt;br /&gt;
&lt;br /&gt;
We know that the turtle belongs to Suzan because every other pet was given a owner, except for the Turtle. &lt;br /&gt;
&lt;br /&gt;
We are given two names of the animals – Suzan the frog and Jun the Cat. From this, we can deduce the names of the other animals – Jaela the turtle, Bianca the Snake and Tosh the Parrot&lt;br /&gt;
&lt;br /&gt;
With this information, we can create a table:&lt;br /&gt;
&lt;br /&gt;
Owner	Gender	Pet	Pet name&lt;br /&gt;
Tosh	Male	Cat	Jun&lt;br /&gt;
Bianca	Female	Frog 	Suzan&lt;br /&gt;
Suzan’s Mom	Female	Turtle	Jaela&lt;br /&gt;
Jun	Male	Snake	Bianca&lt;br /&gt;
Jaela	Female	Parrot 	Tosh&lt;br /&gt;
&lt;br /&gt;
The question is asking us which pet belongs to Suzan’s mother. The answer is Jaela the Turtle. &lt;br /&gt;
&lt;br /&gt;
3)&lt;br /&gt;
Use a graph where the column headings are for the names of the people and the row headings are for the different positions. Then use the given information to eliminate and deduce what positions each person will play.&lt;br /&gt;
&lt;br /&gt;
Ed- pitcher &lt;br /&gt;
&lt;br /&gt;
Pascal - left outfielder &lt;br /&gt;
&lt;br /&gt;
Mathieu - right outfielder &lt;br /&gt;
&lt;br /&gt;
Sung - first base &lt;br /&gt;
&lt;br /&gt;
Adam - centre outfielder &lt;br /&gt;
&lt;br /&gt;
Hasan - second base &lt;br /&gt;
&lt;br /&gt;
Bobo - catcher &lt;br /&gt;
&lt;br /&gt;
Jason - third base &lt;br /&gt;
&lt;br /&gt;
Charles - shortstop &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
4) &lt;br /&gt;
Understand the problem&lt;br /&gt;
&lt;br /&gt;
The question is asking me who will be playing with Fernanda on the fifth day.  The question state that 3 matches are played simultaneously during each day and each player plays each of the others once. &lt;br /&gt;
1st	Carla vs. Petra	&lt;br /&gt;
2nd	Carla vs. Janet&lt;br /&gt;
3rd	Janet vs. Li&lt;br /&gt;
4th	Petra vs. Sandra&lt;br /&gt;
5th 	Fernanda vs. ?&lt;br /&gt;
&lt;br /&gt;
Plan a strategy for solving the problem &lt;br /&gt;
Since the question already gives us some of the teams during these five days and we already know that each player plays each of the other once, 3 games per day, and played simultaneously which means that one player will only play one game per day. By knowing all these circumstances, we can create a chart to list out all the possible teams that can play on each day by elimination. &lt;br /&gt;
&lt;br /&gt;
Execute the strategy&lt;br /&gt;
&lt;br /&gt;
1st	Carla vs. Petra 	J vs. S	 F vs. L	or	J vs. F	S vs. L	&lt;br /&gt;
Janet can’t play with Li → they will be playing on the 3rd day, we eliminate the second choice since on the 4th day F vs. J will be playing.&lt;br /&gt;
&lt;br /&gt;
2nd	Carla vs. Janet 	P vs. L	 S vs. F	or	P vs. F	S vs. L&lt;br /&gt;
Petra can’t play with Sandra → they will be playing on the 4th day; we eliminate the second choice since on the 3rd day will be P vs. F.&lt;br /&gt;
&lt;br /&gt;
3rd	Janet vs. Li	        C vs. S  P vs. F	or	C vs. F	S vs. P&lt;br /&gt;
Carla can’t play with Petra → they played on the 1st day; we eliminate the second choice since we can tell from the table they will be playing on the 4th day. &lt;br /&gt;
&lt;br /&gt;
4th	Petra vs. Sandra	C vs. Li F vs. J	or	C vs. F	J vs. Li&lt;br /&gt;
Carl can’t play with Janet → they played on the 2nd day, and we eliminate the second choice since we can tell from the table Janet vs. Li played on the 3rd day. &lt;br /&gt;
&lt;br /&gt;
→ from the table above we can list out the teams will be playing on the 5th day&lt;br /&gt;
5th 	Fernanda vs. Carla	Petra vs. Janet		Li vs. Sandra&lt;br /&gt;
The table above list out all the possible arrangement of each teams, and we eliminate all the impossible guesses.  Each of the players only played with each other once during 5 days. &lt;br /&gt;
&lt;br /&gt;
[[on the 5th day, Fernanda is playing with Carla]]&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 13=&lt;br /&gt;
&lt;br /&gt;
Question 1:&lt;br /&gt;
&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
&lt;br /&gt;
Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
&lt;br /&gt;
Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
&lt;br /&gt;
Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
&lt;br /&gt;
Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
&lt;br /&gt;
Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
&lt;br /&gt;
Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
&lt;br /&gt;
a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
&lt;br /&gt;
Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
&lt;br /&gt;
Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
&lt;br /&gt;
Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
&lt;br /&gt;
Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
&lt;br /&gt;
Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;br /&gt;
&lt;br /&gt;
16. 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;
Let x represent the number of daughters&lt;br /&gt;
Let y represent the number of brothers&lt;br /&gt;
&lt;br /&gt;
From the information given in the question, we can deduce the following equations.&lt;br /&gt;
1)	y = x-1&lt;br /&gt;
&lt;br /&gt;
2)	x = 2(y-1)&lt;br /&gt;
   = 2y - 2&lt;br /&gt;
&lt;br /&gt;
Now let us substitute equation 1 into equation 2 to solve for x.&lt;br /&gt;
x = 2y – 2&lt;br /&gt;
x = 2(x-1) – 2&lt;br /&gt;
x = 2x – 2 – 2&lt;br /&gt;
x = 2x – 4&lt;br /&gt;
x = 4&lt;br /&gt;
&lt;br /&gt;
Substitute x = 4 into equation 1.&lt;br /&gt;
	y = x – 1&lt;br /&gt;
	y = 4 – 1&lt;br /&gt;
	y = 3&lt;br /&gt;
&lt;br /&gt;
	Therefore, the family has 4 girls and 3 boys&lt;br /&gt;
&lt;br /&gt;
17. 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;
Let x represent Dan’s actual weight.&lt;br /&gt;
Let y represent Sarah’s actual weight.&lt;br /&gt;
Let z represent the error on the scale.&lt;br /&gt;
&lt;br /&gt;
Then using the information given in the question we can come up with the following 3 equations.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
3)	105kg = (x + y) + z&lt;br /&gt;
Now we will reorganize the variables in equations 1 and 2.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
            x = 60 – z&lt;br /&gt;
&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
      y = 50kg – z&lt;br /&gt;
&lt;br /&gt;
Now substitute x = 60 – z and y = 50kg – z into equation 3&lt;br /&gt;
105kg = (x + y) + z&lt;br /&gt;
105kg = [(60-z) + (50kg – z )] + z&lt;br /&gt;
105kg = (60 – z + 50kg – z) + z&lt;br /&gt;
105kg = 110kg – z&lt;br /&gt;
	z = 5kg&lt;br /&gt;
&lt;br /&gt;
	Therefore, the scale reads too high by 5kg.&lt;br /&gt;
&lt;br /&gt;
18. 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;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar. &lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following 4 equations.&lt;br /&gt;
1)	y = x/3 &lt;br /&gt;
2)	z = (1/3)(x - x/3)&lt;br /&gt;
3)	w = (1/3)[x – (1/3)(x - x/3) – x/3]&lt;br /&gt;
4)	x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Substitute the values of y, z and w into equation 4.&lt;br /&gt;
x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40&lt;br /&gt;
	      x = (x/3) + (2/9)x + (4/27)x + 40&lt;br /&gt;
	      x = (19/27)x + 40&lt;br /&gt;
        (8/27)x = 40	      &lt;br /&gt;
      	      x = 135&lt;br /&gt;
	&lt;br /&gt;
	Therefore, there are 135 pennies in total at the start.&lt;br /&gt;
&lt;br /&gt;
19. 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;
Let b represent the total amount of milk.&lt;br /&gt;
Let a represent the total amount of coffee.&lt;br /&gt;
Let n represent the number of people in Angela’s family.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following equation.&lt;br /&gt;
&lt;br /&gt;
(a/6 + b/4)n = a + b&lt;br /&gt;
2a(6-n) = 3b(n-4)&lt;br /&gt;
Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.&lt;br /&gt;
&lt;br /&gt;
Therefore, there are at least 5 people in the family.&lt;br /&gt;
	&lt;br /&gt;
20. 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;
If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour&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;
Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.&lt;br /&gt;
&lt;br /&gt;
22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&lt;br /&gt;
&lt;br /&gt;
Let R represent a rainy day.&lt;br /&gt;
Let S represent a sunny day.&lt;br /&gt;
It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons.&lt;br /&gt;
Morning - R S S S S R R S S S S S R R S S R R &lt;br /&gt;
Noon - S S R R S S S R R S R S S S R S S S&lt;br /&gt;
There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.&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;
Let x represent the oldest child.&lt;br /&gt;
Let y represent the second child.&lt;br /&gt;
Let z represent the third child.&lt;br /&gt;
&lt;br /&gt;
Given that the produce of their ages is 36. We can deduce equation 1.&lt;br /&gt;
1)	36 = xyz&lt;br /&gt;
&lt;br /&gt;
Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.&lt;br /&gt;
&lt;br /&gt;
Ages		Product	Sum&lt;br /&gt;
	1 x 1 x 36	36		38&lt;br /&gt;
2 x 3 x 6	36		11&lt;br /&gt;
	1 x 3 x 12	36		16	&lt;br /&gt;
2 x 2 x 9	36		13&lt;br /&gt;
	1 x 4 x 9	36		14&lt;br /&gt;
3 x 3 x 4	36		10&lt;br /&gt;
	1 x 2 x 18	36		21&lt;br /&gt;
	1 x 6 x 6	36		13&lt;br /&gt;
&lt;br /&gt;
	The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.&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 A and B represent the 2 candles.&lt;br /&gt;
Let A represent the candle that takes 6 hours to burn out.&lt;br /&gt;
Let B represent the candle that takes 3 hours to burn out.&lt;br /&gt;
&lt;br /&gt;
Candle A takes twice as long to burn out than Candle B. We can form equation 1.&lt;br /&gt;
1)	A = 2B&lt;br /&gt;
&lt;br /&gt;
Assume that time starts at 0 and that the burning of the candle is constant.&lt;br /&gt;
Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour.&lt;br /&gt;
Let x represent the time in hours. We can form equation 2 and 3.&lt;br /&gt;
2)	A = (6-x)/6&lt;br /&gt;
3)	B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
Substitute equation 2 and 3 into equation 1.&lt;br /&gt;
A = 2B&lt;br /&gt;
(6-x)/6 = 2[(3-x)/6]&lt;br /&gt;
x = 2&lt;br /&gt;
&lt;br /&gt;
Therefore, it takes 2 hours for one candle to be half of the other.&lt;br /&gt;
&lt;br /&gt;
25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L. &lt;br /&gt;
&lt;br /&gt;
Given:&lt;br /&gt;
Candle X has length L and was lit at 6 hour.&lt;br /&gt;
Candle Y has length L + 1 and was lit at 4.5 hour.&lt;br /&gt;
At 8.5 hour, the length of Candle A is equal to the length of Candle B.&lt;br /&gt;
Candle A burns out at 10 hours.&lt;br /&gt;
Candle B burns out at 10.5 hours.&lt;br /&gt;
&lt;br /&gt;
Then the total burning time of Candle A is: 10 – 6 = 4 hours&lt;br /&gt;
Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours&lt;br /&gt;
&lt;br /&gt;
Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours.&lt;br /&gt;
Burning time till 8.5h for Candle B  is: 8.5 - 4.5 = 4 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Let x represent the length of Candle A&lt;br /&gt;
Let y represent the length of Candle B&lt;br /&gt;
&lt;br /&gt;
y = x + 1&lt;br /&gt;
&lt;br /&gt;
Let B1 represent the burning constant for x&lt;br /&gt;
Let B2 represent the burning constant for y&lt;br /&gt;
We know that x=y at 8.5 hours.&lt;br /&gt;
&lt;br /&gt;
 (x-4) B1 = x + 1 – 2.5B2&lt;br /&gt;
&lt;br /&gt;
y = 6 B2 at end time&lt;br /&gt;
x = 4B1 at end time&lt;br /&gt;
6B2 = 4B1 + 1 &lt;br /&gt;
B1 = (6B1 -1)/4&lt;br /&gt;
&lt;br /&gt;
Using the above equations and substituting back into the original.&lt;br /&gt;
4B2 = 1 – 2.5B1&lt;br /&gt;
4B2 = 1 – 2.5((6B1 -1)/4)&lt;br /&gt;
B2 = 1.5&lt;br /&gt;
&lt;br /&gt;
Substitute B2 into y = 6B2&lt;br /&gt;
4((4B1 + 1)/6) = 1 – 2.5B1&lt;br /&gt;
B1 = 2&lt;br /&gt;
Substitute B1 into x = 4B1&lt;br /&gt;
&lt;br /&gt;
y = 6B2&lt;br /&gt;
y = 6(1.5) = 9&lt;br /&gt;
&lt;br /&gt;
x = 4B1&lt;br /&gt;
x = 4(2) = 8&lt;br /&gt;
&lt;br /&gt;
Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=56511</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=56511"/>
		<updated>2010-10-20T07:27:52Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Infobox MATH110 Groups&lt;br /&gt;
| group number = 4&lt;br /&gt;
| member 1 = Anoud Al-Moqbil&lt;br /&gt;
| member 2 = Owen Johnston&lt;br /&gt;
| member 3 = Charlene Ng&lt;br /&gt;
| member 4 = Ling Sun&lt;br /&gt;
| member 5 = [[User:Dominicjansunga|Dominic Sunga]]&lt;br /&gt;
| member 6 = &lt;br /&gt;
}}&lt;br /&gt;
=Homework - Due October 20=&lt;br /&gt;
&lt;br /&gt;
1)&lt;br /&gt;
&lt;br /&gt;
We know that the turtle belongs to Suzan because every other pet was given a owner, except for the Turtle. &lt;br /&gt;
&lt;br /&gt;
We are given two names of the animals – Suzan the frog and Jun the Cat. From this, we can deduce the names of the other animals – Jaela the turtle, Bianca the Snake and Tosh the Parrot&lt;br /&gt;
&lt;br /&gt;
With this information, we can create a table:&lt;br /&gt;
&lt;br /&gt;
Owner	Gender	Pet	Pet name&lt;br /&gt;
Tosh	Male	Cat	Jun&lt;br /&gt;
Bianca	Female	Frog 	Suzan&lt;br /&gt;
Suzan’s Mom	Female	Turtle	Jaela&lt;br /&gt;
Jun	Male	Snake	Bianca&lt;br /&gt;
Jaela	Female	Parrot 	Tosh&lt;br /&gt;
&lt;br /&gt;
The question is asking us which pet belongs to Suzan’s mother. The answer is Jaela the Turtle. &lt;br /&gt;
&lt;br /&gt;
4) &lt;br /&gt;
Understand the problem&lt;br /&gt;
&lt;br /&gt;
The question is asking me who will be playing with Fernanda on the fifth day.  The question state that 3 matches are played simultaneously during each day and each player plays each of the others once. &lt;br /&gt;
1st	Carla vs. Petra	&lt;br /&gt;
2nd	Carla vs. Janet&lt;br /&gt;
3rd	Janet vs. Li&lt;br /&gt;
4th	Petra vs. Sandra&lt;br /&gt;
5th 	Fernanda vs. ?&lt;br /&gt;
&lt;br /&gt;
Plan a strategy for solving the problem &lt;br /&gt;
Since the question already gives us some of the teams during these five days and we already know that each player plays each of the other once, 3 games per day, and played simultaneously which means that one player will only play one game per day. By knowing all these circumstances, we can create a chart to list out all the possible teams that can play on each day by elimination. &lt;br /&gt;
&lt;br /&gt;
Execute the strategy&lt;br /&gt;
&lt;br /&gt;
1st	Carla vs. Petra 	J vs. S	 F vs. L	or	J vs. F	S vs. L	&lt;br /&gt;
Janet can’t play with Li → they will be playing on the 3rd day, we eliminate the second choice since on the 4th day F vs. J will be playing.&lt;br /&gt;
&lt;br /&gt;
2nd	Carla vs. Janet 	P vs. L	 S vs. F	or	P vs. F	S vs. L&lt;br /&gt;
Petra can’t play with Sandra → they will be playing on the 4th day; we eliminate the second choice since on the 3rd day will be P vs. F.&lt;br /&gt;
&lt;br /&gt;
3rd	Janet vs. Li	        C vs. S  P vs. F	or	C vs. F	S vs. P&lt;br /&gt;
Carla can’t play with Petra → they played on the 1st day; we eliminate the second choice since we can tell from the table they will be playing on the 4th day. &lt;br /&gt;
&lt;br /&gt;
4th	Petra vs. Sandra	C vs. Li F vs. J	or	C vs. F	J vs. Li&lt;br /&gt;
Carl can’t play with Janet → they played on the 2nd day, and we eliminate the second choice since we can tell from the table Janet vs. Li played on the 3rd day. &lt;br /&gt;
&lt;br /&gt;
→ from the table above we can list out the teams will be playing on the 5th day&lt;br /&gt;
5th 	Fernanda vs. Carla	Petra vs. Janet		Li vs. Sandra&lt;br /&gt;
The table above list out all the possible arrangement of each teams, and we eliminate all the impossible guesses.  Each of the players only played with each other once during 5 days. &lt;br /&gt;
&lt;br /&gt;
[[on the 5th day, Fernanda is playing with Carla]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=Homework - Due October 13=&lt;br /&gt;
&lt;br /&gt;
Question 1:&lt;br /&gt;
&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
&lt;br /&gt;
Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
&lt;br /&gt;
Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
&lt;br /&gt;
Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
&lt;br /&gt;
Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
&lt;br /&gt;
Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
&lt;br /&gt;
Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
&lt;br /&gt;
a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
&lt;br /&gt;
Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
&lt;br /&gt;
Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
&lt;br /&gt;
Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
&lt;br /&gt;
Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
&lt;br /&gt;
Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;br /&gt;
&lt;br /&gt;
16. 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;
Let x represent the number of daughters&lt;br /&gt;
Let y represent the number of brothers&lt;br /&gt;
&lt;br /&gt;
From the information given in the question, we can deduce the following equations.&lt;br /&gt;
1)	y = x-1&lt;br /&gt;
&lt;br /&gt;
2)	x = 2(y-1)&lt;br /&gt;
   = 2y - 2&lt;br /&gt;
&lt;br /&gt;
Now let us substitute equation 1 into equation 2 to solve for x.&lt;br /&gt;
x = 2y – 2&lt;br /&gt;
x = 2(x-1) – 2&lt;br /&gt;
x = 2x – 2 – 2&lt;br /&gt;
x = 2x – 4&lt;br /&gt;
x = 4&lt;br /&gt;
&lt;br /&gt;
Substitute x = 4 into equation 1.&lt;br /&gt;
	y = x – 1&lt;br /&gt;
	y = 4 – 1&lt;br /&gt;
	y = 3&lt;br /&gt;
&lt;br /&gt;
	Therefore, the family has 4 girls and 3 boys&lt;br /&gt;
&lt;br /&gt;
17. 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;
Let x represent Dan’s actual weight.&lt;br /&gt;
Let y represent Sarah’s actual weight.&lt;br /&gt;
Let z represent the error on the scale.&lt;br /&gt;
&lt;br /&gt;
Then using the information given in the question we can come up with the following 3 equations.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
3)	105kg = (x + y) + z&lt;br /&gt;
Now we will reorganize the variables in equations 1 and 2.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
            x = 60 – z&lt;br /&gt;
&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
      y = 50kg – z&lt;br /&gt;
&lt;br /&gt;
Now substitute x = 60 – z and y = 50kg – z into equation 3&lt;br /&gt;
105kg = (x + y) + z&lt;br /&gt;
105kg = [(60-z) + (50kg – z )] + z&lt;br /&gt;
105kg = (60 – z + 50kg – z) + z&lt;br /&gt;
105kg = 110kg – z&lt;br /&gt;
	z = 5kg&lt;br /&gt;
&lt;br /&gt;
	Therefore, the scale reads too high by 5kg.&lt;br /&gt;
&lt;br /&gt;
18. 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;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar. &lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following 4 equations.&lt;br /&gt;
1)	y = x/3 &lt;br /&gt;
2)	z = (1/3)(x - x/3)&lt;br /&gt;
3)	w = (1/3)[x – (1/3)(x - x/3) – x/3]&lt;br /&gt;
4)	x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Substitute the values of y, z and w into equation 4.&lt;br /&gt;
x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40&lt;br /&gt;
	      x = (x/3) + (2/9)x + (4/27)x + 40&lt;br /&gt;
	      x = (19/27)x + 40&lt;br /&gt;
        (8/27)x = 40	      &lt;br /&gt;
      	      x = 135&lt;br /&gt;
	&lt;br /&gt;
	Therefore, there are 135 pennies in total at the start.&lt;br /&gt;
&lt;br /&gt;
19. 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;
Let b represent the total amount of milk.&lt;br /&gt;
Let a represent the total amount of coffee.&lt;br /&gt;
Let n represent the number of people in Angela’s family.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following equation.&lt;br /&gt;
&lt;br /&gt;
(a/6 + b/4)n = a + b&lt;br /&gt;
2a(6-n) = 3b(n-4)&lt;br /&gt;
Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.&lt;br /&gt;
&lt;br /&gt;
Therefore, there are at least 5 people in the family.&lt;br /&gt;
	&lt;br /&gt;
20. 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;
If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour&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;
Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.&lt;br /&gt;
&lt;br /&gt;
22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&lt;br /&gt;
&lt;br /&gt;
Let R represent a rainy day.&lt;br /&gt;
Let S represent a sunny day.&lt;br /&gt;
It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons.&lt;br /&gt;
Morning - R S S S S R R S S S S S R R S S R R &lt;br /&gt;
Noon - S S R R S S S R R S R S S S R S S S&lt;br /&gt;
There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.&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;
Let x represent the oldest child.&lt;br /&gt;
Let y represent the second child.&lt;br /&gt;
Let z represent the third child.&lt;br /&gt;
&lt;br /&gt;
Given that the produce of their ages is 36. We can deduce equation 1.&lt;br /&gt;
1)	36 = xyz&lt;br /&gt;
&lt;br /&gt;
Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.&lt;br /&gt;
&lt;br /&gt;
Ages		Product	Sum&lt;br /&gt;
	1 x 1 x 36	36		38&lt;br /&gt;
2 x 3 x 6	36		11&lt;br /&gt;
	1 x 3 x 12	36		16	&lt;br /&gt;
2 x 2 x 9	36		13&lt;br /&gt;
	1 x 4 x 9	36		14&lt;br /&gt;
3 x 3 x 4	36		10&lt;br /&gt;
	1 x 2 x 18	36		21&lt;br /&gt;
	1 x 6 x 6	36		13&lt;br /&gt;
&lt;br /&gt;
	The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.&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 A and B represent the 2 candles.&lt;br /&gt;
Let A represent the candle that takes 6 hours to burn out.&lt;br /&gt;
Let B represent the candle that takes 3 hours to burn out.&lt;br /&gt;
&lt;br /&gt;
Candle A takes twice as long to burn out than Candle B. We can form equation 1.&lt;br /&gt;
1)	A = 2B&lt;br /&gt;
&lt;br /&gt;
Assume that time starts at 0 and that the burning of the candle is constant.&lt;br /&gt;
Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour.&lt;br /&gt;
Let x represent the time in hours. We can form equation 2 and 3.&lt;br /&gt;
2)	A = (6-x)/6&lt;br /&gt;
3)	B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
Substitute equation 2 and 3 into equation 1.&lt;br /&gt;
A = 2B&lt;br /&gt;
(6-x)/6 = 2[(3-x)/6]&lt;br /&gt;
x = 2&lt;br /&gt;
&lt;br /&gt;
Therefore, it takes 2 hours for one candle to be half of the other.&lt;br /&gt;
&lt;br /&gt;
25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L. &lt;br /&gt;
&lt;br /&gt;
Given:&lt;br /&gt;
Candle X has length L and was lit at 6 hour.&lt;br /&gt;
Candle Y has length L + 1 and was lit at 4.5 hour.&lt;br /&gt;
At 8.5 hour, the length of Candle A is equal to the length of Candle B.&lt;br /&gt;
Candle A burns out at 10 hours.&lt;br /&gt;
Candle B burns out at 10.5 hours.&lt;br /&gt;
&lt;br /&gt;
Then the total burning time of Candle A is: 10 – 6 = 4 hours&lt;br /&gt;
Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours&lt;br /&gt;
&lt;br /&gt;
Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours.&lt;br /&gt;
Burning time till 8.5h for Candle B  is: 8.5 - 4.5 = 4 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Let x represent the length of Candle A&lt;br /&gt;
Let y represent the length of Candle B&lt;br /&gt;
&lt;br /&gt;
y = x + 1&lt;br /&gt;
&lt;br /&gt;
Let B1 represent the burning constant for x&lt;br /&gt;
Let B2 represent the burning constant for y&lt;br /&gt;
We know that x=y at 8.5 hours.&lt;br /&gt;
&lt;br /&gt;
 (x-4) B1 = x + 1 – 2.5B2&lt;br /&gt;
&lt;br /&gt;
y = 6 B2 at end time&lt;br /&gt;
x = 4B1 at end time&lt;br /&gt;
6B2 = 4B1 + 1 &lt;br /&gt;
B1 = (6B1 -1)/4&lt;br /&gt;
&lt;br /&gt;
Using the above equations and substituting back into the original.&lt;br /&gt;
4B2 = 1 – 2.5B1&lt;br /&gt;
4B2 = 1 – 2.5((6B1 -1)/4)&lt;br /&gt;
B2 = 1.5&lt;br /&gt;
&lt;br /&gt;
Substitute B2 into y = 6B2&lt;br /&gt;
4((4B1 + 1)/6) = 1 – 2.5B1&lt;br /&gt;
B1 = 2&lt;br /&gt;
Substitute B1 into x = 4B1&lt;br /&gt;
&lt;br /&gt;
y = 6B2&lt;br /&gt;
y = 6(1.5) = 9&lt;br /&gt;
&lt;br /&gt;
x = 4B1&lt;br /&gt;
x = 4(2) = 8&lt;br /&gt;
&lt;br /&gt;
Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=54464</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=54464"/>
		<updated>2010-10-13T08:21:39Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Anoud Al-Moqbil&lt;br /&gt;
* Owen Johnston&lt;br /&gt;
* Charlene Ng&lt;br /&gt;
* Ling Sun&lt;br /&gt;
* Dominic Sunga&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
HOMEWORK - DUE OCTOBER 13&lt;br /&gt;
&lt;br /&gt;
Question 1:&lt;br /&gt;
&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
&lt;br /&gt;
Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
&lt;br /&gt;
Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
&lt;br /&gt;
Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
&lt;br /&gt;
Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
&lt;br /&gt;
Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
&lt;br /&gt;
Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
&lt;br /&gt;
a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
&lt;br /&gt;
Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
&lt;br /&gt;
Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
&lt;br /&gt;
Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
&lt;br /&gt;
Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
&lt;br /&gt;
Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;br /&gt;
&lt;br /&gt;
16. 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;
Let x represent the number of daughters&lt;br /&gt;
Let y represent the number of brothers&lt;br /&gt;
&lt;br /&gt;
From the information given in the question, we can deduce the following equations.&lt;br /&gt;
1)	y = x-1&lt;br /&gt;
&lt;br /&gt;
2)	x = 2(y-1)&lt;br /&gt;
   = 2y - 2&lt;br /&gt;
&lt;br /&gt;
Now let us substitute equation 1 into equation 2 to solve for x.&lt;br /&gt;
x = 2y – 2&lt;br /&gt;
x = 2(x-1) – 2&lt;br /&gt;
x = 2x – 2 – 2&lt;br /&gt;
x = 2x – 4&lt;br /&gt;
x = 4&lt;br /&gt;
&lt;br /&gt;
Substitute x = 4 into equation 1.&lt;br /&gt;
	y = x – 1&lt;br /&gt;
	y = 4 – 1&lt;br /&gt;
	y = 3&lt;br /&gt;
&lt;br /&gt;
	Therefore, the family has 4 girls and 3 boys&lt;br /&gt;
&lt;br /&gt;
17. 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;
Let x represent Dan’s actual weight.&lt;br /&gt;
Let y represent Sarah’s actual weight.&lt;br /&gt;
Let z represent the error on the scale.&lt;br /&gt;
&lt;br /&gt;
Then using the information given in the question we can come up with the following 3 equations.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
3)	105kg = (x + y) + z&lt;br /&gt;
Now we will reorganize the variables in equations 1 and 2.&lt;br /&gt;
&lt;br /&gt;
1)	60kg = x + z&lt;br /&gt;
            x = 60 – z&lt;br /&gt;
&lt;br /&gt;
2)	50kg = y + z&lt;br /&gt;
      y = 50kg – z&lt;br /&gt;
&lt;br /&gt;
Now substitute x = 60 – z and y = 50kg – z into equation 3&lt;br /&gt;
105kg = (x + y) + z&lt;br /&gt;
105kg = [(60-z) + (50kg – z )] + z&lt;br /&gt;
105kg = (60 – z + 50kg – z) + z&lt;br /&gt;
105kg = 110kg – z&lt;br /&gt;
	z = 5kg&lt;br /&gt;
&lt;br /&gt;
	Therefore, the scale reads too high by 5kg.&lt;br /&gt;
&lt;br /&gt;
18. 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;
Let x represent the total number of pennies in the jar.&lt;br /&gt;
Let y represent the number of pennies Alice took from the jar.&lt;br /&gt;
Let z represent the number of pennies Bret took from the jar. &lt;br /&gt;
Let w represent the number of pennies Carla took from the jar.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following 4 equations.&lt;br /&gt;
1)	y = x/3 &lt;br /&gt;
2)	z = (1/3)(x - x/3)&lt;br /&gt;
3)	w = (1/3)[x – (1/3)(x - x/3) – x/3]&lt;br /&gt;
4)	x = y + z + w + 40&lt;br /&gt;
&lt;br /&gt;
Substitute the values of y, z and w into equation 4.&lt;br /&gt;
x = (x/3) + (1/3)(x - x/3) + (1/3)[x – (1/3)(x - x/3) – x/3] + 40&lt;br /&gt;
	      x = (x/3) + (2/9)x + (4/27)x + 40&lt;br /&gt;
	      x = (19/27)x + 40&lt;br /&gt;
        (8/27)x = 40	      &lt;br /&gt;
      	      x = 135&lt;br /&gt;
	&lt;br /&gt;
	Therefore, there are 135 pennies in total at the start.&lt;br /&gt;
&lt;br /&gt;
19. 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;
Let b represent the total amount of milk.&lt;br /&gt;
Let a represent the total amount of coffee.&lt;br /&gt;
Let n represent the number of people in Angela’s family.&lt;br /&gt;
&lt;br /&gt;
From the information given in the question we can deduce the following equation.&lt;br /&gt;
&lt;br /&gt;
(a/6 + b/4)n = a + b&lt;br /&gt;
2a(6-n) = 3b(n-4)&lt;br /&gt;
Since both a and b are positive values, it follows that 6-n and n-4 are also positive which is only possible when n=5.&lt;br /&gt;
&lt;br /&gt;
Therefore, there are at least 5 people in the family.&lt;br /&gt;
	&lt;br /&gt;
20. 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;
If one clock is 5 minutes faster per hour, and the other is 5 minutes slower per hour, we can deduce that after 6 hours have passed, they will be 1 hour apart. The first clock would be at 6:30, while the other at 5:30. This is because in 6 hours, one clock would have gone ahead by half an hour while the other slowed down by half an hour&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;
Given that: Sven is the median of the sequence, Dan is the 10th and Lars is the 16th. We know that there must be at least 16 runners. But as 16 is an even number, an exact median cannot be found. Therefore, there should be 17 placements for the answer to be reasonable where Sven is the 9th.&lt;br /&gt;
&lt;br /&gt;
22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?&lt;br /&gt;
&lt;br /&gt;
Let R represent a rainy day.&lt;br /&gt;
Let S represent a sunny day.&lt;br /&gt;
It is given that it rained on 13 days. There were 11 sunny morning and 12 sunny afternoons. I will now list out the combinations of rainy/sunny mornings and afternoons.&lt;br /&gt;
Morning - R S S S S R R S S S S S R R S S R R &lt;br /&gt;
Noon - S S R R S S S R R S R S S S R S S S&lt;br /&gt;
There are 11 sunny mornings, 12 sunny afternoons and it rained on 13 days. Therefore, there were 18 total days of vacation.&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;
Let x represent the oldest child.&lt;br /&gt;
Let y represent the second child.&lt;br /&gt;
Let z represent the third child.&lt;br /&gt;
&lt;br /&gt;
Given that the produce of their ages is 36. We can deduce equation 1.&lt;br /&gt;
1)	36 = xyz&lt;br /&gt;
&lt;br /&gt;
Given that the sum of their ages is the same as today’s date. I will now make a list listing out every possible value for x, y and z which will result in a product of 36.&lt;br /&gt;
&lt;br /&gt;
Ages		Product	Sum&lt;br /&gt;
	1 x 1 x 36	36		38&lt;br /&gt;
2 x 3 x 6	36		11&lt;br /&gt;
	1 x 3 x 12	36		16	&lt;br /&gt;
2 x 2 x 9	36		13&lt;br /&gt;
	1 x 4 x 9	36		14&lt;br /&gt;
3 x 3 x 4	36		10&lt;br /&gt;
	1 x 2 x 18	36		21&lt;br /&gt;
	1 x 6 x 6	36		13&lt;br /&gt;
&lt;br /&gt;
	The only possible ages that meet the conditions given in the question are the ages of 1, 6, 6 and 2, 2, 9. But if we assume that the oldest child and the second child are not the same age, the only possible ages of the children are 2, 2, and 9.&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 A and B represent the 2 candles.&lt;br /&gt;
Let A represent the candle that takes 6 hours to burn out.&lt;br /&gt;
Let B represent the candle that takes 3 hours to burn out.&lt;br /&gt;
&lt;br /&gt;
Candle A takes twice as long to burn out than Candle B. We can form equation 1.&lt;br /&gt;
1)	A = 2B&lt;br /&gt;
&lt;br /&gt;
Assume that time starts at 0 and that the burning of the candle is constant.&lt;br /&gt;
Then we can say that Candle A burns 1/6 per hour and Candle B burns 1/3 per hour.&lt;br /&gt;
Let x represent the time in hours. We can form equation 2 and 3.&lt;br /&gt;
2)	A = (6-x)/6&lt;br /&gt;
3)	B = (3-x)/6&lt;br /&gt;
&lt;br /&gt;
Substitute equation 2 and 3 into equation 1.&lt;br /&gt;
A = 2B&lt;br /&gt;
(6-x)/6 = 2[(3-x)/6]&lt;br /&gt;
x = 2&lt;br /&gt;
&lt;br /&gt;
Therefore, it takes 2 hours for one candle to be half of the other.&lt;br /&gt;
&lt;br /&gt;
25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L. &lt;br /&gt;
&lt;br /&gt;
Given:&lt;br /&gt;
Candle X has length L and was lit at 6 hour.&lt;br /&gt;
Candle Y has length L + 1 and was lit at 4.5 hour.&lt;br /&gt;
At 8.5 hour, the length of Candle A is equal to the length of Candle B.&lt;br /&gt;
Candle A burns out at 10 hours.&lt;br /&gt;
Candle B burns out at 10.5 hours.&lt;br /&gt;
&lt;br /&gt;
Then the total burning time of Candle A is: 10 – 6 = 4 hours&lt;br /&gt;
Then the total burning time of Candle B is: 10.5 – 4.6 = 6 hours&lt;br /&gt;
&lt;br /&gt;
Burning time till 8.5 hours for Candle A is: 8.5 – 6 = 2.5 hours.&lt;br /&gt;
Burning time till 8.5h for Candle B  is: 8.5 - 4.5 = 4 hours.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Let x represent the length of Candle A&lt;br /&gt;
Let y represent the length of Candle B&lt;br /&gt;
&lt;br /&gt;
y = x + 1&lt;br /&gt;
&lt;br /&gt;
Let B1 represent the burning constant for x&lt;br /&gt;
Let B2 represent the burning constant for y&lt;br /&gt;
We know that x=y at 8.5 hours.&lt;br /&gt;
&lt;br /&gt;
 (x-4) B1 = x + 1 – 2.5B2&lt;br /&gt;
&lt;br /&gt;
y = 6 B2 at end time&lt;br /&gt;
x = 4B1 at end time&lt;br /&gt;
6B2 = 4B1 + 1 &lt;br /&gt;
B1 = (6B1 -1)/4&lt;br /&gt;
&lt;br /&gt;
Using the above equations and substituting back into the original.&lt;br /&gt;
4B2 = 1 – 2.5B1&lt;br /&gt;
4B2 = 1 – 2.5((6B1 -1)/4)&lt;br /&gt;
B2 = 1.5&lt;br /&gt;
&lt;br /&gt;
Substitute B2 into y = 6B2&lt;br /&gt;
4((4B1 + 1)/6) = 1 – 2.5B1&lt;br /&gt;
B1 = 2&lt;br /&gt;
Substitute B1 into x = 4B1&lt;br /&gt;
&lt;br /&gt;
y = 6B2&lt;br /&gt;
y = 6(1.5) = 9&lt;br /&gt;
&lt;br /&gt;
x = 4B1&lt;br /&gt;
x = 4(2) = 8&lt;br /&gt;
&lt;br /&gt;
Therefore, the length of Candle B is 9 units longer than the length of Candle A which is 8 units.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=54415</id>
		<title>Course:MATH110/Archive/2010-2011/003/Groups/Group 04</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Groups/Group_04&amp;diff=54415"/>
		<updated>2010-10-13T07:46:04Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Group members:&lt;br /&gt;
* Anoud Al-Moqbil&lt;br /&gt;
* Owen Johnston&lt;br /&gt;
* Charlene Ng&lt;br /&gt;
* Ling Sun&lt;br /&gt;
* Dominic Sunga&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
HOMEWORK - DUE OCTOBER 13&lt;br /&gt;
&lt;br /&gt;
Question 1:&lt;br /&gt;
&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the times are different if the route and speed for travelling to and from the airport is the same&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question, we must look at how the time is being measured for the travel to the airport and the travel back&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It is actually the same time, because 80 mins is equivalent to 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
1 hour = 60 mins, so&lt;br /&gt;
80 mins = 1 hour and 20 mins&lt;br /&gt;
&lt;br /&gt;
Question 2:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me why the policeman didn’t stop the lady when she didn’t stop at the stop sign and went three blocks in the wrong direction on a one way street&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
TO solve this question, we must look at the specifics of the words being used and discern what the really mean&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The lady was not actually driving a car. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
The question never stated that the woman was driving when she didn’t stop at the stop sign and went three blocks the wrong way. &lt;br /&gt;
&lt;br /&gt;
Question 3:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how to discern which box has which fruit if we only pick one fruit from one box&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this question we need to look at the fact that all labels are incorrect and that there are only two different types of fruits&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
If you pick one fruit out of the box that is labeled “apples and oranges,” whatever fruit you picked out, that box is supposed to be labeled that fruit. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If you picked an apple from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is the box full of apples, the “apples” box is full of oranges, and the “oranges” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
If you picked an orange from the box that is incorrectly labeled “apples and oranges,” the “apples and oranges” box is full of oranges, the “oranges” box is full of apples, and the “apples” box is apples and oranges. &lt;br /&gt;
&lt;br /&gt;
Question 4:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how I can be the brother of a blind fiddler, but have no brothers myself&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to determine who the blind fiddler is&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The blind fiddler is actually my sister&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
To be the brother of a blind fiddler but have no brothers means that the blind fiddler is my sister&lt;br /&gt;
&lt;br /&gt;
Question 5:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many revolutions it would take for a quarter to completely roll around the edges of another&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this equation we need to look at the sizes of the coin, the starting postion, the halfway position and the final position&lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
The coin will need to make three revolutions&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because we know that the coins are the same size. When the coin makes one revolution, it means that it is in the halfway point. So to go back to the original position, it has to make one more revolution. Therefore, the coin has to make two revolutions&lt;br /&gt;
&lt;br /&gt;
Question 6:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how many apples I would have to pick in order to absolutely have two of a kind&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
To solve this problem we need probability, logic and reasoning. &lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick 4 apples&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there is a chance that if you pick 3, you could get three different types. But if you pick four, at least two of the four will definitely be two of a kind&lt;br /&gt;
&lt;br /&gt;
Question 7:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me times I would have to pick a sock to be sure of getting one pair that is the same color and one pair that is different colors&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need the same things as we did from the previous question&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
You would need to pick a sock 3 three times.&lt;br /&gt;
&lt;br /&gt;
Step 4: &lt;br /&gt;
This is because when you pick three, two of them of will definitely be the same color, and two will be different&lt;br /&gt;
&lt;br /&gt;
Question 8:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me how Reuben can be 21 two das ago, but 23 later next year. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to determine his birthday and his present day. &lt;br /&gt;
&lt;br /&gt;
Step 3: &lt;br /&gt;
His birthday is on December 31st, and his present day is January 1st. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if his birthday was on December 31st, 2010, and his present day was January 1st, 2011, then two days ago he would have been 20 at December 30, 2010. So when he says that later next year he will be 23, he means that December 30, 2012, he will turn 23, which is accurate. &lt;br /&gt;
&lt;br /&gt;
Question 9:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how many rungs will be shown if the tide rises 5 feet.&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this problem, we have to look at it in two different ways&lt;br /&gt;
&lt;br /&gt;
a)	if the rung is at sea level&lt;br /&gt;
b)	if the rung is one foot above sea level&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
If the rung is at sea level, 4 rungs will be shown when the tide rises 5 feet&lt;br /&gt;
If the rung is one foot above sea leve, then five rungs will be shown when the tide rises 5 feet. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because when the rung is at sea level, two rungs will be covered when the tide rises one foot, three rungs will be covered when it rises another foot, four rungs will be covered when it rises another foot, five rungs will be covered when it rises another food, and 6 rungs will be covered in total when the tide rises 5 feet, therefore only 4 rungs will be shown. &lt;br /&gt;
&lt;br /&gt;
When the last rung is one foot above sea level, one rung will be covered for every foot the tide rises. So when the tide rises 5 feet, 5 rungs will be covered and 5 rung will be shown. &lt;br /&gt;
&lt;br /&gt;
Question 10:&lt;br /&gt;
Step 1:&lt;br /&gt;
The question is asking me if half the people are chocolate eaters, whether or not a quarter of the world are women chocolate eaters and half the men are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question, we need to use fractions&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
Yes, it does mean that a quarter of all people are women chocolate eaters and half the men are chocolate eaters, if we follow this very simplified logic&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if all people are comprised of one half being men and one half being women, then half of each gender would be a chocolate eater of half of all people are chocolate eaters. &lt;br /&gt;
&lt;br /&gt;
Question 11:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me who the worst chess player is&lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
 To solve this, need to look closely at the adjectives being used to describe the ages of all players&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
The worst player is either the son or the daughter&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
It cannot be the mother or her older brother, because they are not of the same age (he is older). Therefore, it has to be either the son or the daughter. &lt;br /&gt;
&lt;br /&gt;
Question 12:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me why the Manhattan man’s Bronx Girlfriend left him. &lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to look at the train schedules (they both arrive regularly every 10 mins)&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
She left him because the Brooklyn train arrives earlier than the Bronx train. And also because he was cheating on her – not cool. &lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
&lt;br /&gt;
Question 13:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me how long it would take for a clock to strike 10&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
To solve this question we need to look at the previous information we’re given&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
It would take 10 seconds&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because if it takes 5 seconds for it to strike 5, then it 10 would just be double. &lt;br /&gt;
&lt;br /&gt;
Question 14:&lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me is how many ways can 2 babies be tagged correctly and 2 babies tagged incorrectly. &lt;br /&gt;
&lt;br /&gt;
Step 2:&lt;br /&gt;
We need to use probability&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
There are 6 ways two babies can be tagged correctly and two babies tagged incorrectly. &lt;br /&gt;
It is impossible to incorrectly tag a baby wrong if three were tagged correctly&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
If three babies are tagged, it is impossible to tag the last baby incorrectly&lt;br /&gt;
&lt;br /&gt;
Question 15: &lt;br /&gt;
Step 1:&lt;br /&gt;
This question is asking me if I should accept this extremely risky bet&lt;br /&gt;
&lt;br /&gt;
Step 2: &lt;br /&gt;
We need to use logic and common sense&lt;br /&gt;
&lt;br /&gt;
Step 3:&lt;br /&gt;
I should take the bet&lt;br /&gt;
&lt;br /&gt;
Step 4:&lt;br /&gt;
This is because there are 26 red cards and 26 black cards. If there are 15 red cards in the first half and 12 black cards in the first half, there will be 11 red cards in the second half and 14 cards. Unless the guy is a magician or something.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Dominicjansunga&amp;diff=48063</id>
		<title>User:Dominicjansunga</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Dominicjansunga&amp;diff=48063"/>
		<updated>2010-09-20T01:30:45Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hey Guys&lt;br /&gt;
&lt;br /&gt;
My Name is Dominic. I am currently studying commerce. I am from and was born in the Philippines, but i grew up and went to high school in the Jakarta International School in Indonesia. I chose UBC because the Campus is amazing and so are the people here. I hope to one day start my own business. &lt;br /&gt;
&lt;br /&gt;
That&#039;s pretty much all I have to say!&lt;br /&gt;
&lt;br /&gt;
Dominic&lt;br /&gt;
&lt;br /&gt;
The Pythagorean theorem is the method that mathematicians traditionally use to find the hypotenuse (the longest side) of a right triangle. The Pythagorean theorem also has applications in finding the other lengths of a right triangle, and can also be used to find angles as well. It also has influences in the unit circle and finding angles. It is an important component in trigonometry and is one of the widely used tools.&lt;br /&gt;
The formula is to take two of the shortest lengths of three in a right triangle, square them and then add them together. The answer would be the remaining and longest length, squared. Its discovery is often attributed to Pythagoras, a Greek Mathematician (570 - 495 BC). &lt;br /&gt;
The Pythagorean theorem is applied in real world situations in things such as finding the dimensions of a building, structure or object. The Pythagorean theorem would be most helpful in trying to find out if the angle of a corner in a four-sided building really is 90 degrees.&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:Dominicjansunga&amp;diff=48021</id>
		<title>User:Dominicjansunga</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:Dominicjansunga&amp;diff=48021"/>
		<updated>2010-09-19T23:21:08Z</updated>

		<summary type="html">&lt;p&gt;Dominicjansunga: Created page with &amp;#039;Hey Guys  My Name is Dominic. I am currently studying commerce. I am from and was born in the Philippines, but i grew up and went to high school in the Jakarta International Scho…&amp;#039;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hey Guys&lt;br /&gt;
&lt;br /&gt;
My Name is Dominic. I am currently studying commerce. I am from and was born in the Philippines, but i grew up and went to high school in the Jakarta International School in Indonesia. I chose UBC because the Campus is amazing and so are the people here. I hope to one day start my own business. &lt;br /&gt;
&lt;br /&gt;
That&#039;s pretty much all I have to say!&lt;br /&gt;
&lt;br /&gt;
Dominic&lt;/div&gt;</summary>
		<author><name>Dominicjansunga</name></author>
	</entry>
</feed>