UBC Wiki - User contributions [en]

Thread:Course talk:MATH110/003/Groups/Group 16/Contact Information

2015-09-22T21:40:25Z

ShamillaBirring: Blanked the page

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2012-10-01T19:53:42Z

ShamillaBirring:

{{Infobox MATH110 Groups
| group number = 16

| member 2 = [[User:MatthewJohnson|Matt Johnson]]
| member 3 = [[User:ChristopherWong|Christopher Wong]]
| member 4 = [[User:GaiaSilvestri|Gaia Silvestri]]
| member 5 = [[User:]]
| member 6 =
}}

[[/Basic Skills Project | Basic Skills Project: Find intersection of graphs]]

GROUP RESHUFFLED: 14 November, '10

[[/Homework 5 | Homework 5]]

[[/Homework 4 | Homework 4]]

[[/Homework 3 | Homework 3]]

Course:MATH110/Archive/2010-2011/003/Teams/Jura

2011-01-28T18:40:14Z

ShamillaBirring:

{{Infobox MATH110 Teams
| team name = Jura
| member 1 = [mailto:adam@premierwestmma.com Adam Nguyen] - Grapple
| member 2 = Christa Bicego - Achieve
| member 3 = [mailto:matthew.hsu@gmail.com Matthew Hsu] - Happy
| member 4 = Shamilla Birring - Smile
}}

== Workshop ==
In workshop L. 

Pick one of the topic offered below and then explain in your own words what it means that these concepts work on a logarithmic scale. Create a wiki page with all your explanations. The length of that page is up to you, but it should feel like the work of four people thinking about a topic and trying to make sense of it. If you drafted something and would like some comments (within 24 hours), send me an email and a link where to look at, I'll post comments). 

Topics: 
Decibels (Christa Bicego) 
Richter magnitude scale 
Brightness of stars 
pH (Shamilla Birring) 
Put your name beside the one you are going to do.

== Homework ==
http://wiki.ubc.ca/Course:MATH110/003/Teams/Jura/Homework/

Course:MATH110/Archive/2010-2011/003/Teams/Jura

2011-01-28T18:39:44Z

ShamillaBirring:

{{Infobox MATH110 Teams
| team name = Jura
| member 1 = [mailto:adam@premierwestmma.com Adam Nguyen] - Grapple
| member 2 = Christa Bicego - Achieve
| member 3 = [mailto:matthew.hsu@gmail.com Matthew Hsu] - Happy
| member 4 = Shamilla Birring - Smile
}}

== Workshop ==
In workshop L. 

Pick one of the topic offered below and then explain in your own words what it means that these concepts work on a logarithmic scale. Create a wiki page with all your explanations. The length of that page is up to you, but it should feel like the work of four people thinking about a topic and trying to make sense of it. If you drafted something and would like some comments (within 24 hours), send me an email and a link where to look at, I'll post comments). 

Topics: 
Decibels (Christa Bicego) 
Richter magnitude scale 
Brightness of stars 
pH(Shamilla Birring) 
Put your name beside the one you are going to do.

== Homework ==
http://wiki.ubc.ca/Course:MATH110/003/Teams/Jura/Homework/

Course:MATH110/Archive/2010-2011/003/Teams/Jura/Homework

2011-01-28T03:40:03Z

ShamillaBirring: /* Homework #11 - Part 3 */

== Homework #12 - Part 3 ==

http://wiki.ubc.ca/Course:MATH110/003/Teams/Jura/Homework/12_Part3

== Homework #11 - Part 3 ==
http://wiki.ubc.ca/Course:MATH110/003/Teams/Jura/Homework/11_Part3

User:ShamillaBirring

2011-01-28T02:17:56Z

ShamillaBirring:

Hi, my name is Shamilla and I am a first year student here at UBC.

Pythagorean theorem:
The pythagorean theorem expresses the relationship between the three sides of a right triangle, it is named after the Greek mathematician Pythagoras. The relationship found is that the area of the two squares of the lines adjacent to the right angle will be equal to the area of the square opposite from the right angle, also referred to as the hypotenuse. This relationship can be represented by the formula <math>a^2 + b^2 = c^2</math>. In this equation a and b represent the length of the lines adjacent to the right angle, and c represents the length of the hypotenuse. Therefore, using this formula one can find any of the lengths of either sides of a right triangle. There are several proofs that validate the formula which include algebraic proofs, proofs using similar triangles and proofs using differentials. The theorem is utilized in several situations such as in determining the distance between two sites, as well as in architecture and research.
[[http://en.wikipedia.org/wiki/Pythagorean_theorem]]

Course:MATH110/Archive/2010-2011/003/Teams/Jura

2011-01-27T20:43:33Z

ShamillaBirring:

{{Infobox MATH110 Teams
| team name = Jura
| member 1 = [mailto:adam@premierwestmma.com Adam Nguyen] - Grapple
| member 2 = Christa Bicego - Achieve
| member 3 = [mailto:matthew.hsu@gmail.com Matthew Hsu] - Happy
| member 4 = Shamilla Birring - Smile
}}

==discussion==

Hey guys, I've been playing with the
 
(1/(1+e^-t))
 
function and I came to a few conclusions

I was able to modify the function with a few added variables. The new functions becomes: 
 
(1(q))/(1+we^(-rt)).
 
q - will determine the horizontal asymptote on the right
 
w - will determine at what y should be when t (or x) is 0
 
r - will determine how quickly or slowly the growth will be.
 
Take a look!
 
- Adam

== Workshop ==
In workshop L.

== Homework ==
http://wiki.ubc.ca/Course:MATH110/003/Teams/Jura/Homework/

Course:MATH110/Archive/2010-2011/003/Groups/Group 16/Basic Skills Project

2010-12-03T08:54:07Z

ShamillaBirring: /* Overview of the Beautiful Descartes Invention between Geometry and Algebra */

=What is the "Graph of a Function"?=

The graph of a function ''f'' is the collection of all possible pairs (x, f(x)) that can be found on the graph of ''f'', in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. <ref>http://en.wikipedia.org/wiki/Graph_of_a_function</ref> This implies that, in order to find the intersection(s) of the graphs ''f'' and ''g'', we need to be able to solve the equation f(x)=g(x), which is not always possible. We shall shown you one such case in the last section of this page. Right now we shall focus on more typical examples of graphs one encounters perhaps with a little bit more frequency. 

The plots below showcase some common power functions we have encountered in the course:<ref>http://www.biology.arizona.edu/biomath/tutorials/power/GraphingPowerFunctions.html</ref>

[[File:Power_family.gif]][[File:Power_odd.gif]]

=Find intersection of graphs=

There are at least three types of intersections of graphs, 

1. Intersection of parabolas 

2. Intersection of linear lines 

3. Intersection of a Parabola and a Line 

4. Overview of the beautiful Descartes invention between the geometry and the algebra(reword) 

5. How to write an equation for two graphs in order to find the intersection points algebraically(reword) 

6. Special case-<math>e^x = −x</math>

==Intersection of power functions==

===Warm-Up Question===

What is/are (a) possible scenario(s) for two parabolas appearing on the same coordinate plane? 

a. 1 point of intersection only 

b. 1 or 2 points of intersection 

c. 2 points of intersection without exception 

d. Depending on the magnitude of the plane in question, there can be infinite points of intersection 

e. Others 

f. Do I care? 

The answer will come soon except perhaps for those that picked f. Right now, let’s look at how to find if two quadratic functions, f(x) and g(x), has any point of intersection. 

<math>F(x) = 1.5 \times 2 - 9x + 11.5</math> 

<math>G(x) = -0.2 \times 2 - 0.4x + 2.8</math> 

===Finding intersection of power functions algebraically===

If you understand the following concept, you will not even have to memorise the technique. ''The points of intersection are where both parabolas share the same (x, y) coordinates. So, at the points of intersection the (x, y) coordinates for f(x) equal the (x, y) coordinates for g(x)''. To put this into practice using ''f'' and ''g'', 

<math>1.5 \times 2 - 9x + 11.5=-0.2 \times 2 - 0.4x + 2.8</math> 

From here find x if the equation is factoreble. If not, use the quadratic formula, and if there are points of intersection, <math>\vartriangle b^2-4ac</math> will render one, or two, solutions. Alternatively, if <math>\vartriangle b^2-4ac</math> render a negative root, that points to the fact that there are no points of interception of ''f'' and ''g''. 

===How to predict the number of intersecting points between two or more power functions===

If you want to find how many points of intersections there are between more than two functions, well, there's no special way. You might as well use the algebraic trick in the previous section to work out the point(s) of intersection, that will give you the number of intersections too. This is not limited to parabolas, but applicable to all power function types.

===Tea Time===

Get up and get something to drink/eat because you know you've been wanting to.

===Answer to warmup question===

If you have been such a star student and actually looked through this section for intersection points of power functions, you should have absolutely no trouble answering the warmup question. The answer is ''e''. Or more precisely, two quadratic graphs (parabolas) can have one, two, or zero point of intersection. Since seeing is believing...

===Visualising points of intersection for two quadratic equations===

* One Single Point of Intersection

[[File:OnePoint1.gif]] [[File:OnePoint2.gif]]

* Two Points of Intersection

[[File:TwoPoints1.gif]] [[File:TwoPoints2.gif]]

* No Intersecting Points

[[File:NoPoints1.gif]] [[File:NoPoints2.gif]]

==Intersection of linear functions==

Here we will cover how to find the point of intersection for two linear functions. That is, finding the (x, y) coordinate pair for the point where two lines cross.

We will be using these two linear functions as examples:

<math>F(x) = 3x + 18</math> 

<math>G(x) = -10x + 25</math> 

===Visualising how Two Lines Intersect on the Cartesian Plane===

Where two lines cross is called their point of intersection. As you may or may not know, this intersection point has (x, y) coordinates. It is the same point for both linear functions, this means the (x, y) coordinates are the same for both linear functions of f(x) and g(x).

As we will be graphing the linear functions, we can call f(x) line 1 and g(x) line 2. We will be graphing the functions on a (x, y) coordinate plane therefore we can express the functions into slope-intercept form. That is, express them in equations of this form:
<math>y = mx + b</math>:

<math>y = 3x + 18</math>

<math>y = -10x + 25</math>

[[File:POIGraph1.jpg]]

===Finding Point of Intersection Algebraically===

Since the (x, y) coordinates are the same for both lines at the point of intersection, we can first equate both functions in order to solve for ‘x’.
By equating the two equations we have something like this:

[[File:POItable.jpg]]

Now that we have the x-coordinate for the point of intersection, we can move on to finding the y-coordinate. In order to find the y-coordinate, we can substitute the x-coordinate we found earlier into either of the equations and then solve for ‘y’.

We can use line 1 here:

<math>x = 7/13</math> 
<math>y = 3x + 18</math> 
<math>y = 3(7/13) +18</math> 
<math>y = (21/13) + 18</math> 
<math>y = 255/13</math> 
<math>y = 19.62</math> 

Therefore, the y-coordinate is 255/13 or 19.62. If we want to make sure our calculations are correct, we can check by substituting ‘x’ into line 2 as it should produce the same answer.

<math>x = 7/13</math> 
<math>y = -10x + 25</math> 
<math>y = -10(7/13) + 25</math> 
<math>y = (-70/13) + 25</math> 
<math>y = 255/13</math> 
<math>y = 19.62</math> 

We see that the answer for the y-coordinate is the same, meaning this is the correct number. Therefore, the point of intersection for these two lines is '''(0.54, 19.62)'''.

==Intersection of a Parabola and a Line==

==Overview of the Beautiful Descartes Invention between Geometry and Algebra==

Why is Descartes a significant part of mathematics history?
René Descartes, often referred to as the father of analytical geometry, was the inventor of the Cartesian coordinate grid system. It is in Descartes system that the first coordinate value (x-coordinate) represents where the point is located along the horizontal axis (the x-axis). The second coordinate value (y-coordinate) represents where the point is in terms of the vertical axis (the y-axis). It is this system that allows us to illustrate the intersection of two lines, visually understand the relationship between coordinate points, equations and functions and much more. In other words, this system allows us to draw geometric shapes that are expressed as algebraic equations.

==How to write an equation for two graphs in order to find the intersection points algebraically==

==Special case <math>-e^x = -x</math>==

References:

<references/>

Course:MATH110/Archive/2010-2011/003/Groups/Group 16/Basic Skills Project

2010-12-03T08:51:40Z

ShamillaBirring: /* Overview of the Beautiful Descartes Invention between Geometry and Algebra */

=What is the "Graph of a Function"?=

The graph of a function ''f'' is the collection of all possible pairs (x, f(x)) that can be found on the graph of ''f'', in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. <ref>http://en.wikipedia.org/wiki/Graph_of_a_function</ref> This implies that, in order to find the intersection(s) of the graphs ''f'' and ''g'', we need to be able to solve the equation f(x)=g(x), which is not always possible. We shall shown you one such case in the last section of this page. Right now we shall focus on more typical examples of graphs one encounters perhaps with a little bit more frequency. 

The plots below showcase some common power functions we have encountered in the course:<ref>http://www.biology.arizona.edu/biomath/tutorials/power/GraphingPowerFunctions.html</ref>

[[File:Power_family.gif]][[File:Power_odd.gif]]

=Find intersection of graphs=

There are at least three types of intersections of graphs, 

1. Intersection of parabolas 

2. Intersection of linear lines 

3. Intersection of a Parabola and a Line 

4. Overview of the beautiful Descartes invention between the geometry and the algebra(reword) 

5. How to write an equation for two graphs in order to find the intersection points algebraically(reword) 

6. Special case-<math>e^x = −x</math>

==Intersection of power functions==

===Warm-Up Question===

What is/are (a) possible scenario(s) for two parabolas appearing on the same coordinate plane? 

a. 1 point of intersection only 

b. 1 or 2 points of intersection 

c. 2 points of intersection without exception 

d. Depending on the magnitude of the plane in question, there can be infinite points of intersection 

e. Others 

f. Do I care? 

The answer will come soon except perhaps for those that picked f. Right now, let’s look at how to find if two quadratic functions, f(x) and g(x), has any point of intersection. 

<math>F(x) = 1.5 \times 2 - 9x + 11.5</math> 

<math>G(x) = -0.2 \times 2 - 0.4x + 2.8</math> 

===Finding intersection of power functions algebraically===

If you understand the following concept, you will not even have to memorise the technique. ''The points of intersection are where both parabolas share the same (x, y) coordinates. So, at the points of intersection the (x, y) coordinates for f(x) equal the (x, y) coordinates for g(x)''. To put this into practice using ''f'' and ''g'', 

<math>1.5 \times 2 - 9x + 11.5=-0.2 \times 2 - 0.4x + 2.8</math> 

From here find x if the equation is factoreble. If not, use the quadratic formula, and if there are points of intersection, <math>\vartriangle b^2-4ac</math> will render one, or two, solutions. Alternatively, if <math>\vartriangle b^2-4ac</math> render a negative root, that points to the fact that there are no points of interception of ''f'' and ''g''. 

===How to predict the number of intersecting points between two or more power functions===

If you want to find how many points of intersections there are between more than two functions, well, there's no special way. You might as well use the algebraic trick in the previous section to work out the point(s) of intersection, that will give you the number of intersections too. This is not limited to parabolas, but applicable to all power function types.

===Tea Time===

Get up and get something to drink/eat because you know you've been wanting to.

===Answer to warmup question===

If you have been such a star student and actually looked through this section for intersection points of power functions, you should have absolutely no trouble answering the warmup question. The answer is ''e''. Or more precisely, two quadratic graphs (parabolas) can have one, two, or zero point of intersection. Since seeing is believing...

===Visualising points of intersection for two quadratic equations===

* One Single Point of Intersection

[[File:OnePoint1.gif]] [[File:OnePoint2.gif]]

* Two Points of Intersection

[[File:TwoPoints1.gif]] [[File:TwoPoints2.gif]]

* No Intersecting Points

[[File:NoPoints1.gif]] [[File:NoPoints2.gif]]

==Intersection of linear functions==

Here we will cover how to find the point of intersection for two linear functions. That is, finding the (x, y) coordinate pair for the point where two lines cross.

We will be using these two linear functions as examples:

<math>F(x) = 3x + 18</math> 

<math>G(x) = -10x + 25</math> 

===Visualising how Two Lines Intersect on the Cartesian Plane===

Where two lines cross is called their point of intersection. As you may or may not know, this intersection point has (x, y) coordinates. It is the same point for both linear functions, this means the (x, y) coordinates are the same for both linear functions of f(x) and g(x).

As we will be graphing the linear functions, we can call f(x) line 1 and g(x) line 2. We will be graphing the functions on a (x, y) coordinate plane therefore we can express the functions into slope-intercept form. That is, express them in equations of this form:
<math>y = mx + b</math>:

<math>y = 3x + 18</math>

<math>y = -10x + 25</math>

[[File:POIGraph1.jpg]]

===Finding Point of Intersection Algebraically===

Since the (x, y) coordinates are the same for both lines at the point of intersection, we can first equate both functions in order to solve for ‘x’.
By equating the two equations we have something like this:

[[File:POItable.jpg]]

Now that we have the x-coordinate for the point of intersection, we can move on to finding the y-coordinate. In order to find the y-coordinate, we can substitute the x-coordinate we found earlier into either of the equations and then solve for ‘y’.

We can use line 1 here:

<math>x = 7/13</math> 
<math>y = 3x + 18</math> 
<math>y = 3(7/13) +18</math> 
<math>y = (21/13) + 18</math> 
<math>y = 255/13</math> 
<math>y = 19.62</math> 

Therefore, the y-coordinate is 255/13 or 19.62. If we want to make sure our calculations are correct, we can check by substituting ‘x’ into line 2 as it should produce the same answer.

<math>x = 7/13</math> 
<math>y = -10x + 25</math> 
<math>y = -10(7/13) + 25</math> 
<math>y = (-70/13) + 25</math> 
<math>y = 255/13</math> 
<math>y = 19.62</math> 

We see that the answer for the y-coordinate is the same, meaning this is the correct number. Therefore, the point of intersection for these two lines is '''(0.54, 19.62)'''.

==Intersection of a Parabola and a Line==

==Overview of the Beautiful Descartes Invention between Geometry and Algebra==

Why is Descartes a significant part of mathematics history?
René Descartes was the inventor of the Cartesian coordinate grid system. It is in Descartes system that the first coordinate value (x-coordinate) represents where the point is located along the horizontal axis (the x-axis). The second coordinate value (y-coordinate) represents where the point is in terms of the vertical axis (the y-axis). It is this system that allows us to illustrate the intersection of two lines and visually understand the relationship between coordinate points, equations and functions.

==How to write an equation for two graphs in order to find the intersection points algebraically==

==Special case <math>-e^x = -x</math>==

References:

<references/>

Course:MATH110/Archive/2010-2011/003/Groups/Group 16/Basic Skills Project

2010-12-03T08:51:21Z

ShamillaBirring: /* Overview of the Beautiful Descartes Invention between Geometry and Algebra */

=What is the "Graph of a Function"?=

The graph of a function ''f'' is the collection of all possible pairs (x, f(x)) that can be found on the graph of ''f'', in the form of a curve on a Cartesian plane, together with Cartesian axes, etc. <ref>http://en.wikipedia.org/wiki/Graph_of_a_function</ref> This implies that, in order to find the intersection(s) of the graphs ''f'' and ''g'', we need to be able to solve the equation f(x)=g(x), which is not always possible. We shall shown you one such case in the last section of this page. Right now we shall focus on more typical examples of graphs one encounters perhaps with a little bit more frequency. 

The plots below showcase some common power functions we have encountered in the course:<ref>http://www.biology.arizona.edu/biomath/tutorials/power/GraphingPowerFunctions.html</ref>

[[File:Power_family.gif]][[File:Power_odd.gif]]

=Find intersection of graphs=

There are at least three types of intersections of graphs, 

1. Intersection of parabolas 

2. Intersection of linear lines 

3. Intersection of a Parabola and a Line 

4. Overview of the beautiful Descartes invention between the geometry and the algebra(reword) 

5. How to write an equation for two graphs in order to find the intersection points algebraically(reword) 

6. Special case-<math>e^x = −x</math>

==Intersection of power functions==

===Warm-Up Question===

What is/are (a) possible scenario(s) for two parabolas appearing on the same coordinate plane? 

a. 1 point of intersection only 

b. 1 or 2 points of intersection 

c. 2 points of intersection without exception 

d. Depending on the magnitude of the plane in question, there can be infinite points of intersection 

e. Others 

f. Do I care? 

The answer will come soon except perhaps for those that picked f. Right now, let’s look at how to find if two quadratic functions, f(x) and g(x), has any point of intersection. 

<math>F(x) = 1.5 \times 2 - 9x + 11.5</math> 

<math>G(x) = -0.2 \times 2 - 0.4x + 2.8</math> 

===Finding intersection of power functions algebraically===

If you understand the following concept, you will not even have to memorise the technique. ''The points of intersection are where both parabolas share the same (x, y) coordinates. So, at the points of intersection the (x, y) coordinates for f(x) equal the (x, y) coordinates for g(x)''. To put this into practice using ''f'' and ''g'', 

<math>1.5 \times 2 - 9x + 11.5=-0.2 \times 2 - 0.4x + 2.8</math> 

From here find x if the equation is factoreble. If not, use the quadratic formula, and if there are points of intersection, <math>\vartriangle b^2-4ac</math> will render one, or two, solutions. Alternatively, if <math>\vartriangle b^2-4ac</math> render a negative root, that points to the fact that there are no points of interception of ''f'' and ''g''. 

===How to predict the number of intersecting points between two or more power functions===

If you want to find how many points of intersections there are between more than two functions, well, there's no special way. You might as well use the algebraic trick in the previous section to work out the point(s) of intersection, that will give you the number of intersections too. This is not limited to parabolas, but applicable to all power function types.

===Tea Time===

Get up and get something to drink/eat because you know you've been wanting to.

===Answer to warmup question===

If you have been such a star student and actually looked through this section for intersection points of power functions, you should have absolutely no trouble answering the warmup question. The answer is ''e''. Or more precisely, two quadratic graphs (parabolas) can have one, two, or zero point of intersection. Since seeing is believing...

===Visualising points of intersection for two quadratic equations===

* One Single Point of Intersection

[[File:OnePoint1.gif]] [[File:OnePoint2.gif]]

* Two Points of Intersection

[[File:TwoPoints1.gif]] [[File:TwoPoints2.gif]]

* No Intersecting Points

[[File:NoPoints1.gif]] [[File:NoPoints2.gif]]

==Intersection of linear functions==

Here we will cover how to find the point of intersection for two linear functions. That is, finding the (x, y) coordinate pair for the point where two lines cross.

We will be using these two linear functions as examples:

<math>F(x) = 3x + 18</math> 

<math>G(x) = -10x + 25</math> 

===Visualising how Two Lines Intersect on the Cartesian Plane===

Where two lines cross is called their point of intersection. As you may or may not know, this intersection point has (x, y) coordinates. It is the same point for both linear functions, this means the (x, y) coordinates are the same for both linear functions of f(x) and g(x).

As we will be graphing the linear functions, we can call f(x) line 1 and g(x) line 2. We will be graphing the functions on a (x, y) coordinate plane therefore we can express the functions into slope-intercept form. That is, express them in equations of this form:
<math>y = mx + b</math>:

<math>y = 3x + 18</math>

<math>y = -10x + 25</math>

[[File:POIGraph1.jpg]]

===Finding Point of Intersection Algebraically===

Since the (x, y) coordinates are the same for both lines at the point of intersection, we can first equate both functions in order to solve for ‘x’.
By equating the two equations we have something like this:

[[File:POItable.jpg]]

Now that we have the x-coordinate for the point of intersection, we can move on to finding the y-coordinate. In order to find the y-coordinate, we can substitute the x-coordinate we found earlier into either of the equations and then solve for ‘y’.

We can use line 1 here:

<math>x = 7/13</math> 
<math>y = 3x + 18</math> 
<math>y = 3(7/13) +18</math> 
<math>y = (21/13) + 18</math> 
<math>y = 255/13</math> 
<math>y = 19.62</math> 

Therefore, the y-coordinate is 255/13 or 19.62. If we want to make sure our calculations are correct, we can check by substituting ‘x’ into line 2 as it should produce the same answer.

<math>x = 7/13</math> 
<math>y = -10x + 25</math> 
<math>y = -10(7/13) + 25</math> 
<math>y = (-70/13) + 25</math> 
<math>y = 255/13</math> 
<math>y = 19.62</math> 

We see that the answer for the y-coordinate is the same, meaning this is the correct number. Therefore, the point of intersection for these two lines is '''(0.54, 19.62)'''.

==Intersection of a Parabola and a Line==

==Overview of the Beautiful Descartes Invention between Geometry and Algebra==

Why is he a significant part of mathematics history?
René Descartes was the inventor of the Cartesian coordinate grid system. It is in Descartes system that the first coordinate value (x-coordinate) represents where the point is located along the horizontal axis (the x-axis). The second coordinate value (y-coordinate) represents where the point is in terms of the vertical axis (the y-axis). It is this system that allows us to illustrate the intersection of two lines and visually understand the relationship between coordinate points, equations and functions.

==How to write an equation for two graphs in order to find the intersection points algebraically==

==Special case <math>-e^x = -x</math>==

References:

<references/>

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-19T23:14:30Z

ShamillaBirring:

{{Infobox MATH110 Groups
| group number = 16
| member 1 = [[User:ShamillaBirring|Shamilla Birring]]
| member 2 = Matt Johnson
| member 3 = [[User:ArabellaCynthiaOlomide|Arabella Olomide]]
| member 4 = [[User:GaiaSilvestri|Gaia Solvestri]]
| member 5 = [[User:ChristopherWong|Christopher Wong]]
| member 6 =
}}

===Contact Information===
* Shamilla Birring | shamilla_birring@hotmail.com
* Arabella Olomide | aco2@sfu.ca

-----

== Homework 4 - due Wednesday October 20 ==

'''Problem 1'''
Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?

-Tosh owns a cat,

-Bianca owns a frog that she loves,

-Jaela owns a parrot which keeps calling her "darling, darling",

-Jun owns a snake, don't mess with him,

-Suzan is the name of the frog,

-The cat is named Jun,

-The name by which they call the turtle is the name of the woman whose pet is Tosh,

-Finally, Suzan's mother's pet is Bianca.

First it is important to state all the information we know:

Pet owners:

-Tosh (male)

-Bianca (female)

-Jaella (female)

-Jun (male)

-Suzan (female)

Note: we are able to conclude the owners genders by looking for gender indicating words.
For example in the phrase "Bianca owns a frog that ''she'' loves", we can can conclude that Bianca is a female because it states that SHE loves her frog.

Pet owners and their pets: (Note: the pets names are in the parentheses)

Tosh -> cat (Jun)

Bianca -> frog (Suzan)

Jaella -> parrot

Jun -> snake

Suzan -> turtle

Note: The only pet not assigned to an owner is the turtle, therefore it is logical that the turtle is Suzan's pet

Note: Keep in mind that Suzan's mother's pets name is Bianca

We can make some further conclusions from the information provided in the question.
Conclusions:

1)

-The name of the turtle will be a female owner of the pet Tosh.
This is concluded from the phrase "The name by which they call the turtle is the name of the woman whose pet is Tosh"

-> because it is a woman, we recognize that the turtles name will be a females name

-> the turtle will have the name of the pet Tosh's owner

In effect of conclusion 1, we can recognize the name of Suzans turtle to be Janella. This can be done because all the friends names have been used as pets names except; Janella and Tosh. Therefore since the turtles name must be a females name we can conclude the turtles name is Janella.

Following we can conclude that Janella's pet parrot's name is Tosh since it states that "The name by which they call the turtle is the name of the woman ''whose pet is Tosh''".

Therefore adding these new discoveries to our previous knowledge we can update our information.

Pet owners and their pets: (Note: the pets names are in the parentheses)

Tosh -> cat (Jun)

Bianca -> frog (Suzan)

Jaella -> parrot (Tosh)

Jun -> snake

Suzan -> turtle (Janella)

Therefore, since the snake is the only pet that has not been named and the name Bianca has not been assigned, we can conclude that Jun's snakes name is Bianca.
Following, it is imperative to remember the Bianca is the snake,meaning the snake is the pet that belongs to Suzan's mother.

Hence, the snake named Bianca is Suzan's mother's pet.

'''Problem 2'''

'''Problem 3'''

This problem is quite difficult to map out so I will phrase it exactly how I did while finding solutions because that is what worked for me. In this exercise, the more we know about a particular person the better; however it could be possible that one person be suitable for more than one position and vice versa.

Adam: not a catcher, battery or infield, not shortstop

Bobo: not a catcher, not 3rd baseman, not shortstop, not 2nd baseman

Charles: not pitcher, could be outfielder, battery or infield

Ed: not 2nd baseman, not outfielder, not catcher, not 3rd baseman, bachelor, could be shortstop or 1st baseman

Hassan: not 3rd baseman, could be battery or infielder, not catcher, not 2nd baseman, could be either 1st baseman, shortstop, pitcher

Jason: not 3rd baseman, not catcher, bachelor. '''Jason is the 2nd baseman.'''

Matthieu: not centre field, shortest in height of the team

Pascal: not pitcher, could be outfielder, not shortstop, not catcher, not 2nd baseman, 3rd baseman, not right fielder, not centre fielder, bachelor. '''Pascal is the left outfielder.'''

Sung: married, not right fielder, not centre fielder, not left fielder, not 2nd baseman.

''Important notes'':

There are 5 bachelor, 4 married men, the catcher, the 3rd baseman and the pitcher are married.

'''Problem 4'''

'''Problem 5'''

== Homework 3 - due Wednesday October 13 ==

Thanks Shamilla,

That sounds like a great idea.

Arabella.

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful 
[[User:ShamillaBirring|ShamillaBirring]]

All you ever wanted to know about adding images can be found [http://wiki.ubc.ca/Help:Adding_Media#Inserting_an_Image_onto_a_UBC_Wiki_Page here]. 
--[[User:DavidKohler|DavidKohler]]

'''25 PROBLEMS'''

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

"QUESTION 10" -
Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.

Being a man or a woman is independent of preference for chocolate.

i) No, it does not follow.
ii) No, it does not follow.

"QUESTION 11" -
A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?

The situation is not possible. The woman's daughter and son must be either the best or worst players. They are of opposite sex and they have the same age, since the woman is older than both of them (she is their mother) . The woman's older brother is the oldest player.

"QUESTION 12" -
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

For example, the Bronx train arrives at 12.00, 12.10. 12.20...and the Brooklyn train arrives at 12.09, 12.19, 12.29...

"QUESTION 13" -
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

Considering the time between chimes: it takes 45/4 = 11.25 seconds to strike 10.00.

"QUESTION 14" -
One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

i) There are six ways that two of the four babies can be correctly tagged.
ii) There are no ways in which three of the four babies correctly tagged.

"QUESTION 15" -
Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

Alex is right, I shouldn't accept his bet. The deck of cards is made up of an equal number of black and red cards (respectively 1/2 of the total number of cards).

'''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?'''

Looking at the question, we are the given positions of the runners as followed:

Sven: Middle place of all runners

Dan: 10th place

Lars: 16th place

From the details of the questions, we can first of all try to determine a range for which place Sven came in as we only know he is in the middle amongst all runners. First we know that Dan, who was slower than Sven came in 10th place thus we can assume that Sven came in places ranging from 1st to 9th. Taking into account that Sven is in the middle we can say that the maximum number of runners is 18 as the Sven can only be placed up to the 9th at max. Next, as we are told that Sven is placed exactly in the middle therefore we can deduce that the number of runners is an odd number, as an even number cannot produce a runner exactly in the middle. Lastly we take into account the placement of Lars who came in 16th, as mentioned earlier we deduced that the maximum number of runners is 18 therefore shortening the range to 16-18.

In conclusion, the only possible number that fits with all the criteria is 17; therefore there were 17 runners in the race.

'''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?'''

'''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'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's children? Explain.
'''
We are told that Paula has three children and that the product of their ages is 36, therefore our first step would be to find the factors of 36 in sets of 3. This gives us the following sets:

• 1,1,36

• 1,2,18

• 1,3,12

• 1,4,9

• 1,6,6

• 2,2,9

• 2,3,6

• 3, 3,4

Next we are told that the sum of their ages is the same as a date, therefore we know the sum cannot exceed 31 as that is the maximum number of days in a month. By calculating the sums of each set of factors we are given the following:

• 1+ 1+36 = 38

• 1+2+18= 21

• 1+3+12 = 16

• 1+4+9 = 14

• 1+6+6 = 13

• 2+2+9= 13

• 2+3+4 = 11

• 3+3+4= 10

From this we see that we can eliminate the set 1,1,36 as its sum is greater than 31.
Next we are told that the information is still not enough, the only reason for this may be that there are sums of the same number thus making Paul unable to determine which the correct set is. Looking at the sums we can see that the number 13 is duplicated making 1,6,6 and 2,2,9 possible answers.

Lastly, we are told that the oldest child has red hair, the clue here is that there is a eldest sibling, ruling out the set 6,6,1 as there is not one single elder sibling. Therefore, this we can determine the ages of Paula’s children to be 9,2 and 2.

'''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?
'''
From the details of the question we are given the following conditions:

1.) Two candles are of equal length

2.) One candle took 6hr to burn out (we can call this Candle A)

3.) One candle took 3hr to burn out (we can call this Canlde B)

Firstly to make this question more clearly, we can give the candles an actual length giving that both candles are of equal length. Let’s assume that the candle length is 60cm.Taking into account the time it takes each candle to burn out we can come up rate of how fast each candle burns by dividing the length by the time, thus giving us the following information.

Candle A: 60cm/6hr = 10cm/hr

Candle B: 60cm/3hr = 20cm/hr

Second, we can shorten our time to under 3hrs as we know that by 3hrs, Candle B would’ve already burned out leaving only one candle left. Now we can use a trial and error method, by trying the times of each hour. After the first hour, the length of Candle A would be 50cm while the length of Candle B would be 40cm; one is not exactly twice as long as the other. Next after the second hour, the length of Candle A is 40cm while the length of Candle B is 20cm; one candle is exactly twice as long as the other.

Therefore after 2 hours, one candle will be exactly twice as long as the other candle.

'''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.'''

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-19T23:13:23Z

ShamillaBirring:

{{Infobox MATH110 Groups
| group number = 16
| member 1 = [[User:ShamillaBirring|Shamilla Birring]]
| member 2 = Matt Johnson
| member 3 = [[User:ArabellaCynthiaOlomide|Arabella Olomide]]
| member 4 = [[User:GaiaSilvestri|Gaia Solvestri]]
| member 5 = [[User:ChristopherWong|Christopher Wong]]
| member 6 =
}}

===Contact Information===
* Shamilla Birring | shamilla_birring@hotmail.com
* Arabella Olomide | aco2@sfu.ca

-----

== Homework 4 - due Wednesday October 20 ==

'''Problem 1'''
Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?

-Tosh owns a cat,

-Bianca owns a frog that she loves,

-Jaela owns a parrot which keeps calling her "darling, darling",

-Jun owns a snake, don't mess with him,

-Suzan is the name of the frog,

-The cat is named Jun,

-The name by which they call the turtle is the name of the woman whose pet is Tosh,

-Finally, Suzan's mother's pet is Bianca.

First it is important to state all the information we know:

Pet owners:

-Tosh (male)

-Bianca (female)

-Jaella (female)

-Jun (male)

-Suzan (female)

Note: we are able to conclude the owners genders by looking for gender indicating words.
For example in the phrase "Bianca owns a frog that ''she'' loves", we can can conclude that Bianca is a female because it states that SHE loves her frog.

Pet owners and their pets: (Note: the pets names are in the parentheses)

-Tosh -> cat (Jun)

-Bianca -> frog (Suzan)

-Jaella -> parrot

-Jun -> snake

-Suzan -> turtle

Note: The only pet not assigned to an owner is the turtle, therefore it is logical that the turtle is Suzan's pet

Note: Keep in mind that Suzan's mother's pets name is Bianca

We can make some further conclusions from the information provided in the question.
Conclusions:

1)

-The name of the turtle will be a female owner of the pet Tosh.
This is concluded from the phrase "The name by which they call the turtle is the name of the woman whose pet is Tosh"

-> because it is a woman, we recognize that the turtles name will be a females name

-> the turtle will have the name of the pet Tosh's owner

In effect of conclusion 1, we can recognize the name of Suzans turtle to be Janella. This can be done because all the friends names have been used as pets names except; Janella and Tosh. Therefore since the turtles name must be a females name we can conclude the turtles name is Janella.

Following we can conclude that Janella's pet parrot's name is Tosh since it states that "The name by which they call the turtle is the name of the woman ''whose pet is Tosh''".

Therefore adding these new discoveries to our previous knowledge we can update our information.

Pet owners and their pets: (Note: the pets names are in the parentheses)

-Tosh -> cat (Jun)

-Bianca -> frog (Suzan)

-Jaella -> parrot (Tosh)

-Jun -> snake

-Suzan -> turtle (Janella)

Therefore, since the snake is the only pet that has not been named and the name Bianca has not been assigned, we can conclude that Jun's snakes name is Bianca.
Following, it is imperative to remember the Bianca is the snake,meaning the snake is the pet that belongs to Suzan's mother.

Hence, the snake named Bianca is Suzan's mother's pet.

'''Problem 2'''

'''Problem 3'''

This problem is quite difficult to map out so I will phrase it exactly how I did while finding solutions because that is what worked for me. In this exercise, the more we know about a particular person the better; however it could be possible that one person be suitable for more than one position and vice versa.

Adam: not a catcher, battery or infield, not shortstop

Bobo: not a catcher, not 3rd baseman, not shortstop, not 2nd baseman

Charles: not pitcher, could be outfielder, battery or infield

Ed: not 2nd baseman, not outfielder, not catcher, not 3rd baseman, bachelor, could be shortstop or 1st baseman

Hassan: not 3rd baseman, could be battery or infielder, not catcher, not 2nd baseman, could be either 1st baseman, shortstop, pitcher

Jason: not 3rd baseman, not catcher, bachelor. '''Jason is the 2nd baseman.'''

Matthieu: not centre field, shortest in height of the team

Pascal: not pitcher, could be outfielder, not shortstop, not catcher, not 2nd baseman, 3rd baseman, not right fielder, not centre fielder, bachelor. '''Pascal is the left outfielder.'''

Sung: married, not right fielder, not centre fielder, not left fielder, not 2nd baseman.

''Important notes'':

There are 5 bachelor, 4 married men, the catcher, the 3rd baseman and the pitcher are married.

'''Problem 4'''

'''Problem 5'''

== Homework 3 - due Wednesday October 13 ==

Thanks Shamilla,

That sounds like a great idea.

Arabella.

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful 
[[User:ShamillaBirring|ShamillaBirring]]

All you ever wanted to know about adding images can be found [http://wiki.ubc.ca/Help:Adding_Media#Inserting_an_Image_onto_a_UBC_Wiki_Page here]. 
--[[User:DavidKohler|DavidKohler]]

'''25 PROBLEMS'''

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

"QUESTION 10" -
Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.

Being a man or a woman is independent of preference for chocolate.

i) No, it does not follow.
ii) No, it does not follow.

"QUESTION 11" -
A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?

The situation is not possible. The woman's daughter and son must be either the best or worst players. They are of opposite sex and they have the same age, since the woman is older than both of them (she is their mother) . The woman's older brother is the oldest player.

"QUESTION 12" -
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

For example, the Bronx train arrives at 12.00, 12.10. 12.20...and the Brooklyn train arrives at 12.09, 12.19, 12.29...

"QUESTION 13" -
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

Considering the time between chimes: it takes 45/4 = 11.25 seconds to strike 10.00.

"QUESTION 14" -
One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

i) There are six ways that two of the four babies can be correctly tagged.
ii) There are no ways in which three of the four babies correctly tagged.

"QUESTION 15" -
Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

Alex is right, I shouldn't accept his bet. The deck of cards is made up of an equal number of black and red cards (respectively 1/2 of the total number of cards).

'''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?'''

Looking at the question, we are the given positions of the runners as followed:

Sven: Middle place of all runners

Dan: 10th place

Lars: 16th place

From the details of the questions, we can first of all try to determine a range for which place Sven came in as we only know he is in the middle amongst all runners. First we know that Dan, who was slower than Sven came in 10th place thus we can assume that Sven came in places ranging from 1st to 9th. Taking into account that Sven is in the middle we can say that the maximum number of runners is 18 as the Sven can only be placed up to the 9th at max. Next, as we are told that Sven is placed exactly in the middle therefore we can deduce that the number of runners is an odd number, as an even number cannot produce a runner exactly in the middle. Lastly we take into account the placement of Lars who came in 16th, as mentioned earlier we deduced that the maximum number of runners is 18 therefore shortening the range to 16-18.

In conclusion, the only possible number that fits with all the criteria is 17; therefore there were 17 runners in the race.

'''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?'''

'''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'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's children? Explain.
'''
We are told that Paula has three children and that the product of their ages is 36, therefore our first step would be to find the factors of 36 in sets of 3. This gives us the following sets:

• 1,1,36

• 1,2,18

• 1,3,12

• 1,4,9

• 1,6,6

• 2,2,9

• 2,3,6

• 3, 3,4

Next we are told that the sum of their ages is the same as a date, therefore we know the sum cannot exceed 31 as that is the maximum number of days in a month. By calculating the sums of each set of factors we are given the following:

• 1+ 1+36 = 38

• 1+2+18= 21

• 1+3+12 = 16

• 1+4+9 = 14

• 1+6+6 = 13

• 2+2+9= 13

• 2+3+4 = 11

• 3+3+4= 10

From this we see that we can eliminate the set 1,1,36 as its sum is greater than 31.
Next we are told that the information is still not enough, the only reason for this may be that there are sums of the same number thus making Paul unable to determine which the correct set is. Looking at the sums we can see that the number 13 is duplicated making 1,6,6 and 2,2,9 possible answers.

Lastly, we are told that the oldest child has red hair, the clue here is that there is a eldest sibling, ruling out the set 6,6,1 as there is not one single elder sibling. Therefore, this we can determine the ages of Paula’s children to be 9,2 and 2.

'''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?
'''
From the details of the question we are given the following conditions:

1.) Two candles are of equal length

2.) One candle took 6hr to burn out (we can call this Candle A)

3.) One candle took 3hr to burn out (we can call this Canlde B)

Firstly to make this question more clearly, we can give the candles an actual length giving that both candles are of equal length. Let’s assume that the candle length is 60cm.Taking into account the time it takes each candle to burn out we can come up rate of how fast each candle burns by dividing the length by the time, thus giving us the following information.

Candle A: 60cm/6hr = 10cm/hr

Candle B: 60cm/3hr = 20cm/hr

Second, we can shorten our time to under 3hrs as we know that by 3hrs, Candle B would’ve already burned out leaving only one candle left. Now we can use a trial and error method, by trying the times of each hour. After the first hour, the length of Candle A would be 50cm while the length of Candle B would be 40cm; one is not exactly twice as long as the other. Next after the second hour, the length of Candle A is 40cm while the length of Candle B is 20cm; one candle is exactly twice as long as the other.

Therefore after 2 hours, one candle will be exactly twice as long as the other candle.

'''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.'''

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-19T23:12:14Z

ShamillaBirring:

{{Infobox MATH110 Groups
| group number = 16
| member 1 = [[User:ShamillaBirring|Shamilla Birring]]
| member 2 = Matt Johnson
| member 3 = [[User:ArabellaCynthiaOlomide|Arabella Olomide]]
| member 4 = [[User:GaiaSilvestri|Gaia Solvestri]]
| member 5 = [[User:ChristopherWong|Christopher Wong]]
| member 6 =
}}

===Contact Information===
* Shamilla Birring | shamilla_birring@hotmail.com
* Arabella Olomide | aco2@sfu.ca

-----

== Homework 4 - due Wednesday October 20 ==

'''Problem 1'''
Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?

-Tosh owns a cat,

-Bianca owns a frog that she loves,

-Jaela owns a parrot which keeps calling her "darling, darling",

-Jun owns a snake, don't mess with him,

-Suzan is the name of the frog,

-The cat is named Jun,

-The name by which they call the turtle is the name of the woman whose pet is Tosh,

-Finally, Suzan's mother's pet is Bianca.

First it is important to state all the information we know:

Pet owners:
-Tosh (male)
-Bianca (female)
-Jaella (female)
-Jun (male)
-Suzan (female)

Note: we are able to conclude the owners genders by looking for gender indicating words.
For example in the phrase "Bianca owns a frog that ''she'' loves", we can can conclude that Bianca is a female because it states that SHE loves her frog.

Pet owners and their pets: (Note: the pets names are in the parentheses)
-Tosh -> cat (Jun)
-Bianca -> frog (Suzan)
-Jaella -> parrot
-Jun -> snake
-Suzan -> turtle

Note: The only pet not assigned to an owner is the turtle, therefore it is logical that the turtle is Suzans pet
Note: Keep in mind that Suzan's mother's pets name is Bianca

We can make some further conclusions from the information provided in the question.
Conclusions:

1)
-The name of the turtle will be a female owner of the pet Tosh.
This is concluded from the phrase "The name by which they call the turtle is the name of the woman whose pet is Tosh"
-> because it is a woman, we recognize that the turtles name will be a females name
-> the turtle will have the name of the pet Tosh's owner

In effect of conclusion 1, we can recognize the name of Suzans turtle to be Janella. This can be done because all the friends names have been used as pets names except; Janella and Tosh. Therefore since the turtles name must be a females name we can conclude the turtles name is Janella.

Following we can conclude that Janella's pet parrot's name is Tosh since it states that "The name by which they call the turtle is the name of the woman ''whose pet is Tosh''".

Therefore adding these new discoveries to our previous knowledge we can update our information.

Pet owners and their pets: (Note: the pets names are in the parentheses)
-Tosh -> cat (Jun)
-Bianca -> frog (Suzan)
-Jaella -> parrot (Tosh)
-Jun -> snake
-Suzan -> turtle (Janella)

Therefore, since the snake is the only pet that has not been named and the name Bianca has not been assigned, we can conclude that Jun's snakes name is Bianca.
Following, it is imperative to remember the Bianca is the snake,meaning the snake is the pet that belongs to Suzan's mother.

Hence, the snake named Bianca is Suzan's mother's pet.

'''Problem 2'''

'''Problem 3'''

This problem is quite difficult to map out so I will phrase it exactly how I did while finding solutions because that is what worked for me. In this exercise, the more we know about a particular person the better; however it could be possible that one person be suitable for more than one position and vice versa.

Adam: not a catcher, battery or infield, not shortstop

Bobo: not a catcher, not 3rd baseman, not shortstop, not 2nd baseman

Charles: not pitcher, could be outfielder, battery or infield

Ed: not 2nd baseman, not outfielder, not catcher, not 3rd baseman, bachelor, could be shortstop or 1st baseman

Hassan: not 3rd baseman, could be battery or infielder, not catcher, not 2nd baseman, could be either 1st baseman, shortstop, pitcher

Jason: not 3rd baseman, not catcher, bachelor. '''Jason is the 2nd baseman.'''

Matthieu: not centre field, shortest in height of the team

Pascal: not pitcher, could be outfielder, not shortstop, not catcher, not 2nd baseman, 3rd baseman, not right fielder, not centre fielder, bachelor. '''Pascal is the left outfielder.'''

Sung: married, not right fielder, not centre fielder, not left fielder, not 2nd baseman.

''Important notes'':

There are 5 bachelor, 4 married men, the catcher, the 3rd baseman and the pitcher are married.

'''Problem 4'''

'''Problem 5'''

== Homework 3 - due Wednesday October 13 ==

Thanks Shamilla,

That sounds like a great idea.

Arabella.

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful 
[[User:ShamillaBirring|ShamillaBirring]]

All you ever wanted to know about adding images can be found [http://wiki.ubc.ca/Help:Adding_Media#Inserting_an_Image_onto_a_UBC_Wiki_Page here]. 
--[[User:DavidKohler|DavidKohler]]

'''25 PROBLEMS'''

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

"QUESTION 10" -
Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.

Being a man or a woman is independent of preference for chocolate.

i) No, it does not follow.
ii) No, it does not follow.

"QUESTION 11" -
A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?

The situation is not possible. The woman's daughter and son must be either the best or worst players. They are of opposite sex and they have the same age, since the woman is older than both of them (she is their mother) . The woman's older brother is the oldest player.

"QUESTION 12" -
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

For example, the Bronx train arrives at 12.00, 12.10. 12.20...and the Brooklyn train arrives at 12.09, 12.19, 12.29...

"QUESTION 13" -
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

Considering the time between chimes: it takes 45/4 = 11.25 seconds to strike 10.00.

"QUESTION 14" -
One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

i) There are six ways that two of the four babies can be correctly tagged.
ii) There are no ways in which three of the four babies correctly tagged.

"QUESTION 15" -
Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

Alex is right, I shouldn't accept his bet. The deck of cards is made up of an equal number of black and red cards (respectively 1/2 of the total number of cards).

'''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?'''

Looking at the question, we are the given positions of the runners as followed:

Sven: Middle place of all runners

Dan: 10th place

Lars: 16th place

From the details of the questions, we can first of all try to determine a range for which place Sven came in as we only know he is in the middle amongst all runners. First we know that Dan, who was slower than Sven came in 10th place thus we can assume that Sven came in places ranging from 1st to 9th. Taking into account that Sven is in the middle we can say that the maximum number of runners is 18 as the Sven can only be placed up to the 9th at max. Next, as we are told that Sven is placed exactly in the middle therefore we can deduce that the number of runners is an odd number, as an even number cannot produce a runner exactly in the middle. Lastly we take into account the placement of Lars who came in 16th, as mentioned earlier we deduced that the maximum number of runners is 18 therefore shortening the range to 16-18.

In conclusion, the only possible number that fits with all the criteria is 17; therefore there were 17 runners in the race.

'''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?'''

'''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'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's children? Explain.
'''
We are told that Paula has three children and that the product of their ages is 36, therefore our first step would be to find the factors of 36 in sets of 3. This gives us the following sets:

• 1,1,36

• 1,2,18

• 1,3,12

• 1,4,9

• 1,6,6

• 2,2,9

• 2,3,6

• 3, 3,4

Next we are told that the sum of their ages is the same as a date, therefore we know the sum cannot exceed 31 as that is the maximum number of days in a month. By calculating the sums of each set of factors we are given the following:

• 1+ 1+36 = 38

• 1+2+18= 21

• 1+3+12 = 16

• 1+4+9 = 14

• 1+6+6 = 13

• 2+2+9= 13

• 2+3+4 = 11

• 3+3+4= 10

From this we see that we can eliminate the set 1,1,36 as its sum is greater than 31.
Next we are told that the information is still not enough, the only reason for this may be that there are sums of the same number thus making Paul unable to determine which the correct set is. Looking at the sums we can see that the number 13 is duplicated making 1,6,6 and 2,2,9 possible answers.

Lastly, we are told that the oldest child has red hair, the clue here is that there is a eldest sibling, ruling out the set 6,6,1 as there is not one single elder sibling. Therefore, this we can determine the ages of Paula’s children to be 9,2 and 2.

'''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?
'''
From the details of the question we are given the following conditions:

1.) Two candles are of equal length

2.) One candle took 6hr to burn out (we can call this Candle A)

3.) One candle took 3hr to burn out (we can call this Canlde B)

Firstly to make this question more clearly, we can give the candles an actual length giving that both candles are of equal length. Let’s assume that the candle length is 60cm.Taking into account the time it takes each candle to burn out we can come up rate of how fast each candle burns by dividing the length by the time, thus giving us the following information.

Candle A: 60cm/6hr = 10cm/hr

Candle B: 60cm/3hr = 20cm/hr

Second, we can shorten our time to under 3hrs as we know that by 3hrs, Candle B would’ve already burned out leaving only one candle left. Now we can use a trial and error method, by trying the times of each hour. After the first hour, the length of Candle A would be 50cm while the length of Candle B would be 40cm; one is not exactly twice as long as the other. Next after the second hour, the length of Candle A is 40cm while the length of Candle B is 20cm; one candle is exactly twice as long as the other.

Therefore after 2 hours, one candle will be exactly twice as long as the other candle.

'''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.'''

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-19T19:38:43Z

ShamillaBirring:

{{Infobox MATH110 Groups
| group number = 16
| member 1 = [[User:ShamillaBirring|Shamilla Birring]]
| member 2 = Matt Johnson
| member 3 = [[User:ArabellaCynthiaOlomide|Arabella Olomide]]
| member 4 = [[User:GaiaSilvestri|Gaia Solvestri]]
| member 5 = [[User:ChristopherWong|Christopher Wong]]
| member 6 =
}}

===Contact Information===
* Shamilla Birring | shamilla_birring@hotmail.com
* Arabella Olomide | aco2@sfu.ca

-----

== Homework 4 - due Wednesday October 20 ==

'''Problem 1'''
Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?

-Tosh owns a cat,

-Bianca owns a frog that she loves,

-Jaela owns a parrot which keeps calling her "darling, darling",

-Jun owns a snake, don't mess with him,

-Suzan is the name of the frog,

-The cat is named Jun,

-The name by which they call the turtle is the name of the woman whose pet is Tosh,

-Finally, Suzan's mother's pet is Bianca.

'''Problem 2'''

'''Problem 3'''

This problem is quite difficult to map out so I will phrase it exactly how I did while finding solutions because that is what worked for me. In this exercise, the more we know about a particular person the better; however it could be possible that one person be suitable for more than one position and vice versa.

Adam: not a catcher, battery or infield, not shortstop

Bobo: not a catcher, not 3rd baseman, not shortstop, not 2nd baseman

Charles: not pitcher, could be outfielder, battery or infield

Ed: not 2nd baseman, not outfielder, not catcher, not 3rd baseman, bachelor, could be shortstop or 1st baseman

Hassan: not 3rd baseman, could be battery or infielder, not catcher, not 2nd baseman, could be either 1st baseman, shortstop, pitcher

Jason: not 3rd baseman, not catcher, bachelor. '''Jason is the 2nd baseman.'''

Matthieu: not centre field, shortest in height of the team

Pascal: not pitcher, could be outfielder, not shortstop, not catcher, not 2nd baseman, 3rd baseman, not right fielder, not centre fielder, bachelor. '''Pascal is the left outfielder.'''

Sung: married, not right fielder, not centre fielder, not left fielder, not 2nd baseman.

''Important notes'':

There are 5 bachelor, 4 married men, the catcher, the 3rd baseman and the pitcher are married.

'''Problem 4'''

'''Problem 5'''

== Homework 3 - due Wednesday October 13 ==

Thanks Shamilla,

That sounds like a great idea.

Arabella.

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful 
[[User:ShamillaBirring|ShamillaBirring]]

All you ever wanted to know about adding images can be found [http://wiki.ubc.ca/Help:Adding_Media#Inserting_an_Image_onto_a_UBC_Wiki_Page here]. 
--[[User:DavidKohler|DavidKohler]]

'''25 PROBLEMS'''

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

"QUESTION 10" -
Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.

Being a man or a woman is independent of preference for chocolate.

i) No, it does not follow.
ii) No, it does not follow.

"QUESTION 11" -
A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?

The situation is not possible. The woman's daughter and son must be either the best or worst players. They are of opposite sex and they have the same age, since the woman is older than both of them (she is their mother) . The woman's older brother is the oldest player.

"QUESTION 12" -
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

For example, the Bronx train arrives at 12.00, 12.10. 12.20...and the Brooklyn train arrives at 12.09, 12.19, 12.29...

"QUESTION 13" -
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

Considering the time between chimes: it takes 45/4 = 11.25 seconds to strike 10.00.

"QUESTION 14" -
One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

i) There are six ways that two of the four babies can be correctly tagged.
ii) There are no ways in which three of the four babies correctly tagged.

"QUESTION 15" -
Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

Alex is right, I shouldn't accept his bet. The deck of cards is made up of an equal number of black and red cards (respectively 1/2 of the total number of cards).

'''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?'''

Looking at the question, we are the given positions of the runners as followed:

Sven: Middle place of all runners

Dan: 10th place

Lars: 16th place

From the details of the questions, we can first of all try to determine a range for which place Sven came in as we only know he is in the middle amongst all runners. First we know that Dan, who was slower than Sven came in 10th place thus we can assume that Sven came in places ranging from 1st to 9th. Taking into account that Sven is in the middle we can say that the maximum number of runners is 18 as the Sven can only be placed up to the 9th at max. Next, as we are told that Sven is placed exactly in the middle therefore we can deduce that the number of runners is an odd number, as an even number cannot produce a runner exactly in the middle. Lastly we take into account the placement of Lars who came in 16th, as mentioned earlier we deduced that the maximum number of runners is 18 therefore shortening the range to 16-18.

In conclusion, the only possible number that fits with all the criteria is 17; therefore there were 17 runners in the race.

'''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?'''

'''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'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's children? Explain.
'''
We are told that Paula has three children and that the product of their ages is 36, therefore our first step would be to find the factors of 36 in sets of 3. This gives us the following sets:

• 1,1,36

• 1,2,18

• 1,3,12

• 1,4,9

• 1,6,6

• 2,2,9

• 2,3,6

• 3, 3,4

Next we are told that the sum of their ages is the same as a date, therefore we know the sum cannot exceed 31 as that is the maximum number of days in a month. By calculating the sums of each set of factors we are given the following:

• 1+ 1+36 = 38

• 1+2+18= 21

• 1+3+12 = 16

• 1+4+9 = 14

• 1+6+6 = 13

• 2+2+9= 13

• 2+3+4 = 11

• 3+3+4= 10

From this we see that we can eliminate the set 1,1,36 as its sum is greater than 31.
Next we are told that the information is still not enough, the only reason for this may be that there are sums of the same number thus making Paul unable to determine which the correct set is. Looking at the sums we can see that the number 13 is duplicated making 1,6,6 and 2,2,9 possible answers.

Lastly, we are told that the oldest child has red hair, the clue here is that there is a eldest sibling, ruling out the set 6,6,1 as there is not one single elder sibling. Therefore, this we can determine the ages of Paula’s children to be 9,2 and 2.

'''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?
'''
From the details of the question we are given the following conditions:

1.) Two candles are of equal length

2.) One candle took 6hr to burn out (we can call this Candle A)

3.) One candle took 3hr to burn out (we can call this Canlde B)

Firstly to make this question more clearly, we can give the candles an actual length giving that both candles are of equal length. Let’s assume that the candle length is 60cm.Taking into account the time it takes each candle to burn out we can come up rate of how fast each candle burns by dividing the length by the time, thus giving us the following information.

Candle A: 60cm/6hr = 10cm/hr

Candle B: 60cm/3hr = 20cm/hr

Second, we can shorten our time to under 3hrs as we know that by 3hrs, Candle B would’ve already burned out leaving only one candle left. Now we can use a trial and error method, by trying the times of each hour. After the first hour, the length of Candle A would be 50cm while the length of Candle B would be 40cm; one is not exactly twice as long as the other. Next after the second hour, the length of Candle A is 40cm while the length of Candle B is 20cm; one candle is exactly twice as long as the other.

Therefore after 2 hours, one candle will be exactly twice as long as the other candle.

'''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.'''

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-13T12:55:08Z

ShamillaBirring:

Group members:
* Shamilla Birring | shamilla_birring@hotmail.com
* Matt Johnson
* Arabella Olomide aco2@sfu.ca
* Gaia Silvestri
* Christopher Wong

Thanks Shamilla,

That sounds like a great idea.

Arabella.

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful 
[[User:ShamillaBirring|ShamillaBirring]]

All you ever wanted to know about adding images can be found [http://wiki.ubc.ca/Help:Adding_Media#Inserting_an_Image_onto_a_UBC_Wiki_Page here]. 
--[[User:DavidKohler|DavidKohler]]

'''25 PROBLEMS'''

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

"QUESTION 10" -
Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.

Being a man or a woman is independent of preference for chocolate.

i) No, it does not follow.
ii) No, it does not follow.

"QUESTION 11" -
A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?

The situation is not possible. The woman's daughter and son must be either the best or worst players. They are of opposite sex and they have the same age, since the woman is older than both of them (she is their mother) . The woman's older brother is the oldest player.

"QUESTION 12" -
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

For example, the Bronx train arrives at 12.00, 12.10. 12.20...and the Brooklyn train arrives at 12.09, 12.19, 12.29...

"QUESTION 13" -
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

Considering the time between chimes: it takes 45/4 = 11.25 seconds to strike 10.00.

"QUESTION 14" -
One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

i) There are six ways that two of the four babies can be correctly tagged.
ii) There are no ways in which three of the four babies correctly tagged.

"QUESTION 15" -
Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

Alex is right, I shouldn't accept his bet. The deck of cards is made up of an equal number of black and red cards (respectively 1/2 of the total number of cards).

'''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?'''

Looking at the question, we are the given positions of the runners as followed:

Sven: Middle place of all runners

Dan: 10th place

Lars: 16th place

From the details of the questions, we can first of all try to determine a range for which place Sven came in as we only know he is in the middle amongst all runners. First we know that Dan, who was slower than Sven came in 10th place thus we can assume that Sven came in places ranging from 1st to 9th. Taking into account that Sven is in the middle we can say that the maximum number of runners is 18 as the Sven can only be placed up to the 9th at max. Next, as we are told that Sven is placed exactly in the middle therefore we can deduce that the number of runners is an odd number, as an even number cannot produce a runner exactly in the middle. Lastly we take into account the placement of Lars who came in 16th, as mentioned earlier we deduced that the maximum number of runners is 18 therefore shortening the range to 16-18.

In conclusion, the only possible number that fits with all the criteria is 17; therefore there were 17 runners in the race.

'''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?'''

'''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'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's children? Explain.
'''
We are told that Paula has three children and that the product of their ages is 36, therefore our first step would be to find the factors of 36 in sets of 3. This gives us the following sets:

• 1,1,36

• 1,2,18

• 1,3,12

• 1,4,9

• 1,6,6

• 2,2,9

• 2,3,6

• 3, 3,4

Next we are told that the sum of their ages is the same as a date, therefore we know the sum cannot exceed 31 as that is the maximum number of days in a month. By calculating the sums of each set of factors we are given the following:

• 1+ 1+36 = 38

• 1+2+18= 21

• 1+3+12 = 16

• 1+4+9 = 14

• 1+6+6 = 13

• 2+2+9= 13

• 2+3+4 = 11

• 3+3+4= 10

From this we see that we can eliminate the set 1,1,36 as its sum is greater than 31.
Next we are told that the information is still not enough, the only reason for this may be that there are sums of the same number thus making Paul unable to determine which the correct set is. Looking at the sums we can see that the number 13 is duplicated making 1,6,6 and 2,2,9 possible answers.

Lastly, we are told that the oldest child has red hair, the clue here is that there is a eldest sibling, ruling out the set 6,6,1 as there is not one single elder sibling. Therefore, this we can determine the ages of Paula’s children to be 9,2 and 2.

'''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?
'''
From the details of the question we are given the following conditions:

1.) Two candles are of equal length

2.) One candle took 6hr to burn out (we can call this Candle A)

3.) One candle took 3hr to burn out (we can call this Canlde B)

Firstly to make this question more clearly, we can give the candles an actual length giving that both candles are of equal length. Let’s assume that the candle length is 60cm.Taking into account the time it takes each candle to burn out we can come up rate of how fast each candle burns by dividing the length by the time, thus giving us the following information.

Candle A: 60cm/6hr = 10cm/hr

Candle B: 60cm/3hr = 20cm/hr

Second, we can shorten our time to under 3hrs as we know that by 3hrs, Candle B would’ve already burned out leaving only one candle left. Now we can use a trial and error method, by trying the times of each hour. After the first hour, the length of Candle A would be 50cm while the length of Candle B would be 40cm; one is not exactly twice as long as the other. Next after the second hour, the length of Candle A is 40cm while the length of Candle B is 20cm; one candle is exactly twice as long as the other.

Therefore after 2 hours, one candle will be exactly twice as long as the other candle.

'''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.'''

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T20:05:18Z

ShamillaBirring:

Group members:
* Shamilla Birring | shamilla_birring@hotmail.com
* Matt Johnson
* Arabella Olomide
* Gaia Silvestri
* Christopher Wong

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful

Shamilla

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T20:03:58Z

ShamillaBirring:

Group members:
* Shamilla Birring - shamilla_birring@hotmail.com
* Matt Johnson
* Arabella Olomide
* Gaia Silvestri
* Christopher Wong

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful

Shamilla

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T19:45:58Z

ShamillaBirring:

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T19:44:59Z

ShamillaBirring:

Group members:
* Shamilla Birring - shamilla_birring@hotmail.com
* Matt Johnson
* Arabella Olomide
* Gaia Silvestri
* Christopher Wong

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful

Shamilla

'''QUESTION 1'''
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2'''
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3'''
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4'''
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T19:43:32Z

ShamillaBirring:

Group members:
* Shamilla Birring - shamilla_birring@hotmail.com
* Matt Johnson
* Arabella Olomide
* Gaia Silvestri
* Christopher Wong

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful

Shamilla

'''Question 1'''
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''Question 2'''
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''Question 3'''
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''Question 4'''
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T19:29:03Z

ShamillaBirring:

Group members:
* Shamilla Birring - shamilla_birring@hotmail.com
* Matt Johnson
* Arabella Olomide
* Gaia Silvestri
* Christopher Wong

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful

Shamilla

'''Question 1'''
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''Question 2'''
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''Question 3'''
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T18:58:05Z

ShamillaBirring:

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T18:53:06Z

ShamillaBirring:

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T18:48:39Z

ShamillaBirring:

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T18:48:13Z

ShamillaBirring:

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T18:44:52Z

ShamillaBirring:

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T17:38:36Z

ShamillaBirring:

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-12T17:38:00Z

ShamillaBirring:

Group members:
* Shamilla Birring - shamilla_birring@hotmail.com
* Matt Johnson
* Arabella Olomide
* Gaia Silvestri
* Christopher Wong

[[User:ArabellaCynthiaOlomide|ArabellaCynthiaOlomide]]

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

User:ShamillaBirring

2010-09-19T19:57:45Z

ShamillaBirring: Pythagorean theorum

Hi, my name is Shamilla and I am a first year Arts student here at UBC.

Pythagorean theorem:
The pythagorean theorem expresses the relationship between the three sides of a right triangle, it is named after the Greek mathematician Pythagoras. The relationship found is that the area of the two squares of the lines adjacent to the right angle will be equal to the area of the square opposite from the right angle, also referred to as the hypotenuse. This relationship can be represented by the formula <math>a^2 + b^2 = c^2</math>. In this equation a and b represent the length of the lines adjacent to the right angle, and c represents the length of the hypotenuse. Therefore, using this formula one can find any of the lengths of either sides of a right triangle. There are several proofs that validate the formula which include algebraic proofs, proofs using similar triangles and proofs using differentials. The theorem is utilized in several situations such as in determining the distance between two sites, as well as in architecture and research.
[[http://en.wikipedia.org/wiki/Pythagorean_theorem]]