UBC Wiki - User contributions [en]

User:VictoriaBass

2013-11-12T14:12:28Z

VictoriaBass: Blanked the page

User:VictoriaBass

2013-11-12T14:02:30Z

VictoriaBass:

Yay! The wiki! I think Math is important because you never know when you're going to find yourself at a party, surrounded by a group of mathematicians, who are deciding if you (the outsider) are worthy of dating their friend and colleague and are asked, "Well, tell me this: what's the quadratic formula?" and you cross your fingers and say "-b + or - the square root of b squared - 4ac over 2 a??????" and breathe a sigh of relief when you realize you were miraculously correct ( and subsequently feel like a math genius when they are impressed that you pulled that out of the recesses of your vast and impressive mind and decide to accept you as a mathematically challenged version of one of their own. )

MINI ESSAY!!!!

If you asked a philosopher what the first thing to come into their mind is when you say, “Rene Descartes” they would most likely answer “Cogito ergo sum” (or, for those of you playing along with the English version, “I think, therefore, I am.”) However, if you were to ask a mathematician the same question their answer would most likely be something to do with analytic geometry. Analytic geometry is a broad concept that encompasses many areas of mathematics. It is most fundamentally divided into two parts: a “classical” (or, more accurately, it’s original consideration) and a modern arena. The aspect of it which we will focus on here, and which is attributed to Descartes, is not the modern counterpart, but the original development/appearance of analytic geometry. In this light analytic geometry is also known as coordinate geometry or Cartesian geometry.

There are two main elements to analytic geometry. First, it is a study of geometry where points are represented in a co-ordinate system (which I will address momentarily). Second, the geometrical problems are approached or solved algebraically. An excellent description of analytic geometry is that “it is concerned with defining geometrical shapes in a numerical way and extracting numerical information from that representation.”1 The attribution to Descartes is that he was the person to combine concepts of Euclidian geometry (which was the geometry of the time) with algebra, thus developing analytic geometry.

The Cartesian co-ordinate system (named after Descartes, of course) that is used in classical analytic geometry is a method of representing specific points in two or three dimensional space. It consists of perpendicular lines (each referred to as an axis) that are evenly divided with consistent units of measurement. Points are labeled with either two or three (depending on how many dimensions are being specified) numbers which correspond to a unique location on each axis. The resulting point being expressed is the meeting of the points on each axis on the Cartesian plane.

1. http://en.wikipedia.org/wiki/Analytic_geometry

Calculus Mini Essay

I have two thoughts on how calculus relates to areas that I am interested in. In general I have noticed that doing more math has really helped my problem solving abilities in other areas. For example, when doing long calculations in Chemistry I find myself thinking about breaking it into smaller steps and what to expect at the end. And even in Biology, when faced with a problem that I don't know how to approach or am not familiar with I try to think "What do I know that could be helpful? What does this mean? How can I approach it in another way?" Though that's not the most direct answer to "How does calculus relate to an area of your interest?" I feel like it's valid because I'm interested in being able to solve problems better and be a more rigorous logician and calculus definitely relates to that.

While investigating how calculus relates to my field of study I've discovered something that I think is pretty cool. Apparently you can model the growth of tumors using a logistic growth model. Then, armed with that information you can figure out chemotherapy approaches to suit it. That's super cool! I think it's really neat that you can turn a tumor into a math function. I guess I shouldn't really be surprised because tumors are just cells and obviously we can model cell growth. But still, it's pretty awesome that we can look at these crazy clumps of rogue cells that have hijacked their own mechanisms and make a model of how they work. You can read a little bit more about it here:
http://en.wikipedia.org/wiki/Logistic_function#In_medicine:_modeling_of_growth_of_tumors

Lately though I've been really interested in why chemical reactions are endothermic or exothermic. Apparently this is not something that can be readily determined just by the structure of the molecules; one must observe reactions and then can predict how others will proceed. I think though, that with all we know about physics and energy, there must be some way to take the math of the chemical interaction and use it to predict whether a reaction would be endothermic or exothermic. I don't know if it would be related to the physics on an atomic level, or on a molecular level, I don't really know since I'm only in first year chemistry. But I'm sure that as I continue researching this problem I'm going to run into calculus all over the place so it will probably be handy for that, too.

User:VictoriaBass

2011-01-28T09:58:44Z

VictoriaBass:

Yay! The wiki! I'm in MATH 110 because I'm restarting my life via a second Bachelor's degree. I used to do other, unrelated things but they made me unhappy so I decided to a) stop doing them and b) start doing something I like (which is how I found myself back in school!) I think Math is important because you never know when you're going to find yourself at a party, surrounded by a group of mathematicians, who are deciding if you (the outsider) are worthy of dating their friend and colleague and are asked, "Well, tell me this: what's the quadratic formula?" and you cross your fingers and say "-b + or - the square root of b squared - 4ac over 2 a??????" and breathe a sigh of relief when you realize you were miraculously correct ( and subsequently feel like a math genius when they are impressed that you pulled that out of the recesses of your vast and impressive mind and decide to accept you as a mathematically challenged version of one of their own. )

MINI ESSAY!!!!

If you asked a philosopher what the first thing to come into their mind is when you say, “Rene Descartes” they would most likely answer “Cogito ergo sum” (or, for those of you playing along with the English version, “I think, therefore, I am.”) However, if you were to ask a mathematician the same question their answer would most likely be something to do with analytic geometry. Analytic geometry is a broad concept that encompasses many areas of mathematics. It is most fundamentally divided into two parts: a “classical” (or, more accurately, it’s original consideration) and a modern arena. The aspect of it which we will focus on here, and which is attributed to Descartes, is not the modern counterpart, but the original development/appearance of analytic geometry. In this light analytic geometry is also known as coordinate geometry or Cartesian geometry.

There are two main elements to analytic geometry. First, it is a study of geometry where points are represented in a co-ordinate system (which I will address momentarily). Second, the geometrical problems are approached or solved algebraically. An excellent description of analytic geometry is that “it is concerned with defining geometrical shapes in a numerical way and extracting numerical information from that representation.”1 The attribution to Descartes is that he was the person to combine concepts of Euclidian geometry (which was the geometry of the time) with algebra, thus developing analytic geometry.

The Cartesian co-ordinate system (named after Descartes, of course) that is used in classical analytic geometry is a method of representing specific points in two or three dimensional space. It consists of perpendicular lines (each referred to as an axis) that are evenly divided with consistent units of measurement. Points are labeled with either two or three (depending on how many dimensions are being specified) numbers which correspond to a unique location on each axis. The resulting point being expressed is the meeting of the points on each axis on the Cartesian plane.

1. http://en.wikipedia.org/wiki/Analytic_geometry

Calculus Mini Essay

I have two thoughts on how calculus relates to areas that I am interested in. In general I have noticed that doing more math has really helped my problem solving abilities in other areas. For example, when doing long calculations in Chemistry I find myself thinking about breaking it into smaller steps and what to expect at the end. And even in Biology, when faced with a problem that I don't know how to approach or am not familiar with I try to think "What do I know that could be helpful? What does this mean? How can I approach it in another way?" Though that's not the most direct answer to "How does calculus relate to an area of your interest?" I feel like it's valid because I'm interested in being able to solve problems better and be a more rigorous logician and calculus definitely relates to that.

While investigating how calculus relates to my field of study I've discovered something that I think is pretty cool. Apparently you can model the growth of tumors using a logistic growth model. Then, armed with that information you can figure out chemotherapy approaches to suit it. That's super cool! I think it's really neat that you can turn a tumor into a math function. I guess I shouldn't really be surprised because tumors are just cells and obviously we can model cell growth. But still, it's pretty awesome that we can look at these crazy clumps of rogue cells that have hijacked their own mechanisms and make a model of how they work. You can read a little bit more about it here:
http://en.wikipedia.org/wiki/Logistic_function#In_medicine:_modeling_of_growth_of_tumors

Lately though I've been really interested in why chemical reactions are endothermic or exothermic. Apparently this is not something that can be readily determined just by the structure of the molecules; one must observe reactions and then can predict how others will proceed. I think though, that with all we know about physics and energy, there must be some way to take the math of the chemical interaction and use it to predict whether a reaction would be endothermic or exothermic. I don't know if it would be related to the physics on an atomic level, or on a molecular level, I don't really know since I'm only in first year chemistry. But I'm sure that as I continue researching this problem I'm going to run into calculus all over the place so it will probably be handy for that, too.

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

2011-01-28T09:11:13Z

VictoriaBass:

{{
Infobox MATH110 Teams
| team name = Glarus
| member 1 = [[User:AndreaMameri|Andrea Mameri]]
| member 2 = Catherine Chen
| member 3 = Supreet Saran
| member 4 = Victoria Bass
}}

Hi.... My name is Andrea Mameri

phone number: 604-716-9520

email address: andreamameri@aim.com

skype name: andrea.mameri

You can find me on facebook as: Andrea Mameri

flabbergasted

Supreet Saran

lethargic

skype: supreetsaran

Hi everyone. I'm Victoria.

My email is bassvm@interchange.ubc.ca Yay!

BANANA!!!!!!

Hi! I'm Catherine.

fb and email: catherine chen (catherine1012_ling@hotmail.com)

Water Polo.

Homework #12

With the function P(t)= 1/1+e^-t we can modify it in a couple of ways. If we consider the function as P(t)= K/1+e^-t we can change the height of the horizontal asymptote on the right by setting K=whatever we want the
horizontal asymptote on the right to be. The logic behind this involves considering the limit at +infinity. Horizontal asymptotes are limits at infinity. If we evaluate the limit at infinity for this function we find that it equals K.

Also, if we want to change the y intercept of the function then we can adjust A if we define the function as P(t)= 1/1+e^(-t+A). At the y intercept t=0 so e^-t=1 so we get 1/1+1 or 1/2. This means that without any adjustment the y intercept will always be exactly half of the horizontal asymptote on the right. Since t is our x co-ordinate we can shift the y intercept to a higher value by adding on to t.

We can also adjust the slope of the curved part of the function by introducing another variable r so that we have P(t)= 1/1+e^(-rt+A).

We can now use this information to model something useful, like a population. We decided to model the population in Vancouver. For the sake of argument lets say that the carrying capacity of Vancouver is 1,000,000 people (a safe assumption since the biological definition of carrying capacity refers to the idea of the maximum population that an environment can support without habitat degradation and many people would argue that we're clearly already degrading the habitat and we're not yet at 1,000,000. If anything 1,000,000 would be well above carrying capacity but I digress...ANYWAY) We want to find out in what year the population will reach 90% of its carrying capacity (in this case 900,000). According to trusty Wikipedia, the population of Vancouver was 578,041 people as of the 2006 census. Let's set that data point as our t=0. So our "Year 0" will be 2006 and so the y intercept will be 578,041 (the population at that time.) To do this we need to figure out how much we need to shift our graph to the right so that the y intercept will be 578,041 and not halfway (500,000). To do this we have to find t when P(t)=578,041:

578,041= 10^6/1+e^-t

578,041 (1+e^-t)=10^6

1+e^-t = 10^6/578,041

e^-t=(10^6/578,041)-1

-t= ln [(10^6/578,041)-1]

-t=ln1.7-1

-t=0.7
-t=-0.3
t=0.3

So we need to shift our x coordinate 0.3 to the right for 578,041 to be our y intercept.

Now let's adjust the slope of the curved part by incorporating a specific rate of increase. According to worldometers.info the global population is increasing at a rate of 1.15%/year. Let's assume that since Vancouver is a smaller population to be drawing the percentage from that it is increasing slightly faster. Let's say that in 2006 the population was increasing at a rate of 2%/year. Since the population of Vancouver at that time was 578,041 the population would be increasing by 11,560.82 people that year (I know you can't have 0.82 of a person but let's just be accurate with our numbers and not round off at this point.) With our last variable r we have to determine what r is when the rate of change is 2%. To do this we have to a) find the derivative of the function so that we can model the rate of change b) set the derivative equal to 11,560.82 and solve for r (assuming t=0 because that's where P(t)=578,041)

P'(0)= [-(-re^(-rt+0.3)(10^6)]/(1+e^-rt+0.3)^2

11560=[-(-re^(-rt+0.3)(10^6)]/(1+e^-rt+0.3)^2

[(11560)(1+e^-rt+0.3)^2]/10^6= re^(-rt+0.3)

11560/10^6= (re^(-rt+0.3)/(1+e^-rt+0.3)^2

11560/10^6= [(r)(e^0.3)]/(1+e^0.3)^2

[(11560)(1+e^0.3)^2]/[(10^6)(e^0.3)]=r

0.047=r

So now our model is:

P(t)= 10^6/(1+e^(-0.047t+0.3))

We can determine when the population will reach 900,000 by evaluating P(900000) and solving for t.

900000=10^6/(1+e^(-0.047t+0.3))

We find that t=55.3 years
Since our t=0 at 2006 we add 55+2006 to determine that the population of Vancouver will reach 90% of its carrying capacity during 2061.

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

2011-01-28T09:09:59Z

VictoriaBass:

{{
Infobox MATH110 Teams
| team name = Glarus
| member 1 = [[User:AndreaMameri|Andrea Mameri]]
| member 2 = Catherine Chen
| member 3 = Supreet Saran
| member 4 = Victoria Bass
}}

Hi.... My name is Andrea Mameri

phone number: 604-716-9520

email address: andreamameri@aim.com

skype name: andrea.mameri

You can find me on facebook as: Andrea Mameri

flabbergasted

Supreet Saran

lethargic

skype: supreetsaran

Hi everyone. I'm Victoria.

My email is bassvm@interchange.ubc.ca Yay!

BANANA!!!!!!

Hi! I'm Catherine.

fb and email: catherine chen (catherine1012_ling@hotmail.com)

Water Polo.

Homework #12

With the function P(t)= 1/1+e^-t we can modify it in a couple of ways. If we consider the function as P(t)= K/1+e^-t we can change the height of the horizontal asymptote on the right by setting K=whatever we want the
horizontal asymptote on the right to be. The logic behind this involves considering the limit at +infinity. Horizontal asymptotes are limits at infinity. If we evaluate the limit at infinity for this function we find that it equals K.

Also, if we want to change the y intercept of the function then we can adjust A if we define the function as P(t)= 1/1+e^(-t+A). At the y intercept t=0 so e^-t=1 so we get 1/1+1 or 1/2. This means that without any adjustment the y intercept will always be exactly half of the horizontal asymptote on the right. Since t is our x co-ordinate we can shift the y intercept to a higher value by adding on to t.

We can also adjust the slope of the curved part of the function by introducing another variable r so that we have P(t)= 1/1+e^(-rt+A).

We can now use this information to model something useful, like a population. We decided to model the population in Vancouver. For the sake of argument lets say that the carrying capacity of Vancouver is 1,000,000 people (a safe assumption since the biological definition of carrying capacity refers to the idea of the maximum population that an environment can support without habitat degradation and many people would argue that we're clearly already degrading the habitat and we're not yet at 1,000,000. If anything 1,000,000 would be well above carrying capacity but I digress...ANYWAY) We want to find out in what year the population will reach 90% of its carrying capacity (in this case 900,000). According to trusty Wikipedia, the population of Vancouver was 578,041 people as of the 2006 census. Let's set that data point as our t=0. So our "Year 0" will be 2006 and so the y intercept will be 578,041 (the population at that time.) To do this we need to figure out how much we need to shift our graph to the right so that the y intercept will be 578,041 and not halfway (500,000). To do this we have to find t when P(t)=578,041:

578,041= 10^6/1+e^-t
578,041 (1+e^-t)=10^6
1+e^-t = 10^6/578,041
e^-t=(10^6/578,041)-1
-t= ln [(10^6/578,041)-1]
-t=ln1.7-1
-t=0.7
-t=-0.3
t=0.3

So we need to shift our x coordinate 0.3 to the right for 578,041 to be our y intercept.

Now let's adjust the slope of the curved part by incorporating a specific rate of increase. According to worldometers.info the global population is increasing at a rate of 1.15%/year. Let's assume that since Vancouver is a smaller population to be drawing the percentage from that it is increasing slightly faster. Let's say that in 2006 the population was increasing at a rate of 2%/year. Since the population of Vancouver at that time was 578,041 the population would be increasing by 11,560.82 people that year (I know you can't have 0.82 of a person but let's just be accurate with our numbers and not round off at this point.) With our last variable r we have to determine what r is when the rate of change is 2%. To do this we have to a) find the derivative of the function so that we can model the rate of change b) set the derivative equal to 11,560.82 and solve for r (assuming t=0 because that's where P(t)=578,041)

P'(0)= [-(-re^(-rt+0.3)(10^6)]/(1+e^-rt+0.3)^2
11560=[-(-re^(-rt+0.3)(10^6)]/(1+e^-rt+0.3)^2
[(11560)(1+e^-rt+0.3)^2]/10^6= re^(-rt+0.3)
11560/10^6= (re^(-rt+0.3)/(1+e^-rt+0.3)^2
11560/10^6= [(r)(e^0.3)]/(1+e^0.3)^2
[(11560)(1+e^0.3)^2]/[(10^6)(e^0.3)]=r
0.047=r

So now our model is:

P(t)= 10^6/(1+e^(-0.047t+0.3))

We can determine when the population will reach 900,000 by evaluating P(900000) and solving for t.

900000=10^6/(1+e^(-0.047t+0.3))

We find that t=55.3 years
Since our t=0 at 2006 we add 55+2006 to determine that the population of Vancouver will reach 90% of its carrying capacity during 2061.

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

2011-01-28T04:49:15Z

VictoriaBass:

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

2011-01-19T13:50:57Z

VictoriaBass:

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

2011-01-19T13:48:45Z

VictoriaBass:

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

2011-01-14T21:15:56Z

VictoriaBass:

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

2011-01-14T21:15:37Z

VictoriaBass:

Course:MATH110/Archive/2010-2011/003/Groups/Group 13/Homework 6

2010-11-12T14:49:20Z

VictoriaBass:

==Basic Skills Suggestion==

A while ago the prof suggested that we create a section in the wiki regarding the basic skills where each person would offer something that they know or have a problem with and people would gradually add to it. The idea is to accumulate these basic skills by ourselves through a combined class effort.
Anyways since a lot of people seemed to have a problem with writing out things on the wiki or spending that much time in general on the wiki, our group thought of making a youtube channel.

Pros:
- Most people will have a camera/video editing system built in into their computer
- OR most people own a digital camera that can take 5 min clips
- No written explanation required
- Verbal explanation is easier/requires less work
- Visual and verbal explanation go hand in hand
- Easier to show examples/visuals through video
- Allows for more examples to be shown, and someone to talk through it step by step

 We propose to make a video study guide for composition of functions that will include a short tutorial video for each of the following topics:

1. A short explanation of the theory behind composition of functions and why it works the way it does (including examples)
2. Step by step process for approaching different types of problems including some helpful tips.
3. Examples worked out from start to finish
4. Suggestions for useful practice problems to work on

Course talk:MATH110/003/Lectures

2010-10-26T15:12:56Z

VictoriaBass:

I'm not sure if this is the right place to write this or not but since it's a question to do with the reading it seems like a reasonable try. It seems like we're moving away from readings in the text (which is good, I think, because I find it to be massively confusing.) But I was wondering what you would suggest as the best thing to do for practice problems? Would you say a) just sort of go through the book and do problems that appear to be related to what we're doing in class? or b) would you be able to suggest problems for us to try in the book even though we're not relying on it so much for reading material? I know that you suggesting problems might cause more work for you so I don't want to ask for it if it is a huge pain. However, we have limited time in class and so I find it helpful to a) work on the problems you normally suggest with the readings and b) to know what types of problems you are thinking I should be able to do so that we are on the same page. Anyway, I'm trying to say- if it's not too much trouble would you mind suggesting practice problems from the text to go along with the reading assignments? If you do I promise to do them.
Thanks-
Victoria

:Hey Victoria,

:This a very fair and actually good question. In my evil plan to reduce the use of the textbook notes, I kinda forgot about that. I'll post suggested problems from the textbook (if there's one good thing about the textbook, it's its massive collection of exercices) so that you know on what you can practice all that stuff.

:I'll try to get this done sometime tomorrow (Tuesday Oct. 26) and will send a mass email once its done.

:Thanks for the feedback, -- [[User:DavidKohler|DavidKohler]]

Thanks! It's greatly appreciated. --[[User:VictoriaBass|VictoriaBass]]

Course talk:MATH110/003/Lectures

2010-10-26T04:54:14Z

VictoriaBass:

j'ai mis une faute d'orthographe au mot "pente" dans la description du devoir de lecture pour lundi

A la fin de chaque semaine, ajouter un partie ''review'' pour le lundi, histoire qu'ils retournent bosser sur une liste de problèmes. On pourrait même leur annoncer que l'un des review problems sera dans le warm-up du lundi.

What does that mean?

[[User:BellaTory|BellaTory]]

There are many path to this answer, I think you could find one yourself, but I'll get back to this maybe later, now is not the time for me unfortunately. [[User:DavidKohler|DavidKohler]] 00:32, 2 October 2010 (UTC)

I'm not sure if this is the right place to write this or not but since it's a question to do with the reading it seems like a reasonable try. It seems like we're moving away from readings in the text (which is good, I think, because I find it to be massively confusing.) But I was wondering what you would suggest as the best thing to do for practice problems? Would you say a) just sort of go through the book and do problems that appear to be related to what we're doing in class? or b) would you be able to suggest problems for us to try in the book even though we're not relying on it so much for reading material? I know that you suggesting problems might cause more work for you so I don't want to ask for it if it is a huge pain. However, we have limited time in class and so I find it helpful to a) work on the problems you normally suggest with the readings and b) to know what types of problems you are thinking I should be able to do so that we are on the same page. Anyway, I'm trying to say- if it's not too much trouble would you mind suggesting practice problems from the text to go along with the reading assignments? If you do I promise to do them.
Thanks-
Victoria

Course:MATH110/Archive/2010-2011/003/Math Forum/Lecture18

2010-10-22T04:49:41Z

VictoriaBass: Created page with 'Has anyone got any suggestions for #4a)? I've come up with so many different answers and none of them are right. Any helpful hints out there? (and is anyone else totally frustrat…'

Has anyone got any suggestions for #4a)? I've come up with so many different answers and none of them are right. Any helpful hints out there? (and is anyone else totally frustrated and feeling completely lost on these?) ANY advice on any of the asymptotes is helpful. I felt like the Wiki notes were helpful for the concept but not for the computations.
Thanks- Victoria

User:VictoriaBass/Midterm musings

2010-10-20T03:06:31Z

VictoriaBass: Created page with 'dear calculus. as i approach my midterm i find i have reached my limit. sadly, your function in my life remains continuous. you derive me crazy. i eagerly await a point at which …'

dear calculus. as i approach my midterm i find i have reached my limit. sadly, your function in my life remains continuous. you derive me crazy. i eagerly await a point at which you do not exist.
love, victoria

Course:MATH110/Archive/2010-2011/003/Groups/Group 13/Homework 4

2010-10-20T01:54:34Z

VictoriaBass:

=Homework #4=

===Question 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. 

The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given. 

Tosh ---> Cat (Jun) 
♀Bianca ---> Frog (Suzan) 
♀Jaela ---> Parrot (No name) 
♂Jun ---> Snake (No name) 
♀Suzan ---> Turtle (No name) 

With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement "Bianca owns a frog that '''she''' loves" and we know that Jaela is also female from the statement "Jaela owns a parrot which keeps calling '''her''' 'darling, darling'". We also know that Jun is male from the statement "Jun owns a snake, don't mess with '''him'''", with him referring to the owner and not the pet, otherwise it would be saying don't mess with '''it'''. 

We can also assume that Suzan's pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets. 

Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan's pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle's name needs to be Suzan which is all very confusing and counter-intuitive. 

♀Jaela ---> Parrot (Tosh) 

This piece of information also allows us to name Suzan's pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: 

♀Suzan ---> Turtle (Jaela) 

That only leaves the last part of the problem, which would be Jun's pet: 

♂Jun ---> Snake (Bianca) 

Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan's '''mother''' cannot possibly be male unless the label "mother" is semantically misleading (i.e. nickname, etc.). 

There are then two possible solutions after this error. The first is that the statement "Jun owns a snake, don't mess with '''him'''" has the pronoun 'him' refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca). This solution ends up with Jun being Suzan's mother and the pet Bianca being her pet. 

The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan's pet to be Tosh and lastly having Jaela's pet to be Bianca (and thus having Jaela as Suzan's mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution, and a more likely one then the previously aforementioned. In this case, it is not realistically possible to know who Suzan's mother is with the given information.
 

===Question 2===

'''Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?'''

First, simplify the information:

Bohao, Stewart, Dylan, Tim and Chan

2 are left-handed, 3 are right-handed

Bohnao and Dylan = Same Hand

Tim and Chan = Different Hand

2 are over 2m, 3 are under 2m

Steward and Chan = Same Height

Tim and Dylan = Different Height

Need to find: player that is over 2 meters tall and left-handed

Now draw a diagram to help solve the first part of the problem (Left/Right Handed).
[[Image:Group Project 4 Question 2.JPG|center]]

Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).

[[Image:Group Project 4 Question 2p2.JPG|center]]

Now, using these two diagrams find a player that is over 2 meters tall and left-handed

By process of elimination we know that it can't be:

'''Bohao''' (right-handed)

'''Dylan''' (right-handed)

'''Stewart''' (Under 2 meters)

'''Chan''' (Under 2 meters)

Therefore it COULD only be '''TIM'''

Therefore the player that is over 2 meters tall and left-handed is '''TIM'''

===Question 3===

We are trying to determine which position everyone plays. I started by making a chart of all the players and all the positions so we could cross off possibilities as we went along. Then we started using the information we are given to figure out which position various people CANNOT play so as to start to figure out positions by process of elimination.

We are told the following (and our interpretation is written in parentheses, which we then marked on the chart accordingly)

1. Adam does not like the catcher, (so Adam cannot be the catcher)

2. Ed's sister is engaged to the second baseman, (so Ed cannot be the second baseman)

3. The centre fielder is taller than the right fielder,

4. Hassan and the third baseman live in the same building (so Hassan cannot be the third baseman)

5. Pascal and Charles each won $20 from the pitcher at a poker game (so neither Pascal nor Charles can be the pitcher)

6. Ed and the outfielders play cards during their free time (Ed cannot play outfield)

7. The pitcher's wife is the third baseman's sister,

8. All the battery and infield except Charles, Hassan and Adam are shorter than Sung (Charles, Hassan, and Adam play infield or battery and therefore do not play any of the outfield positions)

9. Pascal, Adam and the shortstop lost $100 each at the race track (neither Pascal nor Adam are the shortstop)

10. The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards (none of Pascal, Hassan, or Bobo play either second base or catcher. And Adam is not going to be the second baseman because he doesn’t like the catcher and so would likely not play billiards with him.)

11. Sung is in the process of getting a divorce (So Sung cannot play second base because the second baseman is engaged.)

12. The catcher and the third baseman each have two legitimate children

13. Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married (none of Ed, Pascal, or Jason play right or center field. Also, none of them can play pitcher because the pitcher has a wife. Also, Sung will not play right or center field because he must be married in order to be getting a divorce and Ed, Pascal, and Jason cannot be catcher or third base because they would have to be married as per #12.)

14. The shortstop, the third baseman and Bobo all attended the fight (Bobo cannot play short stop or third base)

15. Mathieu is the shortest player of the team

At this point we see that Center and Right must be played by either Mathieu or Bobo because they are the only two possibilities left. #3 says the Center is taller than the Right and #15 says Mathieu is the shortest player so Mathieu must play Right and Bobo must play center. We then eliminate them from playing any other positions.
We can then see that only Charles or Jason can play second base. Since the second baseman is engaged (#2) Jason has to be the second baseman (because #13 says the only bachelors by name are Ed, Pascal, and Jason ie. not Charles. So if Charles were to be a bachelor he would have to play right or center field.) So Jason plays second base and we eliminate him from all other positions.
Left field is either Pascal or Sung. Since #6 says Ed plays cards with the outfielders and #5 mentions Pascal playing cards, let’s assume Pascal plays outfield. So Pascal plays left field and we eliminate him from all other positions.
#4 says Hassan and the third baseman live in the same building and maybe they do because the pitcher’s wife is the third baseman’s sister (#7). Since Hassan can’t be the third baseman it might be reasonable to assume that he is the pitcher.
#10 says that the catcher got beat at billiards. Since the catcher is either Sung or Charles but we’ve never heard anything else about Sung doing anything social with the others we’re going to assume that the catcher is Charles (who was playing poker in #5). Following the same logic the shortstop is mentioned to do quite a bit of gambling so I’m going to assume that that is Ed (who is mentioned to be a card player in #6 and he’s also a bachelor so it’s a little more likely for him to be out carousing than Sung.)
That leaves just Adam and Sung and first and third base. Since we haven’t heard anything about first base but we know that the third baseman has legitimate children (#12) and we know Sung is married (because he’s getting divorced #11) it seems more reasonable to peg him as the third baseman, leaving Adam to play first base.

So the positions are:

Adam- First Base

Ed- Short Stop

Bobo- Center Field

Charles – Catcher

Hassan- Pitcher

Jason – Second Base

Mathieu – Right Field

Pascal – Left Field

Sung – Third Base

===Question 4===

'''Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?

First we have to look at each piece of information that is given to us.
We know that there are 6 players, and that each player plays each other once in 5 days:

This means that there are 3 pairs playing against each other everyday

We are given enough information to determine a set of pairs for each day. Using this information, we can determine the other four players and who they play during the 5 days.

First we will group all the pairs that we know has played during the five days

'''Day 1: CP''' We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)
'''Day 2: CJ''' (L, P, S, and F)
'''Day 3: JL''' (F, P, L, and S)
'''Day 4: PS''' (F, J, L, and C)
'''Day 5: F ?'''

Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.

'''Day 1: "CP" "JS" "LF"'''
'''Day 2: "CJ" "LP" "SF"''' (We cannot group F with L again because they have played each other on the first day)
'''Day 3: "JL" "FP" "LS"''' (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)
'''Day 4: "PS" "FJ" "LC"'''(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once)

The grouping for the following day is as follows:
'''Day 5: "PS" "FJ" and "LC"'''

By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.

Therefore, we can come to the conclusion that Fernanda played Janet on the last day.

Course:MATH110/Archive/2010-2011/003/Groups/Group 13/Homework 4

2010-10-20T01:51:34Z

VictoriaBass: Created page with '=Homework #4= ===Question 1=== '''Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?''' - Tos…'

=Homework #4=

===Question 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. 

The best way to tackle a problem such as this would be to go through each piece of information given and organise the data into a form much easier to go back to and make adjustments to. Below is a simple person to pet chart I drew up from the information given. 

Tosh ---> Cat (Jun) 
♀Bianca ---> Frog (Suzan) 
♀Jaela ---> Parrot (No name) 
♂Jun ---> Snake (No name) 
♀Suzan ---> Turtle (No name) 

With ♀ denoting female and ♂ denoting male, we can see in the table that the information has given us enough to start off with. We know Bianca is female from the statement "Bianca owns a frog that '''she''' loves" and we know that Jaela is also female from the statement "Jaela owns a parrot which keeps calling '''her''' 'darling, darling'". We also know that Jun is male from the statement "Jun owns a snake, don't mess with '''him'''", with him referring to the owner and not the pet, otherwise it would be saying don't mess with '''it'''. 

We can also assume that Suzan's pet is the turtle since both are the only owner and pet that do not have a match and since the problem states that there are only five people then we can assume that there are also only five pets. 

Next, we can assign the pet name Tosh to the only other female aside from Suzan, Jaela. We can do this because we know that Suzan's pet is the turtle and if the name by which they call the turtle is the name of the woman whose pet is Tosh then naming the turtle Tosh would mean that the turtle's name needs to be Suzan which is all very confusing and counter-intuitive. 

♀Jaela ---> Parrot (Tosh) 

This piece of information also allows us to name Suzan's pet because again the name of the turtle would be the name of the person whose pet is called Tosh. Thus we receive the next bit of information: 

♀Suzan ---> Turtle (Jaela) 

That only leaves the last part of the problem, which would be Jun's pet: 

♂Jun ---> Snake (Bianca) 

Though the logic may fit, this solution is wrong as Jun is supposedly male and Susan's '''mother''' cannot possibly be male unless the label "mother" is semantically misleading (i.e. nickname, etc.). 

There are then two possible solutions after this error. The first is that the statement "Jun owns a snake, don't mess with '''him'''" has the pronoun 'him' refer to the snake, which can therefore make the solution of Jun owning the snake Bianca correct (although the male snake would be awkwardly named Bianca). This solution ends up with Jun being Suzan's mother and the pet Bianca being her pet. 

The second solution could be that the turtle is not owned by Suzan. This would allow any other person to own the turtle, allowing the turtle to then be named Suzan and therefore allowing Suzan's pet to be Tosh and lastly having Jaela's pet to be Bianca (and thus having Jaela as Suzan's mother). This scenario assumes that Jun is male and assumes that since the definition of the problem never assigns a finite number of pets within the group or never limits the amount of pets that the group can have (or limits the doubling of names within those pets, etc.) that there is the possibility that each person can own more than one pet. The problem never states that the pets listed are the only pets that these individuals own and thus can be seen as a possible solution, and a more likely one then the previously aforementioned. In this case, it is not realistically possible to know who Suzan's mother is with the given information.
 

===Question 2===

'''Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?'''

First, simplify the information:

Bohao, Stewart, Dylan, Tim and Chan

2 are left-handed, 3 are right-handed

Bohnao and Dylan = Same Hand

Tim and Chan = Different Hand

2 are over 2m, 3 are under 2m

Steward and Chan = Same Height

Tim and Dylan = Different Height

Need to find: player that is over 2 meters tall and left-handed

Now draw a diagram to help solve the first part of the problem (Left/Right Handed).
[[Image:Group Project 4 Question 2.JPG|center]]

Now draw another diagram to help solve the final part of the problem (Above/Under 2 meters in height).

[[Image:Group Project 4 Question 2p2.JPG|center]]

Now, using these two diagrams find a player that is over 2 meters tall and left-handed

By process of elimination we know that it can't be:

'''Bohao''' (right-handed)

'''Dylan''' (right-handed)

'''Stewart''' (Under 2 meters)

'''Chan''' (Under 2 meters)

Therefore it COULD only be '''TIM'''

Therefore the player that is over 2 meters tall and left-handed is '''TIM'''

===Question 3===

'''Six players - Petra, Carla, Janet, Sandra, Li and Fernanda - are competing in a chess tournament over a period of five days. Each player plays each of the others once. Three matches are played simultaneously during each of the five days. The first day, Carla beats Petra after 36 moves. The second day, Carla was again victorious when Janet failed to complete 40 moves within the required time limit. The third day had the most exciting match of all when Janet declared that she would checkmate Li in 8 moves and succeeded in doing so. On the fourth day, Petra defeated Sandra. Who played against Fernanda on the fifth day?

First we have to look at each piece of information that is given to us.
We know that there are 6 players, and that each player plays each other once in 5 days:

This means that there are 3 pairs playing against each other everyday

We are given enough information to determine a set of pairs for each day. Using this information, we can determine the other four players and who they play during the 5 days.

First we will group all the pairs that we know has played during the five days

'''Day 1: CP''' We have Janet, Sandra, Li and Fernanda leftover (J, S, L, and F)
'''Day 2: CJ''' (L, P, S, and F)
'''Day 3: JL''' (F, P, L, and S)
'''Day 4: PS''' (F, J, L, and C)
'''Day 5: F ?'''

Since we know the pairs that play each other, we can begin to group the leftover individuals together at random, while making sure that they only play each other ONCE, therefore we have to take into consideration the given pairs and the groupings we have assigned at random for the day BEFORE.

'''Day 1: "CP" "JS" "LF"'''
'''Day 2: "CJ" "LP" "SF"''' (We cannot group F with L again because they have played each other on the first day)
'''Day 3: "JL" "FP" "LS"''' (Make sure each player only plays each other once, F has already played S and L, therefore we can group her with P)
'''Day 4: "PS" "FJ" "LC"'''(Following the above format, for these four days, we have grouped each player with another player that they have not been grouped with before, we can see that Fernanda has played each player once, which leaves us with the last player, the player we are trying to determine. We will still group the rest of the individuals together to double check that they have indeed played each other once)

The grouping for the following day is as follows:
'''Day 5: "PS" "FJ" and "LC"'''

By following the information we have, we can follow the steps that are given to us and logically figure out who Fernanda played against on the last day. Not only does this give us the answer, but we have a listing of all the possible combinations, and who played against who for all five days.

Therefore, we can come to the conclusion that Fernanda played Janet on the last day. 

===Question 4===

We are trying to determine which position everyone plays. I started by making a chart of all the players and all the positions so we could cross off possibilities as we went along. Then we started using the information we are given to figure out which position various people CANNOT play so as to start to figure out positions by process of elimination.

We are told the following (and our interpretation is written in parentheses, which we then marked on the chart accordingly)

1. Adam does not like the catcher, (so Adam cannot be the catcher)

2. Ed's sister is engaged to the second baseman, (so Ed cannot be the second baseman)

3. The centre fielder is taller than the right fielder,

4. Hassan and the third baseman live in the same building (so Hassan cannot be the third baseman)

5. Pascal and Charles each won $20 from the pitcher at a poker game (so neither Pascal nor Charles can be the pitcher)

6. Ed and the outfielders play cards during their free time (Ed cannot play outfield)

7. The pitcher's wife is the third baseman's sister,

8. All the battery and infield except Charles, Hassan and Adam are shorter than Sung (Charles, Hassan, and Adam play infield or battery and therefore do not play any of the outfield positions)

9. Pascal, Adam and the shortstop lost $100 each at the race track (neither Pascal nor Adam are the shortstop)

10. The second baseman beat Pascal, Hassan, Bobo and the catcher at billiards (none of Pascal, Hassan, or Bobo play either second base or catcher. And Adam is not going to be the second baseman because he doesn’t like the catcher and so would likely not play billiards with him.)

11. Sung is in the process of getting a divorce (So Sung cannot play second base because the second baseman is engaged.)

12. The catcher and the third baseman each have two legitimate children

13. Ed, Pascal Jason, the right fielder and the centre fielder are bachelors, the others are all married (none of Ed, Pascal, or Jason play right or center field. Also, none of them can play pitcher because the pitcher has a wife. Also, Sung will not play right or center field because he must be married in order to be getting a divorce and Ed, Pascal, and Jason cannot be catcher or third base because they would have to be married as per #12.)

14. The shortstop, the third baseman and Bobo all attended the fight (Bobo cannot play short stop or third base)

15. Mathieu is the shortest player of the team

At this point we see that Center and Right must be played by either Mathieu or Bobo because they are the only two possibilities left. #3 says the Center is taller than the Right and #15 says Mathieu is the shortest player so Mathieu must play Right and Bobo must play center. We then eliminate them from playing any other positions.
We can then see that only Charles or Jason can play second base. Since the second baseman is engaged (#2) Jason has to be the second baseman (because #13 says the only bachelors by name are Ed, Pascal, and Jason ie. not Charles. So if Charles were to be a bachelor he would have to play right or center field.) So Jason plays second base and we eliminate him from all other positions.
Left field is either Pascal or Sung. Since #6 says Ed plays cards with the outfielders and #5 mentions Pascal playing cards, let’s assume Pascal plays outfield. So Pascal plays left field and we eliminate him from all other positions.
#4 says Hassan and the third baseman live in the same building and maybe they do because the pitcher’s wife is the third baseman’s sister (#7). Since Hassan can’t be the third baseman it might be reasonable to assume that he is the pitcher.
#10 says that the catcher got beat at billiards. Since the catcher is either Sung or Charles but we’ve never heard anything else about Sung doing anything social with the others we’re going to assume that the catcher is Charles (who was playing poker in #5). Following the same logic the shortstop is mentioned to do quite a bit of gambling so I’m going to assume that that is Ed (who is mentioned to be a card player in #6 and he’s also a bachelor so it’s a little more likely for him to be out carousing than Sung.)
That leaves just Adam and Sung and first and third base. Since we haven’t heard anything about first base but we know that the third baseman has legitimate children (#12) and we know Sung is married (because he’s getting divorced #11) it seems more reasonable to peg him as the third baseman, leaving Adam to play first base.

So the positions are:

Adam- First Base

Ed- Short Stop

Bobo- Center Field

Charles – Catcher

Hassan- Pitcher

Jason – Second Base

Mathieu – Right Field

Pascal – Left Field

Sung – Third Base

Course:MATH110/Archive/2010-2011/003/Groups/Group 13

2010-10-13T07:45:22Z

VictoriaBass: /* Question 16 */

Group members:
* Victoria Bass | email: bassvm@interchange.ubc.ca|
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242
* Fiona Ma
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987
* Una Vuckovic | email: v_una@hotmail.com | cell: 778 829 8862

Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don't have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!

ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!

----
=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.

[[Image:Group_Project_question_1.JPG‎]]

'''60min = 1hour'''

'''60min + 20min = 80min'''

'''1hour + 20min = 80min'''

Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same
as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because...
'''1hour and 20min = 80min'''

----

===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.

[[Image:Group_Project_question_2.JPG]]

The simple explanation to this question is that the '''lady is a pedestrian'''. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)

Therefore the witness policeman did not stop her because she did NOT break the law.

----

===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.
[[Image:Group Project question -3.jpg]]

The problem can be solved by taking out the contents of '''Box #1'''. If an APPLE is pulled
out of Box #1, you know that the contents of '''Box #1''' is just APPLES. This is because the two options
(labelled in blue) are either just APPLES or just ORANGES.

Given that '''Box #1''' has only APPLES you know that '''Box #3''' has APPLES & ORANGES. This is because the two
options are just APPLES or APPLES & ORANGES and '''Box #1''' already contains just APPLES.

By process of elimination '''Box #2''' contains just oranges.

The same logic applies if an ORANGE is initially pulled out of '''Box #1'''. Then '''Box #2''' would contain
APPLES & ORANGES and '''Box #3''' would contain just ORANGES.

Therefore the solution can be solved by opening '''Box #1'''.

----

===Question 4===

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

[[Image:Group_Project_question_4.JPG‎]]

The key to this question is to avoid gender bias. The logical solution is that the boy is the brother
of a blind fiddler, who is his sister, therefore he has no brothers.

----

===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?

[[Image:Group Project question 5.JPG]]

Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.

===Question 6===

'''6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?
'''
 If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.

http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg

===Question 7===

'''7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?'''

'''i) pair of the same colour'''

If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.

'''ii) a pair with different colours?'''

If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.

http://www.wholesalefootballkits.com/images/navyblue_socks.jpg

===Question 8===

'''8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.''' 

If Reuben's birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year. 

Dec 30th 2010 - 20 years old 
Dec 31st 2011 - turns 21 
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. 
Dec 31st 2011 (same year) - turns 22 
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year 

===Question 9===

'''9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?''' 

If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.

===Question 10===

'''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.''' 

'''i and ii''') It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.

===Question 11===

'''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?

First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible.

===Question 12===

'''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.

Because each of the trains arrive 10 min apart, we can come up with a train schedule:
Bronx train arrives at 10:10, 10:20, 10:30 etc.
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.

===Question 13===

'''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)?

For each 5 seconds the clock chimes 5 times with equal lengths of space in between.
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5.
5 chimes / 4 spaces = 1.25seconds
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes.
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00

===Question 14===

'''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). First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.
AB = ABCD, ABDC
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.

ii). Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.

===Question 15===

'''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?

Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards.
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).

-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.

We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.

===Question 16===

"Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?"

Trying to figure out how many S (sons) and D (daughters) so I tried to represent the information in an expression.

Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)

So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).

===Question 17===

" The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain."

If the scale read too low then we know that D’s real weight would be >60 kg and S’s real weight would be >50kg and so we would expect their combined weight to be >than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is <60 and S’s real weight is <50 then we would expect the outcome to be <their combined weight (110kg).

===Question 18===

" Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?"

I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135.

===Question 19===

"One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?"

Angela’s cup can be expressed as:
¼ Total Milk + 1/6 Total Coffee = 8 oz
We can also say that:
Total Milk + Total Coffee/8 = # of family members (because they each had an 8 oz cup)

If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:

8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn't really count as a family) that this works with.

===Question 21===

"Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?"

For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30)

===Question 21===

'''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?''' 

First we have to recognise Sven's place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: 

Total = 2(Sven) - 1 

We also know that Sven came in before Dan, who is 10th. Thus giving the definition: 

Sven < 10 

Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: 

Total = 2(Sven) - 1 > 16 

The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.

===Question 22===

'''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?''' 

Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement "every rainy afternoon was preceded by a sunny morning". 

Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement "when it rained in the morning, the afternoon was sunny". 

Let z be the number of days where it didn't rain at all. 

From these definitions we can come up with equations to represent these pieces of information. 

For one we can say that <math> x + 7 = 11 </math> since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). 

Next we can say that <math> y + z = 13 </math> since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. 

Lastly we can say that <math> x + y = 13 </math> since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. 

Then we can simply solve the system of equations. 

<math> y + x = 13 </math> 
<math> x = 13 - y </math> 
<math> x + z = 11 </math> 
<math> 13 - y + z = 11 </math> and <math> y + z = 12 </math> 
<math> 12 - 11 = y + z - (13 - y + z)</math> 
<math> 1 = y + z - 13 + y - z</math> 
<math> 1 = 2y - 13</math> 
<math> y = 7 </math> 

<math> y + x = 13</math> --> <math> 7 + x = 13</math> --> <math> x = 6</math> 

<math> y + z = 12</math> --> <math> 7 + z = 12</math> --> <math> z = 5</math> 

Thus <math> x + y + z = 7 + 6 + 5 = 18 </math>, with 18 being the length of the entire vacation.

===Question 23===

'''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.''' 

First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. 

To start off, you can easily list down all of the 3-set factors of 36 which are: 

{ ( 36, 1, 1) 
(18, 2, 1) 
(12, 3, 1) 
(9, 4, 1) 
(9, 2, 2) 
(6, 6, 1) 
(6, 3, 2) 
(4, 3, 3) } 

The next clue is that the sum of these ages would equal today's date, thus the sum of their ages cannot exceed 31. 

Adding all the ages of the 3-set factors would yield the following: 

36 + 1 + 1 = 38 
18 + 2 + 1 = 21 
12 + 3 + 1 = 16 
9 + 4 + 1 = 14 
9 + 2 + 2 = 13 
6 + 6 + 1 = 13 
6 + 3 + 2 = 11 
4 + 3 + 3 = 10 

Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. 

Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like "my oldest children have red hair". Thus leaving only the set of (9, 2, 2) left.

===Question 24===

'''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?''' 

Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: 

<math>L=((6-t)/6)</math> with t being the number of hours elapsed since the candle was lit 

The second candle's equation is as follows: 

<math>L=((3-t)/3)</math> with t again being the number of hours after the candle was lit 

Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: 

<math>((6-1)/6) = 5/6</math> <----> <math>((3-1)/6) = 2/3</math> 
<math>((6-2)/6) = 4/6</math> <----> <math>((3-2)/6) = 1/3</math> 

<math> 4/6 = 2/3</math>
<math> 2/3 = 2 (1/3)</math> with 1/3 being the length of the second candle. 

Thus giving the answer as after 2 hours.

===Question 25===

'''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.''' 

Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). 

Since when L is at half its length, namely <math> (1/2)L</math>, it equals 4 or 4 hours after 0, then we can put up the equation: 

<math> (1/2) L = 4 </math> 
<math> L = 8 </math> 

L then equals 8.

Course:MATH110/Archive/2010-2011/003/Groups/Group 13

2010-10-13T07:44:40Z

VictoriaBass: /* Question 16 */

Group members:
* Victoria Bass | email: bassvm@interchange.ubc.ca|
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242
* Fiona Ma
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987
* Una Vuckovic | email: v_una@hotmail.com | cell: 778 829 8862

Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don't have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!

ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!

----
=Problems=

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===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.

[[Image:Group_Project_question_1.JPG‎]]

'''60min = 1hour'''

'''60min + 20min = 80min'''

'''1hour + 20min = 80min'''

Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same
as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because...
'''1hour and 20min = 80min'''

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===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.

[[Image:Group_Project_question_2.JPG]]

The simple explanation to this question is that the '''lady is a pedestrian'''. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)

Therefore the witness policeman did not stop her because she did NOT break the law.

----

===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.
[[Image:Group Project question -3.jpg]]

The problem can be solved by taking out the contents of '''Box #1'''. If an APPLE is pulled
out of Box #1, you know that the contents of '''Box #1''' is just APPLES. This is because the two options
(labelled in blue) are either just APPLES or just ORANGES.

Given that '''Box #1''' has only APPLES you know that '''Box #3''' has APPLES & ORANGES. This is because the two
options are just APPLES or APPLES & ORANGES and '''Box #1''' already contains just APPLES.

By process of elimination '''Box #2''' contains just oranges.

The same logic applies if an ORANGE is initially pulled out of '''Box #1'''. Then '''Box #2''' would contain
APPLES & ORANGES and '''Box #3''' would contain just ORANGES.

Therefore the solution can be solved by opening '''Box #1'''.

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===Question 4===

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

[[Image:Group_Project_question_4.JPG‎]]

The key to this question is to avoid gender bias. The logical solution is that the boy is the brother
of a blind fiddler, who is his sister, therefore he has no brothers.

----

===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?

[[Image:Group Project question 5.JPG]]

Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.

===Question 6===

'''6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?
'''
 If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.

http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg

===Question 7===

'''7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?'''

'''i) pair of the same colour'''

If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.

'''ii) a pair with different colours?'''

If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.

http://www.wholesalefootballkits.com/images/navyblue_socks.jpg

===Question 8===

'''8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.''' 

If Reuben's birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year. 

Dec 30th 2010 - 20 years old 
Dec 31st 2011 - turns 21 
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. 
Dec 31st 2011 (same year) - turns 22 
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year 

===Question 9===

'''9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?''' 

If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.

===Question 10===

'''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.''' 

'''i and ii''') It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.

===Question 11===

'''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?

First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible.

===Question 12===

'''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.

Because each of the trains arrive 10 min apart, we can come up with a train schedule:
Bronx train arrives at 10:10, 10:20, 10:30 etc.
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.

===Question 13===

'''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)?

For each 5 seconds the clock chimes 5 times with equal lengths of space in between.
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5.
5 chimes / 4 spaces = 1.25seconds
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes.
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00

===Question 14===

'''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). First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.
AB = ABCD, ABDC
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.

ii). Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.

===Question 15===

'''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?

Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards.
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).

-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.

We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.

===Question 16===

"Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?"

Trying to figure out how many S (sons) and D (daughters).
I tried to represent the problem in an expression
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).

===Question 17===

" The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain."

If the scale read too low then we know that D’s real weight would be >60 kg and S’s real weight would be >50kg and so we would expect their combined weight to be >than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is <60 and S’s real weight is <50 then we would expect the outcome to be <their combined weight (110kg).

===Question 18===

" Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?"

I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135.

===Question 19===

"One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?"

Angela’s cup can be expressed as:
¼ Total Milk + 1/6 Total Coffee = 8 oz
We can also say that:
Total Milk + Total Coffee/8 = # of family members (because they each had an 8 oz cup)

If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:

8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn't really count as a family) that this works with.

===Question 21===

"Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?"

For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30)

===Question 21===

'''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?''' 

First we have to recognise Sven's place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: 

Total = 2(Sven) - 1 

We also know that Sven came in before Dan, who is 10th. Thus giving the definition: 

Sven < 10 

Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: 

Total = 2(Sven) - 1 > 16 

The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.

===Question 22===

'''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?''' 

Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement "every rainy afternoon was preceded by a sunny morning". 

Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement "when it rained in the morning, the afternoon was sunny". 

Let z be the number of days where it didn't rain at all. 

From these definitions we can come up with equations to represent these pieces of information. 

For one we can say that <math> x + 7 = 11 </math> since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). 

Next we can say that <math> y + z = 13 </math> since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. 

Lastly we can say that <math> x + y = 13 </math> since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. 

Then we can simply solve the system of equations. 

<math> y + x = 13 </math> 
<math> x = 13 - y </math> 
<math> x + z = 11 </math> 
<math> 13 - y + z = 11 </math> and <math> y + z = 12 </math> 
<math> 12 - 11 = y + z - (13 - y + z)</math> 
<math> 1 = y + z - 13 + y - z</math> 
<math> 1 = 2y - 13</math> 
<math> y = 7 </math> 

<math> y + x = 13</math> --> <math> 7 + x = 13</math> --> <math> x = 6</math> 

<math> y + z = 12</math> --> <math> 7 + z = 12</math> --> <math> z = 5</math> 

Thus <math> x + y + z = 7 + 6 + 5 = 18 </math>, with 18 being the length of the entire vacation.

===Question 23===

'''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.''' 

First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. 

To start off, you can easily list down all of the 3-set factors of 36 which are: 

{ ( 36, 1, 1) 
(18, 2, 1) 
(12, 3, 1) 
(9, 4, 1) 
(9, 2, 2) 
(6, 6, 1) 
(6, 3, 2) 
(4, 3, 3) } 

The next clue is that the sum of these ages would equal today's date, thus the sum of their ages cannot exceed 31. 

Adding all the ages of the 3-set factors would yield the following: 

36 + 1 + 1 = 38 
18 + 2 + 1 = 21 
12 + 3 + 1 = 16 
9 + 4 + 1 = 14 
9 + 2 + 2 = 13 
6 + 6 + 1 = 13 
6 + 3 + 2 = 11 
4 + 3 + 3 = 10 

Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. 

Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like "my oldest children have red hair". Thus leaving only the set of (9, 2, 2) left.

===Question 24===

'''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?''' 

Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: 

<math>L=((6-t)/6)</math> with t being the number of hours elapsed since the candle was lit 

The second candle's equation is as follows: 

<math>L=((3-t)/3)</math> with t again being the number of hours after the candle was lit 

Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: 

<math>((6-1)/6) = 5/6</math> <----> <math>((3-1)/6) = 2/3</math> 
<math>((6-2)/6) = 4/6</math> <----> <math>((3-2)/6) = 1/3</math> 

<math> 4/6 = 2/3</math>
<math> 2/3 = 2 (1/3)</math> with 1/3 being the length of the second candle. 

Thus giving the answer as after 2 hours.

===Question 25===

'''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.''' 

Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). 

Since when L is at half its length, namely <math> (1/2)L</math>, it equals 4 or 4 hours after 0, then we can put up the equation: 

<math> (1/2) L = 4 </math> 
<math> L = 8 </math> 

L then equals 8.

Course:MATH110/Archive/2010-2011/003/Groups/Group 13

2010-10-13T07:43:53Z

VictoriaBass:

Group members:
* Victoria Bass | email: bassvm@interchange.ubc.ca|
* Miguel Caruncho | email: migcar429@yahoo.com | cell: 778 323 4242
* Fiona Ma
* Adam Nguyen | email: avnguyen213@yahoo.com or adam@premierwestmma.com | cell: 778 868 6987
* Una Vuckovic | email: v_una@hotmail.com | cell: 778 829 8862

Hey guys, this is Adam. I think that it would be best if we put our contact beside our names. You don't have to put down your numbers of course, but an email would be very helpful to the other members of this group. Thanks!

ps. We are the first group (if not then one of the first) to post up answers to the Pyola questions. Good Job group!

----
=Problems=

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===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.

[[Image:Group_Project_question_1.JPG‎]]

'''60min = 1hour'''

'''60min + 20min = 80min'''

'''1hour + 20min = 80min'''

Therefore to travel from the terminal to the airport at an average speed of 30mi/h in an hour and 20min is the same
as traveling from the airport back to the terminal at the average speed of 30mi/h in 80min because...
'''1hour and 20min = 80min'''

----

===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.

[[Image:Group_Project_question_2.JPG]]

The simple explanation to this question is that the '''lady is a pedestrian'''. Since she is not driving a car she does not need to have her license present with her, does not have to stop at STOP signs and may walk down a one way street going the wrong way. (On the sidewalk of course)

Therefore the witness policeman did not stop her because she did NOT break the law.

----

===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.
[[Image:Group Project question -3.jpg]]

The problem can be solved by taking out the contents of '''Box #1'''. If an APPLE is pulled
out of Box #1, you know that the contents of '''Box #1''' is just APPLES. This is because the two options
(labelled in blue) are either just APPLES or just ORANGES.

Given that '''Box #1''' has only APPLES you know that '''Box #3''' has APPLES & ORANGES. This is because the two
options are just APPLES or APPLES & ORANGES and '''Box #1''' already contains just APPLES.

By process of elimination '''Box #2''' contains just oranges.

The same logic applies if an ORANGE is initially pulled out of '''Box #1'''. Then '''Box #2''' would contain
APPLES & ORANGES and '''Box #3''' would contain just ORANGES.

Therefore the solution can be solved by opening '''Box #1'''.

----

===Question 4===

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

[[Image:Group_Project_question_4.JPG‎]]

The key to this question is to avoid gender bias. The logical solution is that the boy is the brother
of a blind fiddler, who is his sister, therefore he has no brothers.

----

===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?

[[Image:Group Project question 5.JPG]]

Just follow the diagram and you will see that the coin is revolved twice when it returns to its original position.

===Question 6===

'''6. Three kinds of apples are all mixed up in a basket. How many apples must you draw (without looking) from the basket to be sure of getting at least two of one kind?
'''
 If we draw four apples from the basket, then we can assure ourselves that the fourth apple must be the same kind as one of the first three that we drew. Therefore, if we draw four apples, we would be sure of getting at least two apples of one kind.

http://static.howstuffworks.com/gif/diet-apples.jpg http://www.garwoodorchard.com/site/images/apple.jpg

===Question 7===

'''7. Suppose you have 40 blue socks and 40 brown socks in a drawer. How many socks must you take from the drawer (without looking) to be sure of getting (i) a pair of the same color, and (ii) a pair with different colors?'''

'''i) pair of the same colour'''

If we draw three socks without looking, we guarantee ourselves of drawing two socks that are of the same colour because there are only two kinds of colours that the socks can be.

'''ii) a pair with different colours?'''

If we draw 41 socks, than we can guarantee ourselves of drawing two socks that are different in colour. A person can have the fortunate (or unfortunate)event of drawing straight 40 socks of the same colour. But on his 41st draw, the sock must be of different color because there are no socks of the other color remaining.

http://www.wholesalefootballkits.com/images/navyblue_socks.jpg

===Question 8===

'''8. Reuben says, “Two days ago I was 20 years old. Later next year I will be 23 years old.” Explain how this is possible.''' 

If Reuben's birthday was on Dec. 31st 2010 and he said this statement on Jan. 1st 2011, than one year later on Jan. 1st 2012, he would turn 23 years old in the same year. 

Dec 30th 2010 - 20 years old 
Dec 31st 2011 - turns 21 
Jan 1st 2011 - Says that two days prior he was 20 years old and that he would turn 23 years old later next year. 
Dec 31st 2011 (same year) - turns 22 
Jan 1st 2012 (year after he makes his statement) - will be turning 23 in this year 

===Question 9===

'''9. A rope ladder hanging over the side of a boat has rungs one foot apart. Ten rungs are showing. If the tide rises five feet, how many rungs will be showing?''' 

If the tide rises by five feet, the tide will also raise the boat by five feet also. Therefore, ten rungs would still be showing.

===Question 10===

'''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.''' 

'''i and ii''') It does not follow that both one-fourth of all people are women chocolate eaters or that one-half of all men are chocolate eaters. The information that we are provided does not say anything about which people are chocolate eaters. It only tells us that half of all people are chocolate eaters and that half of all peopole are women.. In an alternate reality, given that half of all people are chocolate eaters and half of all people are women, it could be that all women in this reality eat chocolate, which negates both numbers i and ii.

===Question 11===

'''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?

First, we have to figure out who the worst and best player are, we know that the worst and best players are of opposite sex and have the same age, therefore the twins are the worst and best player. But because we have limited knowledge, ie we do not know the ages of the four players, or what gender the worst and best player is, we can technically conclude that all four can be the worst player. The woman and her older brother can be twins OR the son and the daughter could be twins. Therefore this situation is not possible.

===Question 12===

'''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.

Because each of the trains arrive 10 min apart, we can come up with a train schedule:
Bronx train arrives at 10:10, 10:20, 10:30 etc.
Brooklyn train arrives at 10:09, 10:29, 10:20 etc.
Because the man arrives at any given time/random time, we can see that unless he arrives exactly between the time the Brooklyn train leaves the and Bronx train arrives, we can see that the Brooklyn train will always arrive first, and that the only chance he will have to take the Bronx train is during that one minute interval. The probability or chances he has of arriving during exactly during that one minute interval is very slim. Therefore we can come to the conclusion that because the Brooklyn train always arrives before the Bronx train, the man usually ends up getting on the Brooklyn train because it is the first to arrive; this makes his visits to his Bronx girlfriend very infrequent and rare.

===Question 13===

'''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)?

For each 5 seconds the clock chimes 5 times with equal lengths of space in between.
Each second there is one space, so in total there are 4 spaces in between each chime. In order to calculate how many seconds it takes for 5 chimes and 4 stops, we can divide 4 by 5.
5 chimes / 4 spaces = 1.25seconds
So when there are 10 equally spaced chimes, we can conclude from the above example that in between each chime there is one stop. 10 chimes has 9 stops in between (because it stops at the 10th chime we do not count the 10th stop). We can then find out how long it takes.
( 9 stops * 5 seconds ) / 4 stops = 11.25 seconds to strike 10:00

===Question 14===

'''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). First we have to start by having two names be labeled correctly, for example A and B are labeled correctly while C and D are mixed up. In order to have A and B be labeled on the correct baby, C and D can only be mixed up twice, there are only two alternatives while keeping A and B on the right baby. Since we can arrive at the solution that for each combination of 2 there are 2 answers. We have to test it out for pair of babies.
AB = ABCD, ABDC
We can go through ABCD and figure out that in that four letter sequence there are six possible pairing combinations AB, AC, AD, BC, BD, CD and that each of these pairs can have two of the other single letters mixed up
therefore, 6 * 2 = 12, the answer gives us that there can be 12 possible combinations.

ii). Logically if we think about the question asked, how many ways could three babies be tagged correctly and one baby be tagged incorrectly, the question does not make sense. Because there are 4 babies, if 3 babies are tagged correctly then the fourth one MUST be tagged correctly because there leaves no other alternative/baby to be tagged. Therefore there are no ways that three of the four babies could be tagged correctly since having three tagged correctly means that all four are tagged correctly.

===Question 15===

'''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?

Each deck of cards has 52 cards, where they are either red or black: 26 red cards and 26 black cards.
No matter how Alex splits the deck, there should always be an equal amount of red cards to black cards. For example, if Alex splits the deck in two (each deck containing 26 cards) we see that there are 12 black cards and 14 red cards. Because there are an equal number of black and red cards, the second deck should have the same number or cards with the reverse color combination. Therefore no matter how many different ways Alex splits the deck, the colored cards in one deck will always equal the opposite colored cards in the other desk because they have to equal to 26 (the number of cards in the deck and the number of red/black cards).

-Alex splits the deck, first deck has 10 BLACK cards and 16 RED cards = second deck has 10 RED cards and 16 BLACK cards
-Alex splits the deck, first deck has 3 BLACK cards and 23 RED cards = second deck has 3 RED cards and 23 BLACK cards etc etc.

We can come to the conclusion that because the cards are equal in value, that no matter how Alex splits the deck, the red cards will equal the black cards in a split deck.

===Question 16===

"Suppose that each daughter in your family has the same number of brothers as she has sisters, and each son in your family has twice as many sisters as he has brothers. How many sons and daughters are in the family?"

Trying to figure out how many S (sons) and D (daughters).
I tried to represent the problem in an expression:
Daughters-1 = Sons (because when you take away the daughter who is counting her siblings the # of daughters will = # of sons) and
Sons = (Daughters/2)-1 (because when you take away the son who is counting there will be twice as many daughters as sons)
So there always needs to be 2 daughters for every son, which will grow pretty rapidly and, I think, exclude the possibility that a daughter could have equal # brothers and sisters (because there will just be too many daughters.) So I think it might only work if there is only one son. Then each daughter has 1 brother and 1 sister, and each son has twice as many sisters as brothers ( ie 0 brothers).

===Question 17===

" The zero point on a bathroom scale is set incorrectly, but otherwise the scale is accurate. It shows 60 kg when Dan stands on the scale, 50 kg when Sarah stands on the scale, but 105 kg when Dan and Sarah both stand on the scale. Does the scale read too high or too low? Explain."

If the scale read too low then we know that D’s real weight would be >60 kg and S’s real weight would be >50kg and so we would expect their combined weight to be >than their weights combined (or 110 kg). Since it is less than this, the scale must read too high. Which makes sense because if D’s real weight is <60 and S’s real weight is <50 then we would expect the outcome to be <their combined weight (110kg).

===Question 18===

" Alice takes one-third of the pennies from a large jar. Then Bret takes one-third of the remaining pennies from the jar. Finally, Carla takes one-third of the remaining pennies from the jar, leaving 40 pennies in the jar. How many pennies were in the jar at the start?"

I thought of this by thinking of what was always left in the jar. There is always 2/3 of the previous jar amount left (because someone removes 1/3). So then I just worked backwards. I asked “40 is 2/3 of what number? That must be how much was in the jar before.” Turns out 40 is 2/3 of60. I continued in this manner until I had calculated back the appropriate removals and determined that the number of pennies in the jar to start with was 135.

===Question 19===

"One morning each member of Angela's family drank an eight-ounce cup of coffee and milk, with the (nonzero) amounts of coffee and milk varying from cup to cup. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. What is the least number of people in the family?"

Angela’s cup can be expressed as:
¼ Total Milk + 1/6 Total Coffee = 8 oz
We can also say that:
Total Milk + Total Coffee/8 = # of family members (because they each had an 8 oz cup)

If the smallest amount of people in a family would be two (because that’s the smallest number bigger than 1) then we could substitute that in to see if it works:

8 x 2 = 16 (total coffee and total milk) Which is a totally reasonable conclusion. So the least number of people in the family is 2 because that is the least number (other than 1, which wouldn't really count as a family) that this works with.

===Question 21===

"Of two clocks next to each other, one runs 5 min per hour fast and the other runs 5 min per hour slow. At midnight the clocks show the same time. At what time are they are one hour apart?"

For each hour of time that passes Clock A gains 1hr + 5 minutes and Clock B gains 1 hr -5 mins. This means that there is always an increasing difference between them and the difference is always increasing by 10 minutes. So if we want to know when they will be an hour apart it’s when they have had their 10 minute difference compounded 6 times, so 6 hours after they started, or 6 o’clock (by real time. Clock A would show 6:30 and Clock B would show 5:30)

===Question 21===

'''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?''' 

First we have to recognise Sven's place, which is exactly in the middle. This definition implies that the number of people in the race is an odd number since in even numbers it is impossible to be exactly in the middle. This gives the following definition of: 

Total = 2(Sven) - 1 

We also know that Sven came in before Dan, who is 10th. Thus giving the definition: 

Sven < 10 

Lastly we know that Lars came in 16th, thus adding another aspect to the equality given for the total number of runners: 

Total = 2(Sven) - 1 > 16 

The only number satisfying both inequalities of Sven being less than 10 and the total being greater than 16 is Sven being ninth place. All numbers under nine make the total runner inequality untrue (i.e. 2(7)-1 = 15), thus making the total number of runners 17.

===Question 22===

'''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?''' 

Let x be the number of rainy afternoons, which can also be seen as the number of sunny mornings, as given by the statement "every rainy afternoon was preceded by a sunny morning". 

Let y be the number of rainy mornings, which can also be seen as the number of sunny afternoons, as given by the statement "when it rained in the morning, the afternoon was sunny". 

Let z be the number of days where it didn't rain at all. 

From these definitions we can come up with equations to represent these pieces of information. 

For one we can say that <math> x + 7 = 11 </math> since every rainy afternoon was preceded by a sunny morning but not every sunny morning led to a rainy afternoon (i.e. some sunny mornings led to a sunny afternoon). 

Next we can say that <math> y + z = 13 </math> since every rainy morning led to a sunny afternoon, but every sunny afternoon does not entail a rainy morning before it. 

Lastly we can say that <math> x + y = 13 </math> since the number of rainy mornings and rainy afternoons obviously equals the number of days it rained. 

Then we can simply solve the system of equations. 

<math> y + x = 13 </math> 
<math> x = 13 - y </math> 
<math> x + z = 11 </math> 
<math> 13 - y + z = 11 </math> and <math> y + z = 12 </math> 
<math> 12 - 11 = y + z - (13 - y + z)</math> 
<math> 1 = y + z - 13 + y - z</math> 
<math> 1 = 2y - 13</math> 
<math> y = 7 </math> 

<math> y + x = 13</math> --> <math> 7 + x = 13</math> --> <math> x = 6</math> 

<math> y + z = 12</math> --> <math> 7 + z = 12</math> --> <math> z = 5</math> 

Thus <math> x + y + z = 7 + 6 + 5 = 18 </math>, with 18 being the length of the entire vacation.

===Question 23===

'''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.''' 

First, the strongest clue in the conversation with Paula is that the three ages of the children have a product of 36. This limits the numbers to being combined factors of 36. 

To start off, you can easily list down all of the 3-set factors of 36 which are: 

{ ( 36, 1, 1) 
(18, 2, 1) 
(12, 3, 1) 
(9, 4, 1) 
(9, 2, 2) 
(6, 6, 1) 
(6, 3, 2) 
(4, 3, 3) } 

The next clue is that the sum of these ages would equal today's date, thus the sum of their ages cannot exceed 31. 

Adding all the ages of the 3-set factors would yield the following: 

36 + 1 + 1 = 38 
18 + 2 + 1 = 21 
12 + 3 + 1 = 16 
9 + 4 + 1 = 14 
9 + 2 + 2 = 13 
6 + 6 + 1 = 13 
6 + 3 + 2 = 11 
4 + 3 + 3 = 10 

Next, since Paul says that giving that clue is not enough information, we can conclude that the ages are either (9, 2, 2) or (6, 6, 1) since if it was any other 3-set combination, the sum is unique leaving no room for uncertainty. 

Lastly, Paula says that the oldest child has red hair, thus implying that there is only one oldest child. Since if the oldest children indeed shared their age as in the case of (6, 6, 1) then the statement would read something like "my oldest children have red hair". Thus leaving only the set of (9, 2, 2) left.

===Question 24===

'''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?''' 

Both candles were at equal length when they were lit, given by L. With one burning out after 6 hours and the other after 3 hours. From this we can see a progression of the first candle having the equation: 

<math>L=((6-t)/6)</math> with t being the number of hours elapsed since the candle was lit 

The second candle's equation is as follows: 

<math>L=((3-t)/3)</math> with t again being the number of hours after the candle was lit 

Following the values given by substituting t with increasing multiples of 1 we achieve the criteria in the question after 2 hours: 

<math>((6-1)/6) = 5/6</math> <----> <math>((3-1)/6) = 2/3</math> 
<math>((6-2)/6) = 4/6</math> <----> <math>((3-2)/6) = 1/3</math> 

<math> 4/6 = 2/3</math>
<math> 2/3 = 2 (1/3)</math> with 1/3 being the length of the second candle. 

Thus giving the answer as after 2 hours.

===Question 25===

'''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.''' 

Let t = 0 be the time at which the longer candle is lit (i.e. 4:30). 

Since when L is at half its length, namely <math> (1/2)L</math>, it equals 4 or 4 hours after 0, then we can put up the equation: 

<math> (1/2) L = 4 </math> 
<math> L = 8 </math> 

L then equals 8.

User:VictoriaBass

2010-09-20T04:08:13Z

VictoriaBass:

User:VictoriaBass

2010-09-20T04:05:37Z

VictoriaBass:

Course:MATH110/Archive/2010-2011/003/FAQ

2010-09-17T05:06:59Z

VictoriaBass: /* File:faq.gif About the course */

This space is available for both the instructor and the students to add questions and/or answers.

==[[File:faq.gif]] General information==

===How do I find the instructor's office?===
Check the [http://www.maps.ubc.ca/PROD/index_detail.php?show=y,n,n,n,y,y&bldg2Search=n&locat1=045&locat2=#showMapLocal campus map].

Here's a [http://maps.google.com/maps/ms?ie=UTF8&hl=en&msa=0&ll=49.266677,-123.255975&spn=0.000905,0.002411&z=19&msid=108297701092887686753.00049064d841e2bb895f8 google map] with extra details I've added.

==[[File:faq.gif]] About the course==

===When is the final exam?===
This course has a December exam. Exams are scheduled on a university-wide level later in the term; you should not arrange to leave Vancouver until you know your exam schedule.

===Is there a specific calculator we are supposed to use for this course?===
I'll let you know asap, [[User:DavidKohler|DavidKohler]] 18:59, 9 September 2010 (UTC)

===Where can I get extra help with my math?===
* The math department has a [http://www.math.ubc.ca/Ugrad/ugradTutorials.shtml Drop-In Tutorial Centre] which is located for this term in the Auditorium Annex. Student will be able to find help from experienced graduate student in this location from 9 am to 4 pm Mondays to Thursdays, and 9 am to 3 pm on Fridays.
* The AMS also has a [http://tutoring.ams.ubc.ca/ Drop-In Tutorial Centre] which is located in the Qualicum Room in the Irving K. Barber Learning Centre. You will find there most days from 3 pm to 7 pm, check their website for more details.

===How, when and where do we register our iClickers?===
To register your clicker, simply go on the [http://vista.ubc.ca Vista webpage of this course]. In there, you will find a link for you to register your clicker in the system. This only needs to be done once for the whole university, so you only need to do it once, not once per course. If you have already used your clicker in the classroom, the system has already started recorder your entries, the registration simply allows the instructor to attach your student number to your clicker ID.

===Can iClickers be associated to several students numbers?===
It is good to know that iClickers can be shared by students, as long as they don't attend the same class of course. The system allows for multiple students numbers to be associated to the clicker ID. This also means that if you got a used iClicker from someone else, you can still use it in the classroom, just register it in Vista the usual way.

===If we can devise a relevant way to incorporate coffee and doughnuts into teaching a concept will David bring us said coffee and doughnuts to class one morning?===
All I'm saying is it's pretty early, and there's got to be a way we can learn calculus with baked goods. [[User:VictoriaBass]]

==[[File:faq.gif]] How to use this wiki==

===How do I get started using the UBC wiki?===
Have a look at the [[Help:Contents | Help]] page, it will tell you all from how to create a page, to how to format your text and even contains a useful [[UBC_Wiki:FAQs | FAQ]] worth checking out!

===Who can edit this wiki?===
Anyone at UBC using their [http://www.it.ubc.ca/cwl/homelink.shtml CWL] login. We usually suggest though that only the instructors and the students involved in this course modifies the pages of the Math 110 wiki.

===How to write math in the wiki?===
The help page of the wiki on [[Help:Formatting | formatting]] is really helpful for this. In short, it works by using the ''math'' tags. For example, the code

<pre><math>x^2</math></pre> produces <math>x^2</math>

The following example should provide you with most of the symbols used at the moment (the code ''quad'' creates a bigger space than the regular space). The code

<pre><math>\frac{3}{4} \quad \div \quad \times \quad \cdot \quad x^n \quad \alpha \quad N^{12} \quad P^{-3} \quad m^{\frac{1}{2}} </math></pre>

produces

: <math>\frac{3}{4} \quad \div \quad \times \quad \cdot \quad x^n \quad \alpha \quad N^{12} \quad P^{-3} \quad m^{\frac{1}{2}} </math>

===I like a page very much, can I print it?===
Yes! As you can see, the menu on the left contains a link called ''Download as PDF'' which is very handy if you want to print a page of the wiki.

User:VictoriaBass

2010-09-17T05:03:23Z

VictoriaBass: Created page with 'Yay! The wiki! I'm in MATH 110 because I'm restarting my life via a second Bachelor's degree. I used to do other, unrelated things but they made me unhappy so I decided to a) sto…'

Course:MATH110/Archive/2010-2011/003/Math Forum

2010-09-13T04:00:31Z

VictoriaBass: /* Exercise 22 */

This space is meant to be organized by the students. The instructor will help organize the wiki, but won't necessarily interfere with the math content.

For example, you could ask more details about the quiz of Lecture 2. Or get some help understanding something from the reading in section 3.1 or 4.2.

==Question from exercise 3.1==
===Exercise 22===
Hello there, I am stuck on question 22. It seems straightforward but I have to be doing something wrong. The question is :

"Find the side of a square such that when each side is increased by 6 in., the area of the square is increased by 156 sq. in."

I tried solving it by

Let x in. be the original length of each side of the square.

<math>(x+6)^2-x^2=156</math>

<math>x^2+36-x^2=156</math>

That didn't make sense. I'd appreciate if somebody can point out my mistake. I know this should be easy, but I can't figure out what I have got wrong there.

Regards, Bella Tory

Hey Bella,
So this is what I did:

If x is the length of the square then the area would be x^2. But the length is increased by 6 so the length is now (x+6). Therefore the area will be (x+6)^2.
The question notes that the area of the square will increase by 156. So if the area to begin with was x^2 then the area will now be x^2+156.
That means your equation is:
(x+6)^2 = x^2 + 156

Then you just solve for x:
(x+6)^2 = x^2 + 156
x^2+12x+36=x^2+156
x^2+12x-x^2=156-36
12x=120
x=120/12
x=10

Then you can check:
If x=10 the original area was 10^2=100.
If the length is increased by 6 or (10+6)=16 then the area would be 16^2 = 256 which is 156 more than 100.

Hope that helps. =) Victoria

==Arithmetic Skills Test==
Here is the original pdf file of the [[Media:Arithmetic Skills Test.pdf | Arithmetic Skills Test]].
===Problem 1 - Solution===

Here's a solution for this problem. First, we rewrite the expression as:

: <math>-\frac{7}{51} \div \frac{3}{12}</math>

Now, we recall that dividing by a fraction is equivalent to multiply by its inverse (have a look [http://cs.gmu.edu/cne/modules/dau/algebra/fractions/frac4_frm.html here] if you want more details). This gives us:

: <math>= -\frac{7}{51} \times \frac{12}{3} = -\frac{7 \times 12}{51 \times 3}</math>

Before multiplying out, we simplify (it makes computations more simple, I guess that's why we call it simplification):

: <math>= -\frac{7 \times 4}{51 \times 1} = - \frac{28}{51}</math>

Which is the correct answer.

===Problem 3 - Solution===

There are a few different ways to solve this problem.

====An important fact about fractions====
In all cases, it is important to remember the following rule:

: <math>x^{-n} = \frac{1}{x^n}</math>

This means that when you see a negative exponent, it is a short notation to mean one divided by the rest of the expression. Now, what can be a bit confusing, is what about the expression:

: <math> \frac{1}{x^{-n}} </math>

Well, let's do this slowly. First, we see a division, let us rewrite it with a nice division symbol for a second:

: <math> =1 \div x^{-n} </math>

Now, we rewrite the exponent without the negative sign:

: <math> =1 \div \frac{1}{x^n} </math>

Well, since we're dividing by a fraction, we remember from problem 1 that it means we need to actually multiply by its inverse, which gives:

: <math> =1 \times \frac{x^n}{1} = x^n </math>

So, putting everything together, we just computed that:

: <math>\frac{1}{x^{-n}} = x^n</math>

Actually, there's a slightly shorter way to do this, using this time the notation instead of getting rid of it. Think like this, the negative exponent notation means something like ''one divided by...'' so actually, we have that

: <math> \frac{1}{x^{-n}} = (x^{-n})^{-1} = x^{(-n) \times (-1)} = x^n</math>

: since <math>\frac{1}{A} = A^{-1} </math> for any term <math>A</math>.

====First solution====
So, taking all this together, we can finally solve this problem. We first start by simplifying the expression inside of the parenthesis:

: <math>\frac{x^{-4}}{x^{-7}} = x^{-4} \div x^{-7} </math>

Then we remove the negative exponents:

: <math> =\frac{1}{x^4} \div \frac{1}{x^7} = \frac{1}{x^4} \times \frac{x^7}{1} = \frac{x^7}{x^4} = x^3</math>

So the problem can be now rewritten as:

: <math> \left( \frac{x^{-4}}{x^{-7}} \right)^{-2} = \left( x^3 \right)^{-2} = x^{3 \times (-2)} = x^{-6}</math>

Which is the answer.

====Second solution====
Another solution for example, consists of using the fact that

: <math>\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n} </math>

This allows to solve this problem as follow:

: <math> \left( \frac{x^{-4}}{x^{-7}} \right)^{-2} = \frac{\left( x^{-4} \right)^{-2}}{\left( x^{-7} \right)^{-2}} = \frac{x^{(-4) \times (-2)}}{x^{(-7) \times (-2)}} = \frac{x^8}{x^{14}} = \frac{1}{x^6} = x^{-6}</math>

Course:MATH110/Archive/2010-2011/003/FAQ

2010-09-09T04:25:00Z

VictoriaBass: /* When is the final exam? */

Course:MATH110/Archive/2010-2011/003

2010-09-09T04:20:52Z

VictoriaBass: /* FAQ */

[[Image:1888086353_715278a4bc.jpg|right|frame|Uploaded to Flickr by [http://www.flickr.com/photos/simeon_barkas/1888086353/ Akbar Simonse]]]

Welcome to the course's wiki for the academic year 2010/2011.

==Basic Informations==
* Course: Math 110, section 003 - Differentiable Calculus
* Instructor: [http://www.math.ubc.ca/~dkohler David Kohler]
* Office: AA 120
* Office hours: TBA
* Class schedule: Monday/Wednesday/Friday from 8:00 am to 8:50
* Classroom: IBLC 261
* Textbook: A custom edition of Calculus: Early Transcendentals, Single Variable by Briggs and Cochran is the required textbook. (The custom edition is simply an abbreviated version of the regular edition.) The accompanying custom Student Solutions Manual is also required, as is a supplementary text, the 3rd edition of Just-In-Time Algebra and Trigonometry for Calculus by Mueller and Brent. All of these resources may be purchased as a package from the UBC Bookstore.
* Clickers: This course will make use of [http://web.ubc.ca/okanagan/ctl/elearning/clickers.html iClickers]. A lot of various courses at UBC make use of this technology. You can get your iClicker at the UBC Bookstore.

==Course description==
Math 110 is a two-term course in differential calculus with additional material designed to strengthen understanding of essential pre-calculus topics. In addition to regular lectures, students in MATH 110 are assigned to workshop sections which meet weekly under the guidance of graduate and undergraduate teaching assistants. Workshops will consist of problem-solving sessions, projects and quizzes. They are an integral part of the course, and attendance is mandatory.

You can find the course outline [http://www.math.ubc.ca/~fsl/110.outline.pdf here]. It contains some useful information about the different components of the course, the grading scheme and and a weekly schedule which contains an approximative breakdown of topics and some important dates, such as assignments and midterms.

==Reading assignments==
Daily reading assignments will be described on the [[/Reading | Reading Assignment]] page.

==Homework assignments==
There will be a assigned homework almost every week. The [[/Homework | Homework Assignement]] page will contain all the relevant information.

==Classroom policies==
The [[/Classroom_policies | Classroom policies]] page contains all the rules that have been agreed on by the class.

==FAQ==
The [[/FAQ | Frequently Asked Questions]] page highlights common questions and provides tips and answers.

Course:MATH110/Archive/2010-2011/003

2010-09-09T04:20:32Z

VictoriaBass: /* FAQ */