Problem Solving Set #4

November 12, 2010
Question 15
A loonie is a $1 coin and a dime is a $0.10 coin.  One loonie has the same mass as 4 dimes.  A bag of dimes has the same mass as a bag of loonies.  The coins in the bag of loonies are worth $400 total.  How much are the coins in the bag of dimes worth?
a) $40  b) $100  c) $160  d) $1000  e) $1600

So we know that there are 400 loonies in the bag, since one loonie is one dollar and we have 400 dollars.  Now if we say that for every loonie we have four dimes, because they have the same mass.  Therrefore we times the fourty cents (the monetary value of the dimes) by the number of loonies, because that's how much we have in dimes for each loonie.  Therefore we get the equation $0.4 * 400 = the value of the bag of dimes.  The answer is c) $160.

I found this problem interesting because it sounds confusing to start, but if you think it through slowly it becomes a lot simpler.

I learned to take things step by step.  Once you find all the information you need written in the problem you simply need to write it out in a formula and solve it.

Who Wants to Be a Mathematician

October 22, 2010
On Thursday October 21 Carver's Honours Math classes went to UBC to attend a math lecture, the game Who Wants to Be a Mathematician, and a workshop.  It was a fun and informative event that was also our first ever math field trip.


Ten new things I learned were:

1. The Pythagorean Theorem may not have originally been created by Pythagoras, but rather by a Chinese mathematician.
2. The Pythagorean Theorem is always true on a flat plain.
3. There are an infinite number of Pythagorean triplets.
4. Mathematical theorems always remain true.
5. There are 7 millennium problems in the United States of America, created in 2000.  For solving one of these problems there is a million dollar reward.
6. Fermat's Last Theorem is not a millennium problem.
7. There is a million dollar reward for the solving of the formula y = x³ + ax +b.
8. The formula x³ + y³ = 1 is not possible.
9. Carl Gauss was a famous mathematician who summed all the integers from 1-100 while in elementary school.
10. The have computer designed to run math programs that can do guess and check.

The first thing we attended was the math lecture.   The talk was interesting and the professor made us all feel comfortable, even if we didn't understand everything.  It was fascinating to have him show us how to prove the Pythagorean Theorem true, especially as we frequently use it in class.  He also showed us how to prove that there are an infinite number of Pythagorean triplets, even if I didn't understand how he did it.  It was rather exciting to see how you don't need to be a brilliant mathematician to prove these theorems.  The professor also invited audience participation, which made it more interesting.  He spoke well, keeping the lecture from becoming dry.  All in all it was a fascinating experience.

After the lecture we watched the game Who Wants to Be a Mathematician in which a Carver student, Kathleen, participated in.  It was a lot of fun to watch her compete.

After lunch we attended a math workshop.  There were university students there to help us with our work.  As a grade ten student I was given a grade eight to ten worksheet and was able to do some of the questions while working with friends.  It was fun to be able just to try out the math, without worrying about it becoming homework.  One of the things I learned was when doing algebra always try to simplify as much as you can.  It also showed me that all the information given is there for a reason and to use it.  It was a learning experience.

I think we all enjoyed this field trip.

Problem Solving Set #3

October 15, 2010
Question 8
An equilateral triangle has a side length of 20.  If a square has the same perimeter as this triangle, the area of the square is:
a) 25  b)400  c)225  d)60  e)100

To solve this problem I found the perimeter of the triangle by 20+20+20= 60, since all sides are equal because it's an equilateral triangle.  Since the square has the same perimeter of 60 and it has four equal sides to find the side length you do the following equation 60/4= 15.  If a square has a side length of 15 you'd use the equation 15² = 225 to find the area of the square.  Therefore the area of the square is c)225.

This question is interesting because it shows how different shapes with the same perimeter can have vastly different areas.  For example an equilateral triangle with a side length of 20 would have an area of around 86.6 square units, whereas the square has 225 square units.  I find this rather fascinating.

In solving this problem I have discovered that problems are easy to solve when you follow logical steps and know your formulas.