Pizza and Problems/Pizza and Problems 2 (3/28/08)
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Use this page to post solutions to our sixth Pizza and Problems session, held 3/28/08. Examine the original problem set.
A great time was had by all at our meeting of Pizza and Problems. We had some very cool problems, good pizza, and enthusiam and chuckles were high. Finally, we had some very good student presentations of solutions. Some students worked together on problems, arguing their case and developing ideas! A very cool strategy.
Contents |
Student Solutions to the Problems in Pizza and Problems Set #2
Problem 1: Solution by: Hans Parshall
Let the number of bricks in the wall be b.
The first bricklayer could lay b bricks in 9 hours, and so has a rate of bricklaying of 
bricks per hour.
The second bricklayer could lay b bricks in 10 hours, and so has a rate of bricklaying of 
bricks per hour.
Working together, they work at the sum of their rates minus 10 bricks per hour. Equivilently, they can together lay 
bricks per hour.
They laid b bricks in 5 hours, and so 
Simplifying,
18b = 19b - 900
b = 900
The wall has 900 bricks.
Problem 2: Solution by:
Problem 3: Solution:by: David Arnold
Unbelievably simple! Can't see why we didn't see it before. Draw EC parallel to AD.
The angle at D is twice the angle at B.
The parallelogram makes it easy to mark the remaining angles. Amazingly, triangle DCB is isosceles!
Hence, it is easy to compute the length AB.
Wow!
