2008 AMC 12B Problems/Problem 10

Problem

Bricklayer Brenda would take $9$ hours to build a chimney alone, and bricklayer Brandon would take $10$ hours to build it alone. When they work together they talk a lot, and their combined output is decreased by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?

$\textbf{(A)}\ 500 \qquad \textbf{(B)}\ 900 \qquad \textbf{(C)}\ 950 \qquad \textbf{(D)}\ 1000 \qquad \textbf{(E)}\ 1900$

Solution

Let $h$ be the number of bricks in the chimney.

Without talking, Brenda and Brandon lay $\frac{h}{9}$ and $\frac{h}{10}$ bricks per hour respectively, so together they lay $\frac{h}{9}+\frac{h}{10}-10$ per hour together.

Since they finish the chimney in $5$ hours, $h=5\left( \frac{h}{9}+\frac{h}{10}-10 \right)$. Thus, $h=900 \Rightarrow B$.

Video Solution

https://youtu.be/p6vVYKrAStc

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png