2008 AMC 10B Problems/Problem 18

Revision as of 18:56, 28 August 2022 by Bearjere (talk | contribs) (LaTeX parenthesis fix)

Problem

Bricklayer Brenda would take nine 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 decreases by $10$ bricks per hour. Working together, they build the chimney in $5$ hours. How many bricks are in the chimney?

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

Solution

Let $x$ be the number of bricks in the chimney. The work done is the rate multiplied by the time.

Using $w = rt$, we get $x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)$. Solving for $x$, we get $\boxed{\textbf{(B)} \text{900}}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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 10 Problems and Solutions

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