Difference between revisions of "2008 AMC 10B Problems/Problem 18"
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<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math> | <math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math> | ||
| − | ==Solution | + | ==Solution== |
Let <math>x</math> be the number of bricks in the chimney. Using <math>d=rt</math>, we get | Let <math>x</math> be the number of bricks in the chimney. Using <math>d=rt</math>, we get | ||
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900} \Rightarrow B</math>. | <math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900} \Rightarrow B</math>. | ||
Revision as of 23:59, 21 July 2020
Problem
Bricklayer Brenda would take nine hours to build a chimney alone, and bricklayer Brandon would take 
hours to build it alone. When they work together, they talk a lot, and their combined output decreases by bricks per hour. Working together, they build the chimney in 
hours. How many bricks are in the chimney?
Solution
Let 
be the number of bricks in the chimney. Using 
, we get
. Solving for , we get 
.
See also
| 2008 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
