Difference between revisions of "2008 AMC 10B Problems/Problem 18"
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==Problem== | ==Problem== | ||
| − | Bricklayer Brenda | + | Bricklayer Brenda takes <math>9</math> hours to build a chimney alone, and bricklayer Brandon takes <math>10</math> hours to build it alone. When they work together, they talk a lot, and their combined output decreases by <math>10</math> bricks per hour. Working together, they build the chimney in <math>5</math> hours. How many bricks are in the chimney? |
<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> | ||
Revision as of 19:10, 23 March 2023
Problem
Bricklayer Brenda takes 
hours to build a chimney alone, and bricklayer Brandon takes 
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. The work done is the rate multiplied by the time.
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. 
