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 1== |
Let <math>x</math> be the number of bricks in the chimney. Using <math>distance = rate \cdot time</math>, we get | Let <math>x</math> be the number of bricks in the chimney. Using <math>distance = rate \cdot time</math>, we get | ||
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>. | <math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>. | ||
Revision as of 19:12, 23 January 2017
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 1
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. 
