Difference between revisions of "2008 AMC 10B Problems/Problem 18"
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==Problem== | ==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 | + | Bricklayer Brenda would take nine hours to build a chimney alone, and bricklayer Brandon would take <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? |
− | A) 500 B) 900 C) 950 D) 1000 E) 1900 | + | <math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math> |
==Solution== | ==Solution== | ||
− | Let x be the number of bricks in the chimney. Using distance = rate | + | Let <math>x</math> be the number of bricks in the chimney. Using <math>distance = rate \cdot time</math>, we get |
− | x = (x/9 + x/10 - 10)(5) | + | <math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>. |
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==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2008|ab=B|num-b=17|num-a=19}} |
Revision as of 15:02, 25 January 2009
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 |