Difference between revisions of "2008 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | Let <math>h</math> be the number of bricks in the | + | Let <math>h</math> be the number of bricks in the chimney. |
Without talking, Brenda and Brandon lay <math>\frac{h}{9}</math> and <math>\frac{h}{10}</math> bricks per hour respectively, so together they lay <math>\frac{h}{9}+\frac{h}{10}-10</math> per hour together. | Without talking, Brenda and Brandon lay <math>\frac{h}{9}</math> and <math>\frac{h}{10}</math> bricks per hour respectively, so together they lay <math>\frac{h}{9}+\frac{h}{10}-10</math> per hour together. | ||
| − | Since they finish the chimney in <math>5</math> hours, <math>h=5\left( \frac{h}{9}+\frac{h}{10}-10 \right)</math>. Thus, <math>h=900 \Rightarrow B</math>. | + | Since they finish the chimney in <math>5</math> hours, <math>h=5\left( \frac{h}{9}+\frac{h}{10}-10 \right)</math>. Thus, <math>h=900 \Rightarrow B</math>. |
==See Also== | ==See Also== | ||
Revision as of 18:47, 31 January 2014
Problem
Bricklayer Brenda would take 
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 is decreased 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.
Without talking, Brenda and Brandon lay 
and 
bricks per hour respectively, so together they lay 
per hour together.
Since they finish the chimney in hours, 
. Thus, 
.
See Also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
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