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>. |
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/p6vVYKrAStc | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2008|ab=B|num-b=9|num-a=11}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:21, 3 January 2020
Contents
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, 
.
Video Solution
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
