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 alot, and their combined output decreases by 10 bricks per hour. Working together, they build the chimney in 5 hours. How many bricks are in the chimney? | |
| + | |||
| + | A) 500 B) 900 C) 950 D) 1000 E) 1900 | ||
==Solution== | ==Solution== | ||
| − | + | Let x be the number of bricks in the chimney. Using distance = rate * time, we get | |
| + | x = (x/9 + x/10 - 10)(5) | ||
| + | solving for x, we get 900 (B) | ||
==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 11:21, 4 January 2009
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 alot, and their combined output decreases by 10 bricks per hour. Working together, they build the chimney in 5 hours. How many bricks are in the chimney?
A) 500 B) 900 C) 950 D) 1000 E) 1900
Solution
Let x be the number of bricks in the chimney. Using distance = rate * time, we get x = (x/9 + x/10 - 10)(5) solving for x, we get 900 (B)
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 | ||
