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Difference between revisions of "2008 AMC 10B Problems/Problem 18"
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==Problem==
 
==Problem==
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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?
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Bricklayer Brenda takes <math>9</math> hours to build a chimney alone, and bricklayer Brandon takes <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?
  
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A) 500 B) 900 C) 950 D) 1000 E) 1900
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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>
  
 
==Solution==
 
==Solution==
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Let x be the number of bricks in the chimney. Using distance = rate * time, we get
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Let <math>x</math> be the number of bricks in the chimney. The work done is the rate multiplied by the time.
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x = (x/9 + x/10 - 10)(5)
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solving for x, we get 900 (B)
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Using <math>w = rt</math>, we get <math>x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{\mathrm{(B)}\ 900}</math>.
  
 
==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}}
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{{MAA Notice}}

Latest revision as of 20:19, 23 March 2023

Problem

Bricklayer Brenda takes $9$ hours to build a chimney alone, and bricklayer Brandon takes $10$ hours to build it alone. When they work together, they talk a lot, 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?

$\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900$

Solution

Let $x$ be the number of bricks in the chimney. The work done is the rate multiplied by the time.

Using $w = rt$, we get $x = \left(\frac{x}{9} + \frac{x}{10} - 10\right)\cdot(5)$. Solving for $x$, we get $\boxed{\mathrm{(B)}\ 900}$.

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

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