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?
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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
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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==
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. Using <math>distance = rate \cdot time</math>, we get
x = (x/9 + x/10 - 10)(5)
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<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>.
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 15:02, 25 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 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. Using $distance = rate \cdot time$, we get $x = (x/9 + x/10 - 10)\cdot(5)$. Solving for $x$, we get $\boxed{900}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions