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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 <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?
+
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?
  
 
<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math>
 
<math>\mathrm{(A)}\ 500\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 950\qquad\mathrm{(D)}\ 1000\qquad\mathrm{(E)}\ 1900</math>
  
 
==Solution==
 
==Solution==
−
Let <math>x</math> be the number of bricks in the chimney. Using <math>d=rt</math>, we get
+
Let <math>x</math> be the number of bricks in the chimney. The work done is the rate multiplied by the time.
−
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{\textbf{(B)} \text{900}}</math>.
+
 
 +
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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:19, 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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