Difference between revisions of "2008 AMC 10B Problems/Problem 18"

m (Solution 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
 
Let <math>x</math> be the number of bricks in the chimney. Using <math>d=vt</math>, we get
 
Let <math>x</math> be the number of bricks in the chimney. Using <math>d=vt</math>, we get
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math>
+
<math>x = (x/9 + x/10 - 10)\cdot(5)</math>. Solving for <math>x</math>, we get <math>\boxed{900}</math> <math>\Rightarrow</math> <math>\boxed{B}</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}}

Revision as of 03:52, 7 February 2018

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 1

Let $x$ be the number of bricks in the chimney. Using $d=vt$, we get $x = (x/9 + x/10 - 10)\cdot(5)$. Solving for $x$, we get $\boxed{900}$ $\Rightarrow$ $\boxed{B}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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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