Difference between revisions of "2009 AMC 12A Problems/Problem 5"

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{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #5]] and [[2009 AMC 10A Problems|2009 AMC 10A #11]]}}
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== Problem ==
 
== Problem ==
 
One dimension of a cube is increased by <math>1</math>, another is decreased by <math>1</math>, and the third is left unchanged. The volume of the new rectangular solid is <math>5</math> less than that of the cube. What was the volume of the cube?
 
One dimension of a cube is increased by <math>1</math>, another is decreased by <math>1</math>, and the third is left unchanged. The volume of the new rectangular solid is <math>5</math> less than that of the cube. What was the volume of the cube?
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{{AMC12 box|year=2009|ab=A|num-b=4|num-a=6}}
 
{{AMC12 box|year=2009|ab=A|num-b=4|num-a=6}}
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{{AMC10 box|year=2009|ab=A|num-b=10|num-a=12}}

Revision as of 04:07, 13 February 2009

The following problem is from both the 2009 AMC 12A #5 and 2009 AMC 10A #11, so both problems redirect to this page.

Problem

One dimension of a cube is increased by $1$, another is decreased by $1$, and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 216$

Solution

Let the original cube have edge length $a$. Then its volume is $a^3$. The new box has dimensions $a-1$, $a$, and $a+1$, hence its volume is $(a-1)a(a+1) = a^3-a$. The difference between the two volumes is $a$. As we are given that the difference is $5$, we have $a=5$, and the volume of the original cube was $5^3 = \boxed{125}$.

See Also

2009 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 12 Problems and Solutions
2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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