2009 AMC 12A Problems/Problem 5

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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
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All AMC 12 Problems and Solutions
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