2009 AMC 12A Problems/Problem 5

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 $5^3 = \boxed{125}$ .

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2009 AMC 12A Problems/Problem 5

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