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Difference between revisions of "2010 AMC 12A Problems/Problem 9"

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<math>(3^3) - 3(2^2*3) + 2(2^3) = \boxed{7\ \textbf{(A)}}</math>
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<math>(3^3) - 3(2^2\cdot{3}) + 2(2^3) = \boxed{7\ \textbf{(A)}}</math>

Revision as of 19:41, 10 February 2010

Problem 9

A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Solution

We can use Principle of Inclusion-Exclusion to find the final volume of the cube.

There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" is $2 \times 2 \times 3$ inches and the overlap of all the cuts is $2 \times 2 \times 2$.


$(3^3) - 3(2^2\cdot{3}) + 2(2^3) = \boxed{7\ \textbf{(A)}}$

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