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2008 AMC 12A Problems/Problem 8

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Problem

What is the volume of a cube whose surface area is twice that of a cube with volume 1?

\mathrm{(A)}\ \sqrt{2}\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 2\sqrt{2}\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 8

Solution

A cube with volume 1 has a side of length \sqrt[3]{1}=1 and thus a surface area of 6 \cdot 1^2=6.

A cube whose surface area is 6\cdot2=12 has a side of length \sqrt{\frac{12}{6}}=\sqrt{2} and a volume of (\sqrt{2})^3=2\sqrt{2}\Rightarrow\mathrm{(C)}.

See Also

2008 AMC 12A (ProblemsResources)
Preceded by
Problem 7
Followed by
Problem 9
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
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