# 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 $(\sqrt{2})^3=2\sqrt{2}\Rightarrow\mathrm{(C)}$.

Alternatively, we can use the fact that the surface area of a cube is directly proportional to the square of its side length. Therefore, if the surface area of a cube increases by a factor of $2$, its side length must increase by a factor of $\sqrt{2}$, meaning the new side length of the cube is $1 * \sqrt{2} = \sqrt{2}$. So, its volume is $({\sqrt{2}})^3 = 2\sqrt{2} \Rightarrow\mathrm{(C)}$.