2016 AMC 10A Problem 5

1. The sides are in the ratio $1 : 3 : 4$ and are whole numbers, so they are $k$, $3k$ and $4k$ for some positive whole number $k$.
2. The volume is
$$k \cdot 3k \cdot 4k = 12 \cdot k \cdot k \cdot k = 12k^3.$$
3. List the possible volumes by trying small $k$:

1. The volumes jump from $96$ straight to $324$, so the only choice on the list is $96$.

The answer is $\boxed{\textbf{(D)}\ 96}$.

A ratio of whole numbers can be written with one scaling number $k$. Since volume multiplies three lengths, it grows like $k^3$, so only a few volumes are possible. Being a multiple of $12$ is not enough; what is left after dividing by $12$ has to be a perfect cube.

Write the side lengths as $k$, $3k$, $4k$ for some positive integer $k$. The volume is
$$k \cdot 3k \cdot 4k = 12k^3.$$

So the volume must be $12$ times a perfect cube: $12 \cdot 1 = 12$, $12 \cdot 8 = 96$, $12 \cdot 27 = 324$, and so on. Among the choices, only $96 = 12 \cdot 2^3$ has this form (a $2 \times 6 \times 8$ box).

The answer is $\boxed{\textbf{(D)}\ 96}$.