2010 AMC 10A Problem 12

The model is a shrunken copy of the real tower. The question is: if the volume shrinks by a certain amount, how much does the height shrink?

1. How volume changes with size. If you shrink every length (height, width, depth) to $\tfrac{1}{10}$, the volume becomes $\tfrac{1}{10} \times \tfrac{1}{10} \times \tfrac{1}{10} = \tfrac{1}{1000}$. In general, the volume shrinks by the length factor multiplied by itself three times.
2. The volume ratio here. The model holds $0.1$ liters and the real one holds $100{,}000$ liters:
$$\frac{0.1}{100{,}000} = \frac{1}{1{,}000{,}000}.$$
3. Find the length factor. Which number times itself three times gives $1{,}000{,}000$? Since $100 \times 100 \times 100 = 1{,}000{,}000$, it is $100$. So every length of the model is $\tfrac{1}{100}$ of the real length.
4. The height. $40 \div 100 = 0.4$ meters.

The shape of the tank (a sphere) never matters, which is why the choices with $\pi$ are traps.

The answer is $\boxed{\textbf{(C)}\ 0.4}$.

For a scaled copy, lengths change by some factor, areas by that factor twice (squared), and volumes by that factor three times (cubed). So to go from a volume ratio back to a length ratio, take the cube root. This works for any shape, so there is no need to use the sphere formula.