2004 AMC 10A Problem 11

1. A cylinder's volume is (area of the circular base) times (height), and the base area is $\pi r^2$.
2. Increasing the diameter by $25\%$ multiplies it, and so the radius, by $\tfrac54$.
3. The radius gets squared in the base area, so the base area is multiplied by
$$\left(\tfrac54\right)^2 = \tfrac{25}{16}.$$
4. For the volume to stay the same, the height must be multiplied by the flipped fraction, $\tfrac{16}{25}$. Check: $\tfrac{25}{16} \times \tfrac{16}{25} = 1$.
5. The new height is $\tfrac{16}{25} = 64\%$ of the old height. So it went down by
$$100\% - 64\% = 36\%.$$

The answer is $\boxed{\textbf{(C)}\ 36}$.

Volume grows with the square of the radius but only with the height itself. So a percent change in width gets squared before it reaches the volume, and the height has to undo that squared change. Working with multipliers like $\tfrac54$ and $\tfrac{16}{25}$ is cleaner than working with percents.

Try easy numbers. Take an old jar with radius $4$ and height $25$. Its volume is $\pi \times 16 \times 25 = 400\pi$. The new radius is $5$, so the new height $h$ must satisfy $\pi \times 25 \times h = 400\pi$, which gives $h = 16$. The height went from $25$ to $16$, a drop of $9$ out of $25$, which is $36\%$.

The volume of a cylinder is $\pi r^2 h$. Increasing the diameter by $25\%$ multiplies the radius by $\tfrac54$, so the factor $r^2$ is multiplied by $\left(\tfrac54\right)^2 = \tfrac{25}{16}$.

To keep $\pi r^2 h$ unchanged, the height must be multiplied by the reciprocal, $\tfrac{16}{25}$. The new height is $\tfrac{16}{25} = 64\%$ of the old one, a decrease of
$$1 - \frac{16}{25} = \frac{9}{25} = 36\%.$$

The answer is $\boxed{\textbf{(C)}\ 36}$.