2004 AMC 10A Problem 11

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 $\boxed{\textbf{(C)}\ 36}$.

Volume depends on the radius squared and on the height to the first power. Any percent change in a linear dimension of the base gets squared when it reaches the volume, and the compensating change in height is the reciprocal of that squared factor. Working with multipliers ($\tfrac54$, $\tfrac{25}{16}$, $\tfrac{16}{25}$) rather than percentages keeps the arithmetic clean.

Plug in numbers: an original jar with radius $4$ and height $25$ has volume $400\pi$. A radius of $5$ needs height $h$ with $25h = 400$, so $h = 16$. The height went from $25$ to $16$, a $36\%$ decrease.

• Treating the change as linear and answering 25 percent, or forgetting to square the 5/4 factor.
• Reporting $64$ (the percent of the height that remains) or $56.25$ (the percent increase in base area) instead of the percent decrease.