2015 AMC 10A Problem 9

Try real numbers first, since percent questions are easy to get backwards.

1. Give the first cylinder radius $10$. The second radius is $10\%$ more, so $11$.
2. The volume of a cylinder is $\pi \times \text{radius}^2 \times \text{height}$. The base areas are $\pi \times 100$ and $\pi \times 121$.
3. The volumes are equal, so
$$100 \times (\text{first height}) = 121 \times (\text{second height}).$$
4. Pick a second height of $100$. Then the first height is $121$.
5. Now read the choices carefully.
- Going from $100$ up to $121$ is an increase of $21$ out of $100$, which is $21\%$. So "the first height is $21\%$ more than the second" is true.
- Going from $121$ down to $100$ is a drop of $21$ out of $121$, about $17\%$. So "the second height is $21\%$ less than the first" is false.

The answer is $\boxed{\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}}$

The radius is squared in the volume, so a $10\%$ bigger radius ($\times 1.1$) makes the base $1.1 \times 1.1 = 1.21$ times as big, and the height has to shrink by the same factor to keep the volume. The second lesson is about wording: "$A$ is $21\%$ more than $B$" is not the same as "$B$ is $21\%$ less than $A$", because the percent is measured from a different starting number.

With letters: let the first cylinder have radius $r$ and height $h_1$, and the second radius $1.1r$ and height $h_2$. Equal volumes give
$$\pi r^2 h_1 = \pi (1.1r)^2 h_2 = 1.21\, \pi r^2 h_2,$$
so $h_1 = 1.21\, h_2$: the first height is $21\%$ more than the second.