2009 AMC 10B Problem 8

Percent changes are easiest to follow with a real price, so pretend gasoline cost $\$100$ at the start of January.

1. January, up $20\%$: $20\%$ of $100$ is $20$, so the price becomes $\$120$.
2. February, down $20\%$: $20\%$ of $120$ is $24$, so the price becomes $\$96$. (Notice it did not go back to $\$100$.)
3. March, up $25\%$: $25\%$ of $96$ is $24$, so the price becomes $\$120$.
4. April must bring the price back to $\$100$. That is a drop of $\$20$.
5. As a percent of the April starting price: $\frac{20}{120} = \frac16 \approx 0.1667$, which is about $16.67\%$.
6. To the nearest integer, $x = 17$.

The answer is $\boxed{\textbf{(B)}\ 17}$.

Each percent change is taken of the current price, not the original one. That is why "up $20\%$, then down $20\%$" does not cancel. Picking $\$100$ as the starting price turns every step into simple arithmetic, and the final percent is the drop divided by the price just before the drop.

Write each change as a multiplier: up $20\%$ means $\times 1.2$, down $20\%$ means $\times 0.8$, up $25\%$ means $\times 1.25$. So after March the price has been multiplied by $1.2 \times 0.8 \times 1.25 = 1.2$, which is $\frac65$. To get back to the start, April must multiply by $\frac56$, which removes $\frac16$ of the price, about $17\%$.

Track the price as a product of monthly factors. Starting from $P$:
$$P \cdot 1.2 \cdot 0.8 \cdot 1.25 = P \cdot 0.96 \cdot 1.25 = 1.2P.$$
At the end of March the price is $1.2P$, or $\frac{6}{5}P$. To return to $P$, April must multiply by $\frac56$, which is a decrease of $1 - \frac56 = \frac16 \approx 16.67\%$.

To the nearest integer, $x = 17$.

The answer is $\boxed{\textbf{(B)}\ 17}$.