# 2009 AMC 10B Problems/Problem 8

The following problem is from both the 2009 AMC 10B #8 and 2009 AMC 12B #7, so both problems redirect to this page.

## Problem

In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$

$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$

## Solution

Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$, the price decreased by $0.2p$ during April, and the percent decrease was $$x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.$$ So to the nearest integer $x$ is $\boxed{17}$. The answer is $\mathrm{(B)}$.

## Solution 2

Without loss of generality, we can assume the price at the beginning of January was $100$.

When it rose by $20\%$, it became $120$, when it fell by $20\%$, it became $96$, and when it rose by $25\%$, it became $120$ again.

In order for the price at the end of April to be the same as it was at the beginning of January ($100$), the price must decrease by $20$.

20 is $\frac{1}{6}th$ of 120, and $\frac{1}{6} \approx 0.167 \approx 17\%$ So to the nearest integer, $x = 17$ and the answer is $\boxed{\textbf{(B) } 17}$. ~azc1027