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2009 AMC 10B Problem 8

2009 AMC 10B Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 2 of 5 · about 2 min · position P1-10

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 $x$ ?

A)
$12$
B)
$17$
C)
$20$
D)
$25$
E)
$35$

Answer

Key insight

Percent changes multiply: 1.2 times 0.8 times 1.25 = 1.2, so April must multiply by 5/6, a drop of one-sixth, about 17 percent.

Solution

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 $\boxed{\textbf{(B)}\ 17}$ .

Why this works

A rise of $p\%$ followed by a fall of $p\%$ does not cancel; each change scales the current price, so the combined effect is the product of the factors. To undo a factor $f$ , multiply by $1/f$ ; the required percent drop is $1 - 1/f$ , not $f - 1$ .

Alternative approach

Use a concrete price of $\$100$ : after January $\$120$ , after February $\$96$ , after March $\$120$ . Dropping from $\$120$ to $\$100$ is a $\$20$ decrease out of $\$120$ , i.e. $\frac16 \approx 17\%$ .

The trap

Adding and subtracting the percents (20 - 20 + 25 = 25) instead of multiplying the growth factors.

Common mistakes

Adding and subtracting the percents (20 - 20 + 25 = 25) instead of multiplying the growth factors.
Computing the drop as a percent of the original price (20 out of 100, choice C) instead of the March price.

Techniques

Set up the equation/formula and compute; no special trick needed