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2017 AMC 10B Problem 12

2017 AMC 10B Problem 12AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 2 of 5 · about 3 min · position P11-15

Elmer's new car gives $50\%$ better fuel efficiency. However, the new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline the old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?

A)
$20\%$
B)
$26\tfrac23\%$
C)
$27\tfrac79\%$
D)
$33\tfrac13\%$
E)
$66\tfrac23\%$

Answer

Key insight

Fuel used scales by 1/1.5 = 2/3 and price by 1.2, so cost scales by (2/3)(1.2) = 0.8, a 20 percent saving.

Solution

Pick convenient numbers. Let the old car travel $10$ km per liter and let gasoline cost $\$1$ per liter, and take a $30$ km trip.

Old car: uses $30/10 = 3$ liters at $\$1$ each, so the trip costs $\$3$ .

New car: $50\%$ better efficiency means $15$ km per liter, so it uses $30/15 = 2$ liters. Diesel costs $20\%$ more, $\$1.20$ per liter, so the trip costs $2 \cdot 1.20 = \$2.40$ .

Savings: $\$3 - \$2.40 = \$0.60$ , which is $\frac{0.60}{3} = 20\%$ of the old cost.

The answer is $\boxed{\textbf{(A)}\ 20\%}$ .

Why this works

Percent changes compose by multiplication, not addition. Better efficiency divides fuel use by $1.5$ (multiplier $\frac23$ ) while the price multiplies by $1.2$ ; the product $\frac23 \cdot 1.2 = 0.8$ is the new cost relative to the old. Concrete numbers protect against inverting the efficiency factor.

The trap

Subtracting the percents (50 - 20 = 30) or thinking 50 percent better efficiency halves the fuel used.

Common mistakes

Subtracting the percents (50 - 20 = 30) or thinking 50 percent better efficiency halves the fuel used.
Computing the saving as a percent of the new cost ( $0.60/2.40 = 25\%$ ) instead of the old cost.

Techniques

Test small/specific values or special cases to find or verify the answer