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2013 AMC 10B Problem 8

2013 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

Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?

A)
$10$
B)
$16$
C)
$25$
D)
$30$
E)
$40$

Answer

Key insight

Combined mileage is total miles over total gallons; with equal distances, choose a convenient distance like 40 miles and count gallons.

Solution

The combined rate is the total distance divided by the total gasoline used. Since the distance is the same for both, pick a convenient one: suppose each drives $40$ miles.

Ray uses $\frac{40}{40} = 1$ gallon. Tom uses $\frac{40}{10} = 4$ gallons.

Together they cover $80$ miles on $5$ gallons, so the combined rate is
$\frac{80}{5} = 16 \text{ miles per gallon}.$

The answer is $\boxed{\textbf{(B)}\ 16}$ .

Why this works

Miles per gallon is a ratio, and ratios combine by adding numerators and denominators separately, not by averaging. For equal distances $d$ , the combined rate is $\frac{2d}{d/40 + d/10}$ , the harmonic mean of $40$ and $10$ ; the $d$ cancels, which is why any convenient distance works.

Alternative approach

Keep $d$ general: total gas $= \frac{d}{40} + \frac{d}{10} = \frac{5d}{40} = \frac{d}{8}$ , so the rate is $\frac{2d}{d/8} = 16$ .

The trap

Averaging 40 and 10 to get 25; the cars burn different amounts of gas for the same distance, so the poorer car dominates.

Common mistakes

Averaging 40 and 10 to get 25; the cars burn different amounts of gas for the same distance, so the poorer car dominates.
Dividing total gallons by total miles, or computing the combined rate for equal gallons (which would give $25$ ).

Techniques

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