1986 AJHSME Problems/Problem 19

Problem

At the beginning of a trip, the mileage odometer read $56,200$ miles. The driver filled the gas tank with $6$ gallons of gasoline. During the trip, the driver filled his tank again with $12$ gallons of gasoline when the odometer read $56,560$. At the end of the trip, the driver filled his tank again with $20$ gallons of gasoline. The odometer read $57,060$. To the nearest tenth, what was the car's average miles-per-gallon for the entire trip?

$\text{(A)}\ 22.5 \qquad \text{(B)}\ 22.6 \qquad \text{(C)}\ 24.0 \qquad \text{(D)}\ 26.9 \qquad \text{(E)}\ 27.5$

Solution

The first six gallons are irrelevant. We start with the odometer at $56,200$ miles, and a full gas tank. The total gas consumed by the car during the trip is equal to the total gas the driver had to buy to make the tank full again, i.e., $12+20=32$ gallons. The distance covered is $57,060 - 56,200 = 860$ miles. Hence the average MPG ratio is $860 / 32 \approx 26.9 \rightarrow \boxed{\text{D}}$.

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

This problem is problematic. The given is very vague and argumentative.