2011 AMC 10A Problems/Problem 15

Problem 15

Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?

$\mathrm{(A)}\ 140 \qquad \mathrm{(B)}\ 240 \qquad \mathrm{(C)}\ 440 \qquad \mathrm{(D)}\ 640 \qquad \mathrm{(E)}\ 840$

Solution 1

We know that $\frac{\text{total miles}}{\text{total gas}}=55$ . Let $x$ be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is $0.02x$ . The total distance traveled is $40+x$ , so we get $\frac{40+x}{0.02x}=55$ . Solving this equation, we get $x=400$ , so the total distance is $400 + 40 = \boxed{440 \ \mathbf{(C)}}$ .

Solution 2 : Using the Answer Choices

The answer has to be divisible by $55$ , and the only answer that is divisible by 55 is $\boxed{440 \ \mathbf{(C)}}$ .

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 AMC 10 Problems and Solutions

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

Page

Toolbox

Search

2011 AMC 10A Problems/Problem 15

Contents

Problem 15

Solution 1

Solution 2 : Using the Answer Choices

See Also