2011 AMC 10A Problems/Problem 15

Revision as of 16:56, 30 December 2017 by Murtagh (talk | contribs) (Solution 2 : Using the Answer Choices)

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

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

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png