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 in miles 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

Let $d$ be the length of the trip in miles. Roy used no gasoline for the 40 first miles, then used 0.02 gallons of gasoline per mile on the remaining $d - 40$ miles, for a total of $0.02 (d - 40)$ gallons. Hence, his average mileage was \[\frac{d}{0.02 (d - 40)} = 55.\] Multiplying both sides by $0.02 (d - 40)$, we get \[d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.\] Then $0.1d = 44$, so $d = \boxed{440}$. The answer is $(C)$.


Solution 3 (Answer choices)

Since Roy had 0 gasoline usage during the first 40 miles, and we know his usage is 0.02 gallons per mile after that, we can look at all 5 individual answers and simply run them through.

If Roy drove (A) 140 miles would mean he spent 140 - 40 = 100 miles on gasoline, so he used 100 * 0.02 = 2 gallons. $2/140$ = 70 miles per gallon which does not equal 55, so (A) is incorrect. Similarly, (B) 240 gives: 240 - 40 = 200 miles on gasoline, 200 * 0.02 gallons = 4 gallons, $4/240$ = 60 m/g which does not equal 55.

Plugging in (C) 440, we get: 440 - 40 = 400, 400 * 0.02 = 8, $8/440$ = 55 miles per gallon, so answer choice $\boxed{440 \ \mathbf{(C)}}$ is correct.

"And we're done."

  • Richard Rusczyk outro*

~@BeepTheSheep954

Video Solution

https://youtu.be/HQmkIPpuIEc

~savannahsolver

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