Difference between revisions of "2011 AMC 10A Problems/Problem 15"
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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? | 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? | ||
− | + | <math> | |
+ | \mathrm{(A)}\ 140 \qquad | ||
+ | \mathrm{(B)}\ 240 \qquad | ||
+ | \mathrm{(C)}\ 440 \qquad | ||
+ | \mathrm{(D)}\ 640 \qquad | ||
+ | \mathrm{(E)}\ 840 | ||
+ | </math> | ||
== Solution == | == Solution == |
Revision as of 19:52, 23 November 2016
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?
Solution
We know that . Let be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is . The total distance traveled is , so we get . Solving this equation, we get , so the total distance is .
Solution 2
The answer has to be divisble by 55, and the only answer that is divisible by 55 is C. so it's C.
See Also
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 |
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