Difference between revisions of "2011 AMC 10A Problems/Problem 15"
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− | + | == Problem 15 == | |
− | ( | + | Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first <math>40</math> miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of <math>0.02</math> gallons per mile. On the whole trip he averaged <math>55</math> 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 1 == | ||
+ | |||
+ | We know that <math>\frac{\text{total miles}}{\text{total gas}}=55</math>. Let <math>x</math> be the distance in miles the car traveled during the time it ran on gasoline, then the amount of gas used is <math>0.02x</math>. The total distance traveled is <math>40+x</math>, so we get <math>\frac{40+x}{0.02x}=55</math>. Solving this equation, we get <math>x=400</math>, so the total distance is <math>400 + 40 = \boxed{440 \ \mathbf{(C)}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Let <math>d</math> 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 <math>d - 40</math> miles, for a total of <math>0.02 (d - 40)</math> gallons. Hence, his average mileage was | ||
+ | <cmath>\frac{d}{0.02 (d - 40)} = 55.</cmath> | ||
+ | Multiplying both sides by <math>0.02 (d - 40)</math>, we get | ||
+ | <cmath>d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.</cmath> | ||
+ | Then <math>0.1d = 44</math>, so <math>d = \boxed{440}</math>. The answer is <math>(C)</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/HQmkIPpuIEc | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | == See Also == | ||
+ | {{AMC10 box|year=2011|ab=A|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Revision as of 15:31, 3 September 2022
Problem 15
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of gallons per mile. On the whole trip he averaged miles per gallon. How long was the trip in miles?
Solution 1
We know that . Let be the distance in miles 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
Let 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 miles, for a total of gallons. Hence, his average mileage was Multiplying both sides by , we get Then , so . The answer is .
Video Solution
~savannahsolver
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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