Difference between revisions of "2011 AMC 10A Problems/Problem 15"
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<cmath>d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.</cmath> | <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>. | Then <math>0.1d = 44</math>, so <math>d = \boxed{440}</math>. The answer is <math>(C)</math>. | ||
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+ | == 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. | ||
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+ | If Roy drove (A) 140 miles would mean he spent 140 - 40 = 100 miles on gasoline, so he used 100 * 0.02 = 2 gallons. <math>2/140</math> = 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, <math>4/240</math> = 60 m/g which does not equal 55. | ||
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+ | Plugging in (C) 440, we get: | ||
+ | 440 - 40 = 400, 400 * 0.02 = 8, <math>8/440</math> = 55 miles per gallon, so answer choice | ||
==Video Solution== | ==Video Solution== |
Revision as of 01:15, 16 July 2024
Contents
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
.
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. = 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,
= 60 m/g which does not equal 55.
Plugging in (C) 440, we get:
440 - 40 = 400, 400 * 0.02 = 8, = 55 miles per gallon, so answer choice
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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