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) == | ||
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| + | 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. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 00:07, 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
![\[\frac{d}{0.02 (d - 40)} = 55.\]](http://latex.artofproblemsolving.com/f/6/a/f6a772dcfc2fb689f01a1999b157ab50bd974810.png)
Multiplying both sides by , we get
![\[d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.\]](http://latex.artofproblemsolving.com/6/6/c/66c249d90e244a5fc4f853986694d554ca878d6b.png)
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.
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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