Difference between revisions of "2007 AMC 12B Problems/Problem 2"
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<math>\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28</math> | <math>\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28</math> | ||
| − | ==Solution== | + | ==Solution 1== |
The trip was <math>240</math> miles long and took <math>\dfrac{120}{30}+\dfrac{120}{20}=4+6=10</math> gallons. Therefore, the average mileage was <math>\dfrac{240}{10}= \boxed{\mathrm{(B) \ } 24}</math> | The trip was <math>240</math> miles long and took <math>\dfrac{120}{30}+\dfrac{120}{20}=4+6=10</math> gallons. Therefore, the average mileage was <math>\dfrac{240}{10}= \boxed{\mathrm{(B) \ } 24}</math> | ||
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| + | ==Solution 2== | ||
| + | Alternatively, we can use the harmonic mean to get <math>\frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{1}{12}} = 24</math> <math>\longrightarrow \boxed{(\textbf{B})}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 22:26, 12 January 2018
- The following problem is from both the 2007 AMC 12B #2 and 2007 AMC 10B #3, so both problems redirect to this page.
Contents
Problem
A college student drove his compact car 
miles home for the weekend and averaged 
miles per gallon. On the return trip the student drove his parents' SUV and averaged only 
miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
Solution 1
The trip was 
miles long and took 
gallons. Therefore, the average mileage was 
Solution 2
Alternatively, we can use the harmonic mean to get 
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 12 Problems and Solutions | |
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
