Difference between revisions of "2009 AMC 8 Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
| − | This question simply asks for the harmonic mean of <math>60</math> and <math>40</math>, regardless of how far Austin and Temple are. | + | This question simply asks for the [[harmonic mean]] of <math>60</math> and <math>40</math>, regardless of how far Austin and Temple are. |
Plugging in, we have: <math>\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}</math> miles per hour. | Plugging in, we have: <math>\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}</math> miles per hour. | ||
Revision as of 19:00, 29 September 2018
Contents
Problem
Austin and Temple are 
miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging 
miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged 
miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
Solution
The way to Temple took 
hours, and the way back took 
for a total of 
hours. The trip is 
miles. The average speed is 
miles per hour.
Solution 2
This question simply asks for the harmonic mean of and , regardless of how far Austin and Temple are.
Plugging in, we have: 
miles per hour.
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
