Difference between revisions of "2009 AMC 8 Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
− | + | Thjsfnvdhkgehiuwqhjhiefakgnar 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 20:23, 10 January 2022
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
Thjsfnvdhkgehiuwqhjhiefakgnar Austin and Temple are. Plugging in, we have: miles per hour.
See Also
2015 Problem 17
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 |
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