2009 AMC 8 Problems/Problem 14

Problem

Austin and Temple are $50$ miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?

$\textbf{(A)}\ 46 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 54$

Solution

The way to Temple took 50/60=5/6 hours, and the way back took 50/40=5/4 for a total of 5/6 + 5/4 = 25/12 hours. The trip is 50*2=100 miles. The average speed is 100 over 25/12 =48 miles per hour.

Solution 2

This question simply asks for the harmonic mean of $60$ and $40$ , regardless of how far Austin and Temple are. Plugging in, we have: $\frac{2ab}{a+b} = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = \boxed{\textbf{(B)}\ 48}$ miles per hour.

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2009 AMC 8 Problems/Problem 14

Contents

Problem

Solution

Solution 2

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