2009 AMC 8 Problems/Problem 14

Revision as of 14:33, 14 August 2021 by Raina0708 (talk | contribs) (Solution)

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.

See Also

2015 Problem 17

2009 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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