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 $\frac{50}{60}=\frac56$ hours, and the way back took $\frac{50}{40}=\frac54$ for a total of $\frac56 + \frac54 = \frac{25}{12}$ hours. The trip is $50\cdot2=100$ miles. The average speed is $\frac{100}{25/12} = \boxed{\textbf{(B)}\ 48}$ miles per hour.

Solution 2

We calculate the harmonic mean of Austin and Temple. 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.

Solution 3

The 50 miles part of this question is redundant. We find that the average speed is skewed towards the lower speed, since you spend more time in the lower speed. So, we take the LCM of 40 and 60, getting 120. So, we do 40 + 40 + 40 + 60 + 60/5, calculation the ratio of the parts and taking the mean. Solving, we get 240/5 = \textbf{(B)}\ 48 \qquad -themathgood

Video Solution

https://youtu.be/CnWuadfT3xA Soo, DRMS, NM

https://www.youtube.com/watch?v=-O5zQT9mp2I ~David

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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