1952 AHSME Problems/Problem 10

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Problem

An automobile went up a hill at a speed of $10$ miles an hour and down the same distance at a speed of $20$ miles an hour. The average speed for the round trip was:

$\textbf{(A) \ }12\frac{1}{2}\text{mph}  \qquad \textbf{(B) \ }13\frac{1}{3}\text{mph} \qquad \textbf{(C) \ }14\frac{1}{2}\text{mph} \qquad \textbf{(D) \ }15\text{mph} \qquad \textbf{(E) \ }\text{none of these}$

Solution

Let $r$, $t$, and $d$ represent the rate, time, and distance (respectively) of each trip. We know that $rt=d$, or $r_1t_1=r_2t_2$. Because it is given that $2r_1=r_2$, $t_1=2t_2$. Furthermore, $d=10t_1=20t_2$. The average speed can be expressed as $\frac{2d}{t_1+t_2}=\frac{2(20t_2)}{2t_2+t_2}=\frac{40}{3}=\boxed{\textbf{(B)}\ 13\frac{1}{3}\text{ mph}}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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