1959 AHSME Problems/Problem 30

Problem

$A$ can run around a circular track in $40$ seconds. $B$ , running in the opposite direction, meets $A$ every $15$ seconds. What is $B$ 's time to run around the track, expressed in seconds? $\textbf{(A)}\ 12\frac12 \qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 27\frac12\qquad\textbf{(E)}\ 55$

Solution

In $15$ seconds, A will complete $\frac{3}{8}$ of the track. This means that B will complete $\frac{5}{8}$ of the track in $15$ seconds, meaning that to complete the whole track (making the fraction 1), it will take $\frac{8}{5} \cdot 15 = 24$ seconds. So the answer is $\boxed{\textbf{(B) }24}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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1959 AHSME Problems/Problem 30

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