1963 AHSME Problems/Problem 20

Problem

Two men at points $R$ and $S$ , $76$ miles apart, set out at the same time to walk towards each other. The man at $R$ walks uniformly at the rate of $4\tfrac{1}{2}$ miles per hour; the man at $S$ walks at the constant rate of $3\tfrac{1}{4}$ miles per hour for the first hour, at $3\tfrac{3}{4}$ miles per hour for the second hour, and so on, in arithmetic progression. If the men meet $x$ miles nearer $R$ than $S$ in an integral number of hours, then $x$ is:

$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 2$

Solution

First, find the number of hours it takes for the two to meet together. After $h$ hours, the person at $R$ walks $4.5h$ miles. In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\frac{h(6.5+0.5(h-1))}{2}$ miles.

In order for both to meet, the sum of both of the distances walked must total $76$ miles, so $4.5h + \frac{h(6+0.5h)}{2} = 76$ $4.5h + 3h + 0.25h^2 = 76$ $0.25h^2 + 7.5h - 76 = 0$ $h^2 + 30h - 304 = 0$ $(h + 38)(h - 8) = 0$ Since $h$ must be positive, $h = 8$ . Because it takes $8$ hours to meet, the person from $R$ traveled $36$ miles while the person from $S$ traveled $40$ miles. Thus, they are $4$ miles closer to $R$ than $S$ , so the answer is $\boxed{\textbf{(D)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1963 AHSME Problems/Problem 20

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