# 1957 AHSME Problems/Problem 39

## Problem

Two men set out at the same time to walk towards each other from $M$ and $N$, $72$ miles apart. The first man walks at the rate of $4$ mph. The second man walks $2$ miles the first hour, $2\tfrac {1}{2}$ miles the second hour, $3$ miles the third hour, and so on in arithmetic progression. Then the men will meet:

$\textbf{(A)}\ \text{in 7 hours} \qquad \textbf{(B)}\ \text{in }{8\frac {1}{4}}\text{ hours}\qquad \textbf{(C)}\ \text{nearer }{M}\text{ than }{N}\qquad\\ \textbf{(D)}\ \text{nearer }{N}\text{ than }{M}\qquad \textbf{(E)}\ \text{midway between }{M}\text{ and }{N}$

## Solution

Let $t$ be the time it takes (in hours) for the men to meet. Then, the distance the first man will have travelled is $4t$. If $t$ is an integer, the second man will have travelled $\tfrac{2+2+0.5(t-1)}{2} \cdot t$, the sum of the first $t$ terms of the arithmetic progression. If $t$ is not an integer, then we should have a good enough approximation to choose an answer choice (given the relatively large distance of $72$ miles compared to their speeds). Because the combined distance that the men travel must be $72$ miles, we can now solve for $t$ in the following formula: \begin{align*} 4t+\frac{2+2+0.5(t-1)}{2} \cdot t &= 72 \\ 4t+\frac{4+0.5t-0.5}{2} \cdot t &= 72 \\ 8t+3.5t+0.5t^2 &= 144 \\ t^2+23t &= 288 \\ t^2+23t-288 &= 0 \\ (t+32)(t-9) &= 0 \end{align*} Because $t>0$, we are left with $t=9$. At $t=9$, the first man will have travelled $36$ miles, which is half of te distance between $M$ and $N$. Thus, our answer is $\boxed{\textbf{(E)} \text{ midway between } M \text{ and } N}$.