1957 AHSME Problems/Problem 39
Problem
Two men set out at the same time to walk towards each other from 
and 
, 
miles apart.
The first man walks at the rate of 
mph. The second man walks 
miles the first hour, 
miles the second hour, 
miles the third hour, and so on in arithmetic progression. Then the men will meet:
Solution
Let 
be the time it takes (in hours) for the men to meet. Then, the distance the first man will have travelled is 
. If is an integer, the second man will have travelled 
, the sum of the first terms of the arithmetic progression. If is not an integer, then we should have a good enough approximation to choose an answer choice (given the relatively large distance of miles compared to their speeds). Because the combined distance that the men travel must be miles, we can now solve for 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 
, we are left with 
. At , the first man will have travelled 
miles, which is half of te distance between and . Thus, our answer is 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
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