1963 AHSME Problems/Problem 20
Problem
Two men at points and , miles apart, set out at the same time to walk towards each other. The man at walks uniformly at the rate of miles per hour; the man at walks at the constant rate of miles per hour for the first hour, at miles per hour for the second hour, and so on, in arithmetic progression. If the men meet miles nearer than in an integral number of hours, then is:
Solution
First, find the number of hours it takes for the two to meet together. After hours, the person at walks miles. In the same amount of time, the person at has been walking at mph for the past hour, so the person walks miles.
In order for both to meet, the sum of both of the distances walked must total miles, so Since must be positive, . Because it takes hours to meet, the person from traveled miles while the person from traveled miles. Thus, they are miles closer to than , so the answer is .
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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