1962 AHSME Problems/Problem 10
Problem
A man drives miles to the seashore in hours and minutes. He returns from the shore to the starting point in hours and minutes. Let be the average rate for the entire trip. Then the average rate for the trip going exceeds in miles per hour, by:
Solution
Since the equation for rate is , you only need to find and . The distance from the seashore to the starting point is miles, and since he makes a round trip, . Also, you know the times it took him to go both directions, so when you add them up ( hours and minutes hours and minutes), you get hours. Since , after plugging in the values of and , you get , or . But what the question asks is for the positive difference between the speed going to the seashore and the average speed. The equation is , so when you plug in and , you get . Simplifying, you get .
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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