1950 AHSME Problems/Problem 28
Contents
[hide]Problem
Two boys and start at the same time to ride from Port Jervis to Poughkeepsie, miles away. travels miles an hour slower than . reaches Poughkeepsie and at once turns back meeting miles from Poughkeepsie. The rate of was:
Solution
Let the speed of boy be , and the speed of boy be . Notice that travels miles per hour slower than boy , so we can replace with .
Now let us see the distances that the boys each travel. Boy travels miles, and boy travels miles. Now, we can use to make an equation, where we set the time to be equal: Cross-multiplying gives . Isolating the variable , we get the equation , so .
Alternate Solution
Note that travels miles in the time it takes to travel miles. Thus, travels more miles in the given time, meaning have passed, as goes miles per hour faster. Thus, travels miles per hours, or miles per hour.
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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