1954 AHSME Problems/Problem 48
Problem
A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at of its former rate and arrives hours late. Had the accident happened miles farther along the line, it would have arrived only hours late. The length of the trip in miles was:
Solution
Let the speed of the train be miles per hour, and let miles be the total distance of the trip, where and are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled miles after the crash. In other words, the train travelled for after the crash. So, not including the half hour detention after the crash, the entire trip took Now let us consider the alternative trip, where the accident happened miles farther along the line. Originally, the accident happened down the line, but in this situation it happens miles from the line. Therefore the accident happens into the trip. The length of the remaining part of the trip is now , so the remaining trip takes . So, not including the half hour detention after the crash, the entire trip took We are given that the trip that actually happened resulted in being hours late, while the alternative trip would have resulted in being only hours late. Therefore the first trip took exactly hour more: After simplification and cancellation, we get the equation Therefore , so . We solve for and have the original speed of the train: .
We must now solve for the length of the actual trip. If the train had gone for the entire trip, then it would have taken , and the train would have been on time. But the trip actually took hours, not including the half-hour detention, and the train was hours late. Therefore The length of the trip in miles was .
See also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 47 |
Followed by Problem 49 | |
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