Difference between revisions of "1954 AHSME Problems/Problem 48"
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Let the speed of the train be <math>x</math> miles per hour, and let <math>D</math> miles be the total distance of the trip, where <math>x</math> and <math>D</math> are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled <math>D-x</math> miles after the crash. In other words, the train travelled for <math>\frac{(D-x)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x}{3x/4}\text{hr}</math> after the crash. So, not including the half hour detention after the crash, the entire trip took | Let the speed of the train be <math>x</math> miles per hour, and let <math>D</math> miles be the total distance of the trip, where <math>x</math> and <math>D</math> are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled <math>D-x</math> miles after the crash. In other words, the train travelled for <math>\frac{(D-x)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x}{3x/4}\text{hr}</math> after the crash. So, not including the half hour detention after the crash, the entire trip took | ||
<cmath>\left(1+\frac{D-x}{3x/4}\right)\text{hours}.</cmath> | <cmath>\left(1+\frac{D-x}{3x/4}\right)\text{hours}.</cmath> | ||
− | Now let us consider the alternative trip, where the accident happened <math>90</math> miles farther along the line. Originally, the accident happened <math>x | + | Now let us consider the alternative trip, where the accident happened <math>90</math> miles farther along the line. Originally, the accident happened <math>x\frac{\text{mi}}{\text{hr}}\cdot(1\text{ hr})=x\text{ mi}</math> down the line, but in this situation it happens <math>x+90</math> miles from the line. Therefore the accident happens <math>\frac{(x+90)\text{ mi}}{x\frac{\text{mi}}{\text{hr}}}=\frac{x+90}{x}\text{hr}</math> into the trip. The length of the remaining part of the trip is now <math>D-x-90</math>, so the remaining trip takes <math>\frac{(D-x-90)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x-90}{3x/4}\text{hr}</math>. So, not including the half hour detention after the crash, the entire trip took |
<cmath>\left(\frac{x+90}{x}+\frac{D-x-90}{3x/4}\right)\text{hours}.</cmath> | <cmath>\left(\frac{x+90}{x}+\frac{D-x-90}{3x/4}\right)\text{hours}.</cmath> | ||
We are given that the trip that actually happened resulted in being <math>3.5</math> hours late, while the alternative trip would have resulted in being only <math>3</math> hours late. Therefore the first trip took exactly <math>\frac{1}{2}</math> hour more: | We are given that the trip that actually happened resulted in being <math>3.5</math> hours late, while the alternative trip would have resulted in being only <math>3</math> hours late. Therefore the first trip took exactly <math>\frac{1}{2}</math> hour more: |
Latest revision as of 17:56, 18 July 2015
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
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