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Question
A goods train leaves a station at 6 p.m., followed by an express train which leaved at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the train remain constant between the two stations; calculate their speeds.
Solution
Let the speed of goods train be x km/hr.
So, the speed of express train will be (x + 20) km/hr.
Distance = 1040 km
We know
Time = `"Distance"/"Speed"`
Time taken by good train to cover a distance of 1040 km = `1040/x` hrs
Time taken by express train to cover a distance of 1040 km = `1040/(x + 20)` hrs
It is given that the express train arrives at a station 36 minutes before the goods train. Also the express train leaves the station 2 hours after the goods train. This means that the express train arrives at the station `(36/60 + 2) "hrs" = 13/5 "hrs"` before the good train.
Therefore, we have
`1040/x - 1040/(x + 20) = 13/5`
`(1040x + 20800 - 1040x)/(x(x + 20)) = 13/5`
`20800/(x^2 + 20x) = 13/5`
`1600/(x^2 + 20x) = 1/5`
x2 + 20x – 8000 = 0
(x – 80)(x + 100) = 0
x = 80, –100
Since, the speed cannot be negative.
So, x = 80.
Thus, the speed of goods train is 80 km/hr and the speed of express train is 100 km/hr.
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