A train which travels at a uniform speed due to some mechanical fault after traveling for an hour goes at 3/5th of the original speed and reaches the destination 2 hrs late.if the fault had occurred after traveling another 50 miles the train would have reached 40 min earlier. what is distance between the two stations.
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Let, the original speed is x and 3/5 th of original speed is 0.6x
50/0.6x - 50/x= 40/60
(250-150)/3x = 2/3
100/3x = 2/3
6x=300
x=50
3/5 of 50 = 30
speed reduces by = 50-30 = 20
It is late by 2 hours and by this time it goes 30*2 = 60
If the train goes with its original speed it took 60/20 = 3 hours after the first hour. That means total 4 hours.
Therefore, total distance covered 4*50 = 200 and speed is 50.
50/0.6x - 50/x= 40/60
(250-150)/3x = 2/3
100/3x = 2/3
6x=300
x=50
3/5 of 50 = 30
speed reduces by = 50-30 = 20
It is late by 2 hours and by this time it goes 30*2 = 60
If the train goes with its original speed it took 60/20 = 3 hours after the first hour. That means total 4 hours.
Therefore, total distance covered 4*50 = 200 and speed is 50.
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