A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to
Started 10 months ago by Shashank in
Explanatory Answer
Original speed = S & Original time = T
New Speed = S/3, speed and time are inversely proportional hence New time = 3T
The difference between times = 3T-T = 30 mins
Original time taken = 15 mins
Assume Distance is 15 Kms and Speed = 1 km/hr
In 5 mins the train travelled 5 kms and 4 mins it stopped.
In 6 mins it has to cover 10 kms
The original speed will be increased by a factor of 1.66 or 67%
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