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 3 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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