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Question 4 : 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

- 58
- 67
- 50
- 61

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Let’s take speed of the train to be x and the time taken be t

Since speed is reduced to \\frac{1}{3}) rd,

New speed = \\frac{x}{3})

Since the speed is one-third, time taken will be tripled. T = 3t

This 3t is after the scheduled time, So extra 2t = 30 mintues

t = 15 minutes

Train travels at x km/hr takes 15 minutes and

Train travels at \\frac{x}{3}) km/hr takes 45 minutes

So, the train usually takes 15 minutes to cover the distance.

It travels 5 minutes at the usual speed. That is, it travels \\frac{1}{3}) rd of the time at the usual speed. So it covers \\frac{1}{3}) rd of the distance in 5 minutes.

To reach it's destination in the on time, the train has to travel the remaining \\frac{2}{3}) rds of the distance in 10 minutes. Since the train halts for 4 minutes, it should now cover the \\frac{2}{3}) rds of the distance in 6 (10 - 4) minutes.

In other words, the train has to cover the same distance in \\frac{6}{10}) th of the usual time.

In order to do so, the speed must be \\frac{10}{6}) ths of the usual speed. Or the increased speed will be \\frac{4}{6}) ths or \\frac{2}{3}) rds of the usual speed. Which is an increase of 66.66% or nearly 67%.

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