CAT 2020 Question Paper | Quants Slot 1

CAT Previous Year Paper | CAT Quants Questions | Question 4

Speed time distance is one of the most commonly tested topics in CAT exam. Questions from speed time distance have appeared consistently in the CAT exam for the last several years. This topic is very interesting as it is relatable in real-life scenarios. If you are in a hurry then have a go at this question from CAT previous years' question paper before you leave. Maybe It could help you with how late you are going to be!

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

  1. 58
  2. 67
  3. 50
  4. 61

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Explanatory Answer

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%.

The answer is, "67"

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