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CAT Previous Year Paper | CAT Quants Questions | Question 22

This question is from Speed time distance. Questions from Speed time distance have appeared in every edition of the CAT Exam. This is a very interesting topic as it can be relatable in real-life scenarios too. If you're in a hurry, then solve this question. This question might help you to calculate how late you're going to be. To explore 1000+ intriguing questions, check out 2IIM CAT Question Bank, and give your CAT Preparation the right direction.

Question 22 : Vimla starts for office every day at 9 am and reaches exactly on time if she drives at her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. One day, she covers two-thirds of her distance to office in one-thirds of her usual time to reach office, and then stops for 8 minutes. The speed, in km/hr, at which she should drive the remaining distance to reach office exactly on time is

  1. 27
  2. 28
  3. 29
  4. 26

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Explanatory Answer

Vimala's speed has decreased from 40kmph to 35 kmph. That means her current speed is \\frac{35}{40}) of the original Speed. That means, her current speed is \\frac{7}{8}) th of the original speed. Therefore, the time taken will be \\frac{8}{7}) th of the original time taken.

Since she is 6 minutes late when travelling at 35kmph, \\frac{1}{7})th of the original time taken is 6 minutes.
Therefore the total time taken originally is 7 × 6 = 42 minutes.

\\frac{1}{3}) rd of the original time taken = \\frac{42}{3}) = 14 minutes.
\\frac{1}{3}) rd of the total distance is generally covered in 14 minutes.
Currently the same distance is to be covered in (42 - 14 - 8 = 20) minutes.

So, to reach the Destination on time, Vimala, must cover the same \\frac{1}{3}) rd of the Distance in 20 minutes instead of 14 minutes.
That means she has to \\frac{20}{14}) of the initial time.
Which is \\frac{10}{7}) of the original time
This implies that, in order to reach at the same point of time. The speed had to \\frac{7}{10}) or 70% of the original speed.

So, she has to travel at 70% of 40kmph = 28kmph.


The answer is, "28"

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