Office commute delay and recovery speed

CAT 2020 Slot 3 · QA · Medium · Time speed and Distance

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Vimla starts for office every day at 9 am and reaches exactly on time if she drives at her usual speed of 40 km/hr40\text{ km/hr}. She is late by 6 minutes if she drives at 35 km/hr35\text{ 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. A.

    27

  2. B.

    28

  3. C.

    29

  4. D.

    26

Answer

B

Explanation

Let usual time be TT hours and distance be D kmD\text{ km}. D=40T=35(T+6/60)  ⟹  40T=35T+3.5  ⟹  5T=3.5  ⟹  T=0.7 hours=42 minutesD = 40T = 35(T + 6/60) \implies 40T = 35T + 3.5 \implies 5T = 3.5 \implies T = 0.7\text{ hours} = 42\text{ minutes}. D=40×0.7=28 kmD = 40 \times 0.7 = 28\text{ km}.

On that day:

  1. She covers 23D\frac{2}{3} D in 13T=14 minutes\frac{1}{3} T = 14\text{ minutes}. Remaining distance = 13D=283 km\frac{1}{3} D = \frac{28}{3}\text{ km}.
  2. Stops for 8 minutes. Time spent = 14+8=22 minutes14 + 8 = 22\text{ minutes}. Remaining time to reach on time = 42−22=20 minutes=13 hour42 - 22 = 20\text{ minutes} = \frac{1}{3}\text{ hour}.

Required speed = 28/3 km1/3 hour=28 km/hr\frac{28/3\text{ km}}{1/3\text{ hour}} = 28\text{ km/hr}.

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