CAT 2025 Slot 3 QA Q6 Time, Speed & Distance

Question 6 of the Quantitative Ability (QA) section, CAT 2025 Slot 3, with CAT's official answer and a worked solution.

Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was
  1. A
    18
  2. B
    12
  3. C
    20
  4. D
    15

Answer: (D) 15 Time, Speed & Distance MCQ

Solution

  1. Let initial speed be \(v\) and let \(T\) be the time he would still need for the remaining distance at speed \(v\) after the stopping point; remaining distance \(= vT\).
  2. 20-minute stop: \(vT = (v + 3)(T - \frac{1}{3}) \Rightarrow 3T = \frac{v}{3} + 1\).
  3. 30-minute stop: \(vT = (v + 5)(T - \frac{1}{2}) \Rightarrow 5T = \frac{v}{2} + \frac{5}{2}\).
  4. From the first, \(T = \frac{v}{9} + \frac{1}{3}\); substituting: \(\frac{5v}{9} + \frac{5}{3} = \frac{v}{2} + \frac{5}{2} \Rightarrow \frac{v}{18} = \frac{5}{6} \Rightarrow v = 15\).

Shortcut. Check options: with \(v = 15\), \(T = 2\) h; \(30/(5/3)=18\) and \(30/1.5=20\) km/h, i.e. +3 and +5.

Same concept in other years

See every question on Delay or stoppage on the way: new speed needed to arrive on time.

More Time, Speed & Distance questions from CAT

All 48 Time, Speed & Distance questions asked since 2017.