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Q30·CSAT · Prelims 2026

Quantitative — Split Distance Average Speed

NumericalTime-Speed-DistanceArithmetic Problem● Medium

Question

X travels 6 km on a bicycle with average speeds of 5 km per hour, 10 km per hour and 4 km per hour during the first 1 km, the next 2 km and the remaining 3 km, respectively. Y travels the same distances with average speeds of 4 km per hour, 10 km per hour and 5 km per hour, respectively. How many minutes early will Y complete the journey if both X and Y start at the same time?

Options

a

3

b

4

Answer
c

5

d

6

Explanation

To find the difference in completion times, compute the total travel duration for both X and Y across the three distinct stages (t = Distance{Speed}) [cite: 4050, 4051, 4052].

Calculate Total Time for X:
Stage 1 (1 km at 5 km/h): t_1 = 1/5 hours = 12 minutes .
Stage 2 (2 km at 10 km/h): t_2 = 2/10 = 1/5 hours = 12 minutes .
Stage 3 (3 km at 4 km/h): t_3 = 3/4 hours = 45 minutes .
Total Time for X = 12 + 12 + 45 = 69 minutes.
Calculate Total Time for Y:
Stage 1 (1 km at 4 km/h): t_1 = 1/4 hours = 15 minutes .
Stage 2 (2 km at 10 km/h): t_2 = 2/10 = 1/5 hours = 12 minutes .
Stage 3 (3 km at 5 km/h): t_3 = 3/5 hours = 36 minutes .
Total Time for Y = 15 + 12 + 36 = 63 minutes.
Calculate the Difference:

Time Difference = Time for X - Time for Y = 69 - 63 = 6 minutes

Let us re-verify the final calculations: 69 - 63 = 6 minutes. The output maps to choice (d).

For multi-stage journeys with varying speeds, convert each independent segment duration into minutes before summing to simplify finding the net time difference.

Answer: (d).

Question details

Year

2026

Paper

CSAT

Question

Q30

Section

Quantitative Aptitude

Sub-topic

Time-Speed-Distance

Type

Arithmetic Problem

Difficulty

Medium

Source hint

Multi-stage travel — discrete time interval summation

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