SSC CGL 2024 Tier-I · 9 September 2024, Shift 3 · Quantitative Aptitude · Question 6 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

A thief noticed a policeman at a distance of 600 metres. The thief started running and the policeman chased him. The thief and policeman are running at speeds of 12 km/h and 15 km/h, respectively. Find the time (in minutes) required for the policeman to catch the thief.

(a)8
(b)10
(c)12
(d)15

Answer

Why

Correct — C. Both men run the same way, so the gap closes at the difference of their speeds, not at either speed on its own.

Relative speed = 15 − 12 = 3 km/h
Gap to close = 600 m = 0.6 km

Time = gap ÷ relative speed
= 0.6 ⁄ 3 = 0.2 hour

0.2 × 60 = 12 minutes → option (c).

Why the others are wrong

(a)
8 — Short by a third. At the closing speed of 3 km/h, 8 minutes shrinks the gap by 3 × 8⁄60 = 0.4 km, leaving 200 m still between them.
(b)
10 — Still short. 10 minutes closes 3 × 10⁄60 = 0.5 km, which is 100 m less than the 600 m head start the thief began with.
(d)
15 — Too long. In 15 minutes the gap shrinks by 0.75 km, so the policeman would already have overtaken the thief. The catch happens at exactly 0.6 km closed.

Concept

In a same-direction chase the gap is governed by the difference of the speeds — here 15 − 12 = 3 km/h — however fast either runner is actually moving.

Had they been running towards each other the speeds would add to 27 km/h, nine times faster, and the meeting would take about 1.3 minutes instead of 12.

The only real care needed is units: the gap is in metres, the speeds in km/h and the answer in minutes.

Convert once at the start — 600 m = 0.6 km — and everything after that is one division and one multiplication by 60.

The stem gives the 600 m as the distance at the moment the thief starts running, so both are treated as moving at constant speed from that instant — no reaction time is modelled.

Key facts

For two bodies moving in the same direction, relative speed is the difference of their speeds.
For two bodies moving towards each other, relative speed is the sum of their speeds.
Multiply by 5⁄18 to convert km/h into m/s, and by 18⁄5 to go back.
Here the gap of 0.6 km closes at 3 km/h, taking 0.2 hour, which is 12 minutes.

Study next

Relative speed in same-direction and opposite-direction motionTrains crossing a pole, a platform, and each otherBoats and streams, where upstream and downstream use the same difference-and-sum idea

Common traps

Dividing 600 m by 15 km/h and forgetting that the thief is moving too.
Leaving the gap in metres while the speeds are in km/h.
Giving the answer as 0.2 when the question asks for minutes.

The chase is dressed up as thief and policeman, dog and hare, or two trains, but the working stays the same — gap divided by the difference of the speeds, with one unit conversion planted somewhere in it.

Related PYQs

No directly related past PYQ was found.