SSC CGL 2024 Tier-I · 11 September 2024, Shift 2 · Quantitative Aptitude · Question 1 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 is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km/h and 11 km/h, respectively. What is the distance (in metres) between them after 9 minutes?

(a)50
(b)40
(c)60
(d)30

Answer

Why

Correct — A. They run the same way, so the speeds subtract. Rule: closing speed = 11 − 10 = 1 km/h.

Put that into metres per minute:
1 km/h = 1000 m in 60 min = 50⁄3 m/min

Ground the policeman gains in 9 minutes:
(50⁄3) × 9 = 150 m

Gap left = 200 − 150 = 50 m → option (a).

Why the others are wrong

(b)
40 — Forty needs the policeman to gain 160 m in 9 minutes, which is a closing speed of about 1.07 km/h. The two speeds differ by exactly 1 km/h, and that is worth 150 m.
(c)
60 — Sixty leaves only 140 m gained. Check the rate: 1 km/h is 16.67 m/min, and 16.67 × 9 = 150, so 60 m is ten metres too generous to the thief.
(d)
30 — Thirty would need 170 m of the head start to disappear. At 50⁄3 m/min the policeman needs a full 12 minutes to close all 200 m, and he has had nine.

Concept

Two bodies moving the same way close on each other at the difference of their speeds; moving towards each other, at the sum.

Here the difference is 1 km/h, so the 200 m head start shrinks by 1 km every hour and by nothing else. The individual speeds are never needed on their own.

The other half of the question is units. The gap is in metres and the time in minutes, so convert once, at the start: multiply km/h by 1000⁄60 to get metres per minute.

The thief is not caught in this question. Closing 200 m at 50⁄3 m/min takes 12 minutes, and you are asked where things stand at 9.

Reading it as a catch-up question is the fastest way to lose the mark.

Key facts

Same direction: the closing speed is the difference of the two speeds.
Opposite directions: the closing speed is the sum of the two speeds.
1 km/h = 1000⁄60 m/min = 50⁄3 m/min, about 16.67 m/min.
A 1 km/h advantage covers exactly 150 m in 9 minutes.

Study next

Time to catch up: head start divided by closing speedTrains crossing each other, where the lengths are added to the distanceBoats and streams, where the current adds to or subtracts from the boat's speed

Common traps

Adding the two speeds instead of subtracting them
Leaving the closing speed in km/h while the answer is wanted in metres
Answering the time to catch the thief rather than the gap after 9 minutes

SSC gives the head start in metres and the speeds in km/h, so the unit conversion is the real step and the arithmetic is trivial once it is done. The same net-rate idea drives the leaking-tank question in this shift at Quant Q.24, where the leak's rate is subtracted from the pipe's.

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