SSC CGL 2024 Tier-I · 24 September 2024, Shift 1 · Quantitative Aptitude · Question 23 of 98

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 10 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 spotted by a policeman from a distance of 100 m. The thief starts running and the policeman chases him. If the speed of thief and policeman are 21 km/h and 23 km/h, respectively, then how far will the thief have to run before he is over taken?

(a)1090 m
(b)1050 m
(c)1080 m
(d)1020 m

Answer

Why

Correct — B. Same direction, so the gap closes at the difference of the speeds, not their sum.

Closing speed = 23 − 21 = 2 km/h
Head start = 100 m = 0.1 km
Time to overtake = 0.1 ÷ 2 = 0.05 h (3 minutes)

Distance the thief covers = his own speed × that time
= 21 × 0.05 = 1.05 km = 1050 m → option (b).

Cross-check: the policeman runs 23 × 0.05 = 1150 m, exactly 100 m more.

Why the others are wrong

(a)
1090 m — Fails the ratio test. Both run for the same time, so their distances are in the ratio of their speeds, 21 : 23 — and 1090 : 1190 is not.
(c)
1080 m — The 100 m head start is the 2-part difference in a 21 : 23 split, so one part is 50 m and the thief covers 21 × 50 = 1050 m. 1080 is not 21 parts of anything here.
(d)
1020 m — 1020 m at 21 km/h takes 0.0486 h, and in that time the policeman gains only 2 × 0.0486 km ≈ 97 m — short of the 100 m he has to make up.

Concept

A chase is a relative-speed problem in one direction. Sit on the thief and the policeman appears to approach at 23 − 21 = 2 km/h, which is the only speed that matters for the catching time.

Time to overtake = head start ÷ closing speed. Everything else follows from that one number.

There is a faster route worth owning. Both run for the same time, so their distances are in the ratio of their speeds, 21 : 23. The 100 m gap is the 2-part difference, one part is 50 m, and the thief runs 21 parts = 1050 m without any unit conversion at all.

Watch the units: the head start is in metres and the speeds in km/h. Either convert 100 m to 0.1 km, as above, or convert the 2 km/h closing speed to 5⁄9 m/s and divide 100 by it — both give 180 seconds.

Key facts

Same direction: relative speed is the difference of the speeds. Opposite directions: the sum.
Two bodies moving for the same time cover distances in the ratio of their speeds.
Here the closing speed is 2 km/h and the chase lasts 0.05 h, or 3 minutes.
1 km/h = 5⁄18 m/s.

Study next

Relative speed for trains crossing each other in the same and opposite directionsRaces with a head start, where A gives B a start of x metresBoat-and-stream problems, the same difference-and-sum idea in different clothes

Common traps

Adding the speeds to 44 km/h, which is the approaching case, not a chase.
Reporting the policeman's 1150 m when the stem asks how far the thief runs.
Leaving the head start in metres while the speeds stay in km/h.

The same chase set-up, with other numbers, is also at 19 Sep 2024, 12:30, Quant Q.18 (200 m head start), 19 Sep 2024, 16:00, Quant Q.21 (195 m) and 23 Sep 2024, 9:00, Quant Q.2, which asks for the time in seconds rather than the distance.

Related PYQs

No directly related past PYQ was found.

Topic: Time and distance — all 40 SSC CGL questions on it, every solved shift.