SSC CGL 2024 Tier-I · 26 September 2024, Shift 2 · Quantitative Aptitude · Question 15 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 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 takes off on his bike at a certain speed, after seeing a police car at a distance of 250 m. The police car starts chasing the thief and catches him. If the thief runs 1.5 km before being caught and the speed of the police car is 70 km/h, then what is the speed of thief’s bike (in km/h)?

(a)65
(b)50
(c)55
(d)60

Answer

Why

Correct — D. Both vehicles move for the same length of time, so their distances are in the ratio of their speeds.

Police car's distance = the 250 m gap + the thief's run = 250 + 1500 = 1750 m
Ratio of distances = 1500 ⁄ 1750 = 6 ⁄ 7
Thief's speed = 70 × 6 ⁄ 7 = 60 km/h → option (d)

Time check: 1.75 ⁄ 70 = 0.025 h for the police car, and 1.5 ⁄ 0.025 = 60 km/h for the thief.

Why the others are wrong

(a)
65 — At 65 km/h the thief needs 1.5 ⁄ 65 h, in which the police car covers 70 × 1.5 ⁄ 65 ≈ 1.615 km. That closes only about 115 m of the 250 m gap, so he is not caught.
(b)
50 — 50 km/h is what you get by subtracting the 250 m from the thief's run instead of adding it to the police car's: 1250 ⁄ 1750 × 70 = 50. The 1.5 km is already the thief's whole run.
(c)
55 — At 55 km/h the police car covers 70 × 1.5 ⁄ 55 ≈ 1.909 km while the thief covers 1.5 km, closing about 409 m. Only 250 m separated them.

Concept

A chase is a relative-speed question in disguise, and two facts settle it.

First, pursuer and pursued are moving for the same time. Second, the pursuer must cover the head start on top of whatever the runner covers, so police distance = head start + thief distance.

Equal times mean the distances are proportional to the speeds, so a single ratio finishes the question and you never have to compute the time at all.

The relative-speed route works too: the gap shuts at (70 − v) km/h and 0.25 km must close in the 1.5 ⁄ v hours the thief runs, giving 0.25v = 105 − 1.5v and v = 60. Same answer, one more line of algebra and the same unit conversion.

Key facts

In a chase the pursuer covers the head start plus the distance the runner covers, in the same elapsed time.
Equal times make distances proportional to speeds, so 1500 : 1750 reduces to 6 : 7 and matches 60 : 70.
250 m is 0.25 km, so the whole question must be brought to one unit before any ratio is taken.
The chase lasts 0.025 hours, which is 1.5 minutes.

Study next

Head-start chase problems solved by the ratio of distances covered in equal timeRelative speed for bodies moving in the same and in opposite directionsConverting between m, km and km/h before forming a ratio

Common traps

Taking the police car's distance as 1.5 km, which simply returns its own 70 km/h.
Subtracting the 250 m from the thief's distance rather than adding it to the police car's.
Comparing 250 against 1.5 without converting metres to kilometres.

SSC frames these as a policeman-and-thief story with the head start in metres and the speeds in km/h, then asks for a speed, a time or a distance.

The time version is at 24 Sep 2024, 16:00, Quant Q.10, where the thief is 1300 m ahead at 64 km/h and the policeman does 82 km/h.

The distance version is at 19 Sep 2024, 12:30, Quant Q.18, with a 200 m head start at 9 km/h against 10 km/h.

Related PYQs

No directly related past PYQ was found.