SSC CGL 2024 Tier-I · 19 September 2024, Shift 2 · Quantitative Aptitude · Question 18 of 99

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 spotted by a policeman from a distance of 200 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief is 9 km/h and that of the policeman 10 km/h, how far will the thief have to run before he is overtaken?

(a)1.8 km
(b)1.4 km
(c)1.6 km
(d)1.2 km

Answer

Why

Correct — A. Both men run the same way, so the gap closes at the relative speed, not at either speed on its own.

Head start = 200 m = 0.2 km
Relative speed = 10 − 9 = 1 km/h
Time to overtake = 0.2 ÷ 1 = 0.2 hour

The question asks how far the thief runs, so use his own speed for that time.
Distance = 9 × 0.2 = 1.8 km → option (a)

Why the others are wrong

(b)
1.4 km — At the moment of overtaking the two distances stand in the ratio of the speeds, 10 : 9. A thief's run of 1.4 km would mean the policeman ran 1.6 km, a ratio of 8 : 7.
(c)
1.6 km — 1.6 km would put the policeman at 1.8 km, a ratio of 9 : 8. The speeds fix the ratio at 10 : 9, which only 1.8 km against 2 km satisfies.
(d)
1.2 km — 1.2 km would make the policeman's run 1.4 km, a ratio of 7 : 6. Reaching only 1.2 km would need the closing gap to be 1.5 km/h, not the 1 km/h this stem gives.

Concept

A chase is a relative-speed problem. Two bodies moving the same way close the gap between them at the difference of their speeds, so a 200 m lead in front of a policeman only 1 km/h faster takes a full 0.2 hour — twelve minutes — to disappear.

Once the time is known, read the question again to see whose distance it wants. Here it is the thief's: 9 km/h for 0.2 hour is 1.8 km. The policeman covers 2 km in the same time, exactly 200 m more.

The stem mixes units on purpose — the head start is in metres and the speeds in km/h. Convert once, at the start, and the arithmetic stays clean.

Key facts

In a same-direction chase the gap closes at the difference of the speeds, here 10 − 9 = 1 km/h.
Overtaking time = head start ÷ relative speed = 0.2 ÷ 1 = 0.2 hour.
Distances covered in the same time are in the ratio of the speeds, so thief : policeman = 9 : 10 = 1.8 km : 2 km.
200 metres is 0.2 km, and 0.2 hour is 12 minutes.

Study next

Relative speed when two bodies approach each other, where the speeds add insteadChase problems that give a ratio of speeds rather than two speedsConverting km/h to m/s using the factor 5/18

Common traps

Dividing the 200 m head start by the policeman's 10 km/h instead of the 1 km/h closing speed.
Reporting the policeman's 2 km when the stem asks how far the thief runs.
Leaving the head start in metres while the speeds are in kilometres per hour.

SSC reprints this chase with the numbers changed. A thief spotted at 195 m with speeds of 22 and 27 km/h is asked for on 19 Sep 2024, 16:00, Quant Q.21.

The same opening can carry a different question: on 11 Sep 2024, 12:30, Quant Q.1 the gap is again 200 m, the speeds 10 and 11 km/h, and what is wanted is the distance still between them after 9 minutes.

Related PYQs

No directly related past PYQ was found.