A thief is spotted by a policeman from a distance of 195 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief is 22 km/h and that of the policeman is 27 km/h, then how far would the thief have run (in m) before he is overtaken?
- (a)858
- (b)958
- (c)951
- (d)825
Answer
Why
Correct — A. Both run the same way, so the gap closes at the difference of the speeds.
Relative speed = 27 − 22 = 5 km/h
Gap = 195 m = 0.195 km
Time to close it = 0.195 ⁄ 5 = 0.039 hour
Thief's run = 22 × 0.039 = 0.858 km = 858 m → option (a)
Check: the policeman runs 27 × 0.039 = 1,053 m, which is 858 + 195.
Why the others are wrong
- (b)958 — 958 m is 100 m more than the working produces. The thief's distance is 22 × (195 ⁄ 5) = 22 × 39, and 22 × 39 is 858.
- (c)951 — 951 m is not a multiple of 39, the metres run per 1 km/h of the thief's speed (195 ⁄ 5). With whole-number speeds the answer has to sit on that 39-metre grid.
- (d)825 — 825 m falls 33 m short of 858 and is also off the 39-metre grid; it would fit a thief running about 21.2 km/h, not the 22 km/h given.
Concept
A chase is a relative-speed problem. Same direction means the gap shrinks at the difference of the speeds, 27 − 22 = 5 km/h; a head-on approach would use the sum.
The gap of 195 m closes in 195 ⁄ 5 = 39 units of time, and because the units cancel you can read that 39 directly as metres run per km/h of a runner's own speed.
So the thief covers 22 × 39 = 858 m and the policeman 27 × 39 = 1,053 m. Their difference is 195 m, the head start, which is the built-in check.
Nothing here needs a conversion to m/s. Keeping the gap in metres and the speeds in km/h works because the same time factor multiplies both runners, and it is the difference of their distances that has to come to 195 m.
Key facts
- A gap closes at the difference of the speeds when both move the same way, and at the sum when they move towards each other.
- A 195 m gap closing at 5 km/h means each runner covers 39 metres per km/h of his own speed.
- The thief runs 22 × 39 = 858 m and the policeman 27 × 39 = 1,053 m.
- 1 km/h = 5⁄18 m/s, needed only if you insist on working the chase in seconds.
Study next
Common traps
- Adding the speeds instead of subtracting, which is the head-on case rather than a chase.
- Answering with the policeman's 1,053 m when the question asks how far the thief ran.
- Mixing units by converting 195 m to km while leaving a time in seconds.
The identical construction — a policeman spotting a thief at a stated distance, both speeds given, asking how far the thief runs — is set at 19 Sep 2024, 12:30, Quant Q.18 (200 m, 9 km/h against 10 km/h, keyed 1.8 km).
SSC also asks the same chase for the time to catch: 23 Sep 2024, 09:00, Quant Q.2 gives 92 m at 90 against 104.4 km/h, keyed 23 seconds.
And it asks for the gap still open after a fixed time, with no capture at all: 11 Sep 2024, 12:30, Quant Q.1 gives 200 m at 10 against 11 km/h and wants the distance between them after 9 minutes, keyed 50 m.
Related PYQs
No directly related past PYQ was found.