Time & distance

Quantitative Aptitude (CHSL)

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A thief is spotted by a policeman from a distance of 480 m. When the policeman starts the chase, the thief also starts running.

If the speed of the thief is 19 km/h and that of the policeman is 23 km/h, then how far would the thief have to run before he is

overtaken?

This questions was previously asked in

SSC CHSL 31st May 2022 Shift-3

2080 m

2280 m

2290 m

2180 m

- A policeman spots a thief from 480 meters away.

- Both start running, with the thief at 19 km/h and the policeman at 23 km/h.

- The speed difference is 23 km/h - 19 km/h = 4 km/h.

- Convert this speed difference to meters per second: 4 km/h = (4000/3600) m/s = 10/9 m/s.

- The initial distance between them is 480 meters.

- Time to catch up: \( \text{Time} = \frac{480 \, \text{m}}{10/9 \, \text{m/s}} = 432 \, \text{seconds} \).

- In this time, the thief runs \( 432 \, \text{seconds} \times \frac{19 \times 1000}{3600} \, \text{m/s} = 2280 \, \text{meters} \).

Option 2: 2280 m