A policeman is chasing a thief at a speed of 82 km/h. The thief is 1300 m away and is running at a speed of 64 km/h. In how much time will the policeman catch the thief if both are riding on bikes?

Correct — C. Both bikes move the same way, so only the gap between them closes.

Relative speed = 82 − 64 = 18 km/h
In m/s: 18 × 5⁄18 = 5 m/s
Gap to close = 1300 m

Time = 1300 ⁄ 5 = 260 seconds → option (c).

Why the others are wrong

• (a)
  300 sec — At the closing speed of 5 m/s, 300 seconds would shut a gap of 1500 m. The thief is only 1300 m ahead, so he is caught well before the 300th second.
• (b)
  350 sec — 350 seconds closes 1750 m at 5 m/s, far beyond the 1300 m head start. Any time longer than 260 s has the policeman already past the thief.
• (d)
  250 sec — 250 seconds closes 5 × 250 = 1250 m, leaving 50 m still between them. The chase is not over, so 250 s is short of the answer.

A chase in the same direction is a relative-speed problem. Subtract the two speeds and treat the head start as the distance that has to be covered at that difference.

Adding the speeds is right only when the two bodies move towards each other.

The second habit being tested is the unit change. km/h × 5⁄18 = m/s, because 1 km = 1000 m and 1 h = 3600 s, and 1000 ⁄ 3600 = 5 ⁄ 18. Mixing 1300 metres with 18 km/h untouched is where the mark goes.

Key facts

• Same-direction chase: closing speed = faster speed − slower speed.
• km/h to m/s: multiply by 5 ⁄ 18, so 18 km/h = 5 m/s.
• A 1300 m gap closing at 5 m/s takes 260 s, which is 4 minutes 20 seconds.

Common traps

• Adding 82 and 64 instead of subtracting, which is the head-on case rather than a chase.
• Leaving the gap in metres while the closing speed is still in km/h.
• Answering in minutes when the options are printed in seconds.

No directly related past PYQ was found.