A thief pursued by a policeman was 150 m ahead at the start. If the ratio of the speed of the policeman to that of the thief was 5 : 4, then how far (distance in metres) could the thief go before he is caught by the policeman?

Correct — B. Both men run for the same time, so their distances are in the ratio of their speeds, 5 : 4.

Policeman covers 5 parts while the thief covers 4 parts
Gap closed = 5 − 4 = 1 part, and that must swallow the 150 m head start
So 1 part = 150 m
Thief's run = 4 parts = 4 × 150 = 600 m → option (b)

Check: the policeman runs 5 × 150 = 750 m, exactly 150 m more.

Why the others are wrong

• (a)
  200 — The gain is always a quarter of the thief's run — 1 part against his 4. At 200 m the policeman has gained only 50 m, so the 150 m lead still stands.
• (c)
  400 — 400 m of running lets the policeman gain 400 ÷ 4 = 100 m, still 50 m short of the head start. The chase is not over.
• (d)
  350 — 350 m closes just 350 ÷ 4 = 87.5 m of the gap, a little over half the 150 m lead.

When two bodies move for the same length of time, their distances stand in the same ratio as their speeds. That one line turns most chase problems into arithmetic with no speeds in it at all.

Split the run into parts: the pursuer covers 5, the runner 4, and the surplus is 1 part. The chase ends when that surplus equals the head start, so 1 part = the head start, and every distance in the problem is a multiple of it.

Key facts

• Equal running time means the two distances are in the same ratio as the two speeds.
• In a 5 : 4 chase the pursuer gains 1 part for every 4 parts the runner covers.
• Here 1 part is 150 m, so the thief runs 600 m and the policeman 750 m.

Common traps

• Reporting the policeman's distance instead of the thief's.
• Reading the ratio parts as metres, which makes the gain look like 1 m.
• Hunting for a unit conversion when the item supplies no speeds to convert.

No directly related past PYQ was found.