A thief and a police officer are 500 meters apart. If they run in the same direction, the police officer catches the thief in 10 minutes. If they run towards each other, they meet in 2 minutes. What is the speed of the thief in meters per minute?

Correct — B. Let the officer's speed be p and the thief's t, in m/min.

Same direction, 500 m closed in 10 min: p − t = 50
Towards each other, 500 m closed in 2 min: p + t = 250

Subtract: 2t = 250 − 50 = 200
Halve: t = 100 m/min → option (b)

Why the others are wrong

• (a)
  150 meters/minute — 150 is the police officer's speed, p = (250 + 50) ÷ 2. The question asks for the thief's.
• (c)
  200 meters/minute — 200 is 2t, the difference 250 − 50 before halving. The thief's speed is half of it.
• (d)
  50 meters/minute — 50 is the chase's closing speed, p − t: the rate at which the officer eats into the 500 m gap. It is not the thief's own speed.

Two runners close a gap at their relative speed: the difference of their speeds when they run the same way, the sum when they run towards each other.

Each scenario gives one equation, gap ÷ time. Adding the two equations isolates the officer's speed, subtracting isolates the thief's.

Key facts

• Same direction: relative speed = difference of the speeds.
• Towards each other: relative speed = sum of the speeds.
• From a sum S and difference D of two speeds: faster = (S + D) ÷ 2, slower = (S − D) ÷ 2.

Common traps

• Stopping at 2t = 200 and forgetting to halve.
• Giving the officer's speed, 150, because it comes out of the same pair of equations.

No directly related past PYQ was found.