A robber steals a man’s bag at 5:00 p.m. and starts running at a speed of 10 km/h. The policeman is informed about the robber at 5:12 p.m. and he starts chasing the thief on a bicycle at a speed of 15 km/h. At what time will the policeman catch the robber?

Solution (Step-by-step)

Robber starts at 5:00 p.m. with speed = 10 km/h
Policeman starts at 5:12 p.m. with speed = 15 km/h

1. Head start of robber

Time gap = 12 minutes = $ \frac{12}{60} = 0.2 \text{ hours} $

Distance covered by robber in this time:
$ 10 \times 0.2 = 2 \text{ km} $

2. Relative speed

$ 15 - 10 = 5 \text{ km/h} $

3. Time to catch

$ \frac{2}{5} = 0.4 \text{ hours} $

Convert into minutes:
$ 0.4 \times 60 = 24 \text{ minutes} $

4. Final time

Policeman starts at 5:12 p.m.
Catches after 24 minutes → 5:36 p.m.