A thief steals a car at 3:30 p.m. and drives it at 80km/h. The theft is discovered at 4:00 p.m. and the owner sets off in another car at 90km/h. When will he catch the thief?

Correct option is A

Given:
The thief steals the car at 3:30 p.m. and drives at 80 km/h.
The theft is discovered at 4:00 p.m., meaning the thief already has a 30-minute head start.
The owner starts chasing at 4:00 p.m. at a speed of 90 km/h.
Formula Used:

$$\text{Time} = \frac{\text{Distance Gap}}{\text{Relative Speed}}$$​

Solution:

Thief drives at 3:30 p.m. = 80 km/h.

When the owner starts at 4:00 p.m., in (30 min or half an hour),

the thief covers = $$80 \times \frac{1}{2}$$​ = 40 km

Time taken to catch thief = $$\frac{40}{90 - 80} = 4 \text{ hours}$$​​

So,

4:00 p.m. + 4 hours = 8:00 p.m.

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