A thief steals a car at 2.30 p.m. and drives it at 60 kmph. The theft is discovered at 3 p.m. and the owner sets off in another car at 75 kmph. When will he overtake the thief?

### Concept & Delayed Pursuit When pursuit begins after a delay, the pursued entity creates a "head start" distance. The pursuer must close this initial distance using their relative speed. $$Head\ Start\ Distance = Speed_{Thief} \times Delay\ Time$$ $$Catch\ Time = \frac{Head\ Start\ Distance}{Relative\ Speed}$$ ### Step-by-Step Solution 1. **Determine the delay time:** The thief starts at 2:30 p.m., owner starts at 3:00 p.m. Delay = 30 minutes = 0.5 hours. 2. **Calculate initial head start:** In 0.5 hours, the distance covered by the thief. Distance = $60 \text{ km/h} \times 0.5 \text{ h} = 30 \text{ km}$. This is the gap the owner needs to close. 3. **Calculate relative speed:** Owner's speed = 75 kmph, Thief's speed = 60 kmph. Relative speed = $75 - 60 = 15 \text{ kmph}$. 4. **Calculate time to overtake:** Time = (Head Start Distance) / (Relative Speed) = $30 \text{ km} / 15 \text{ kmph} = 2 \text{ hours}$. 5. **Determine the clock time:** The owner starts chasing at 3:00 p.m. Overtake time = 3:00 p.m. + 2 hours = 5:00 p.m. ### Exam Strategy & Shortcut Equate the distances. Let $t$ be the time (in hours) the owner travels until catching the thief. Distance by Owner = Distance by Thief $75 \times t = 60 \times (t + 0.5)$ $75t = 60t + 30$ $15t = 30 \implies t = 2$ hours. Add 2 hours to the owner's start time (3 p.m.) to get 5 p.m. ### Common Pitfall A frequent mistake is adding the calculated pursuit time (2 hours) to the *thief's* start time (2:30 p.m.) instead of the *owner's* start time (3:00 p.m.). ### Final Answer Therefore, the correct answer is **5 p.m.**