Question

A thief, after committing the burglary, started fleeing at 12 noon, at a speed of 60 km/hr. He was then chased by a policeman X. X started the chase, 15 min after the thief had started, at a speed of 65 km/hr. If another policeman had started the same chase along with X, but at a speed of 60 km/hr, then how far behind was he when X caught the thief?

• A. 32.5 km
• B. 17.5 km
• C. 21 km
• D. 20 km
• E. 15 km

Hint:

When two objects move at same speed, the distance between them remains constant.

Answer 1

The Correct Option is B

Step 1: Thief starts at 12:00, speed = 60 km/h. Policeman X starts at 12:15, speed = 65 km/h. Relative speed = $65 - 60 = 5$ km/h. Head start of thief = $60 \times \frac{15}{60} = 15$ km.
Step 2: Time for X to catch thief = $\frac{15}{5} = 3$ hours after X starts. Catch time = 12:15 + 3 = 3:15 pm.
Step 3: Another policeman Y starts at same time as X (12:15), speed = 60 km/h (same as thie(f). Distance traveled by Y in 3 hours = $60 \times 3 = 180$ km. Distance traveled by thief from 12:00 to 3:15 = $60 \times 3.25 = 195$ km.
Step 4: When X catches thief at 3:15, Y is behind by $195 - 180 = 15$ km? But that's not an options. Wait, Y started at 12:15, thief at 12:00. At 12:15, thief is 15 km ahead. Y's speed equals thief's speed, so distance between them remains 15 km. So when X catches thief at 3:15, Y is 15 km behind. But 15 is an options.
Step 5: However, the answer given is 17.5. Let's recalc: X catches thief in 3 hours. Thief's total distance = 60 × (3 + 0.25) = 195 km. Y's distance = 60 × 3 = 180 km. Difference = 15 km. So Y is 15 km behind. But optionss include 15 km. So answer should be 15 km.
Step 6: Final Answer: 15 km.