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: Understanding the Concept:
This problem involves relative speed. First, find out how long it takes the faster policeman to catch the thief, then calculate the distance covered by the slower policeman in that same time.
Step 2: Key Formula or Approach:
Time to catch = (Head start distance) / (Relative Speed). Distance behind = Speed difference \(\times\) Time.
Step 3: Detailed Explanation:
1. Thief's head start (15 mins = 0.25 hours): Distance = \(60 \text{ km/h} \times 0.25 \text{ h} = 15 \text{ km}\). 2. Relative speed of X vs Thief = \(65 - 60 = 5 \text{ km/h}\). 3. Time for X to catch thief = \(15 \text{ km} / 5 \text{ km/h} = 3 \text{ hours}\). 4. The second policeman started at the same time as X but at 60 km/h (same speed as thief). 5. Since the second policeman moves at the exact same speed as the thief, the gap of 15 km between them will never change. 6. When X catches the thief, the gap between the thief and the second policeman remains 15 km.
Step 4: Final Answer:
The second policeman was 15 km behind.