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. At what time did X catch the thief?

• A. 3.30 p.m.
• B. 3 p.m.
• C. 3.15 p.m.
• D. None of these
• A. 18.75 km
• B. 15 km
• C. 21 km
• D. 37.5 km

Answer 1

The Correct Option is B

The thief gets a head start of $15$ minutes = $\frac{1}{4}$ hour. In this time, thief covers = $60 \times \frac{1}{4} = 15$ km. Relative speed of X with respect to the thief = $65 - 60 = 5$ km/hr. Time taken for X to catch the thief = $\frac{\text{head start distance}}{\text{relative speed}} = \frac{15}{5} = 3$ hours. X started at 12:15 p.m., so he catches the thief at $12:15 + 3$ hours = 3:15 p.m. But wait — rechecking: Thief starts at 12:00 noon, X starts at 12:15 p.m. and catches up in 3 hours from his own start → catch time = $12:15 + 3 = 3:15$ p.m. So correct = 3:15 p.m. (Option c).

Answer 2

From Q144, X took 3 hours to catch the thief after starting. If another policeman started with X at 60 km/hr, he would cover = $60 \times 3 = 180$ km in that time. The thief’s total distance when caught = X’s distance from his start point = $65 \times 3 = 195$ km. Thus, the slower policeman is behind by $195 - 180 = 15$ km relative to the thief’s location. But relative to X’s location, still the same = 15 km. Wait — question asks "how far behind he was when X caught the thief": that’s exactly this 15 km. However, rechecking initial head start: This extra policeman starts same time as X, so no extra gap beyond speed difference. So final = 15 km.