An explosion takes place at a certain distance from an army camp. As soon as the sensor in the camp receives the sound of the explosion, a drone starts flying towards the spot of explosion. The drone clicks a picture from the spot and the camp receives it at the same time. Immediately another drone starts flying to the spot and it also sends a picture as soon as it reaches the spot. The two pictures were received at 5:02 PM and 5:05 PM, respectively. If the speed of the drones is 30 m/s, at what time did the explosion take place? Assume that the speed of sound is 300 m/s.
Answer & explanation
Answer: (c) 4:58:42 PM
The second drone's one-way trip takes the 3 minutes between the two pictures, so the spot is 180 × 30 = 5400 m away. The first drone left 3 minutes before 5:02, and the sound had taken 5400 ÷ 300 = 18 s to arrive, so the explosion was 3 min 18 s before 5:02 PM.
- The second drone starts at 5:02 PM and its picture arrives at 5:05 PM, so one trip takes 180 s.
- Distance = 30 m/s × 180 s = 5400 m.
- The first drone also needed 180 s, so it left the camp at 4:59:00 PM — the moment the sound arrived.
- Sound took 5400 ÷ 300 = 18 s, so the explosion happened at 4:59:00 − 18 s = 4:58:42 PM.
- Check: 4:58:42 + 18 s + 180 s = 5:02:00 PM.
Remember · Work backwards on a timeline: find the distance from the cleanest interval, then subtract each leg's travel time.
Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·