UPSC Prelims 2026 · CSAT
Question 41: Basic Numeracy
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.
AnswerUPSC’s answer is (c).
Let d be the distance from the camp to the explosion. The first picture arrives after the sound's travel (d/300) plus the first drone's flight (d/30). The second drone leaves when the first picture arrives, at 5:02, and takes d/30 to reach the spot, so its picture arrives at 5:05: d/30 = 3 minutes = 180 s, so d = 5400 m. Sound took 5400/300 = 18 s. The explosion was 180 + 18 = 198 s before 5:02:00, which is 4:58:42 PM.
Why each option is right or wrong
- (a) Wrong: 4:59:00 is only 180 s before 5:02 and ignores the 18 s sound delay.
- (b) Wrong: 4:59:02 is 178 s before 5:02, not 198 s.
- (c) Right: 5:02:00 minus 198 s is 4:58:42.
- (d) Wrong: 4:56:32 is 328 s before 5:02, not 198 s.
Only the provisional key (27 May 2026) was found; the final key replaces it when found.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2026, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.