UPSC Prelims 2026 CSAT, question 41

Topic: Arithmetic

Q41

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 $300 m/s$ .

(a) 4:59:00 PM
(b) 4:59:02 PM
(c) 4:58:42 PMCorrect answer
(d) 4:56:32 PM

Solution.

Answer: (c) 4:58:42 PM

Since the second drone starts at 5:02 PM (when the first picture arrives) and reaches the explosion spot at 5:05 PM, the drone flight time for distance $D$ at $30$ m/s is $T = 3$ minutes $= 180$ seconds.
Because the speed of sound ( $300$ m/s) is $10$ times the speed of the drone ( $30$ m/s), the sound of the explosion took $\frac{T}{10} = \frac{180}{10} = 18$ seconds to reach the camp sensor before the first drone launched.
The first drone then took $180$ seconds to reach the spot at 5:02:00 PM, so the explosion occurred $180 + 18 = 198$ seconds ( $3$ minutes $18$ seconds) before 5:02:00 PM, which is 4:58:42 PM.