QUESTION

CSAT

Hard

Maths

Prelims 2026

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.

Select an option to attempt

Explanation

Answer: C — 4:58:42 PM

Let $T$ be the time of the explosion and $d$ the distance from the camp to the explosion site.

Time interval between the two pictures.

The second drone leaves at 5:02 PM (when the first picture arrives) and reaches the spot in $\frac{d}{30}$ seconds, with its picture received the moment it lands. So:

$\frac{d}{30} = (5{:}05 - 5{:}02) = 180 \text{ s} \;\Rightarrow\; d = 5400 \text{ m}$

Working backward from the first picture (5:02 PM).

After the explosion at $T$ :

Sound takes $\frac{d}{300} = 18$ s to reach the camp.
The first drone then takes $\frac{d}{30} = 180$ s to reach the spot (picture received instantly).