Two guns are fired from the same place at an interval of 9 minutes. A person approaching the place observes that 8 minutes 48 seconds have elapsed between the hearing of the sound of the two guns. If the velocity of the sound is 330 m/s, at what speed was the person approaching that place (in km/hr) ? 

Given:

Time interval between firing of two guns = 9 minutes = 540 seconds.

The person hears the second gunshot after 8 minutes 48 seconds = 528 seconds.

Velocity of sound = 330 m/s.

Formula used:

The difference in time is due to the person's approach. The time difference between the hearing of the two sounds is:

Time difference = 540 - 528 = 12 seconds.

Let the speed of the person be v (in m/s).

The effective speed at which the sound seems to approach the person is (330 + v).

Using the relation: Distance = Speed × Time, the distance covered by sound in both cases is the same, so:

330 × 540 = (330 + v) × 528

Calculation:

Simplify the equation:

330 × 540 = (330 + v) × 528

⇒ 178200 = (330 + v) × 528

Divide both sides by 528:

⇒ 337.5 = 330 + v

⇒ v = 337.5 - 330 = 7.5 m/s

Convert to km/hr:

⇒ v = 7.5 × 18/5 = 27 km/hr

∴ The speed of the person is 27 km/hr.