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

The interval between two gunshots is $9\text{ minutes}$.
The person hears the second shot after $8\text{ minutes }48\text{ seconds}$ because they are moving towards the source.
This implies that the distance covered by the sound in $(9\text{ minutes}-8\text{ minutes }48\text{ seconds})$ is equal to the distance covered by the person in $8\text{ minutes }48\text{ seconds}$.

Time difference for sound:

$$9\text{ min}-8\text{ min }48\text{ sec}=12\text{ seconds}$$

Time during which the person travels:

$$8\text{ min }48\text{ sec}=(8\times 60)+48=480+48=528\text{ seconds}$$

Let the speed of the person be $v\text{m/s}$.

$$\text{Distance covered by sound in }12\text{ sec}=\text{Distance covered by person in }528\text{ sec}$$

$$330\times 12=v\times 528$$

$$v=\frac{330\times 12}{528}=\frac{3960}{528}=7.5\text{m/s}$$

Convert the speed into $\text{km/hr}$:

$$\text{Speed}=7.5\times \frac{18}{5}=1.5\times 18=27\text{km/hr}$$