Two guns are fired from the same place at an interval of 6 minutes. A person approaching the place observes that 5 minutes 52 seconds have elapsed between the hearings of the sound of the two guns. If the velocity of the sound is 330 m/sec, the man was approaching that place at what speed (in km/h)?

Difference of time
= 6 min - 5 mins. 52 secs.
= 8 secs.
Distance covered by man in 5 mins. 52 secs.
= Distance covered by sound in 8 secs.
= 330 × 8 = 2640 m.
∴ Speed of man
$$\eqalign{ & = \frac{2640\text{ m}}{\text{5 min. 52 secs.}} \cr & = \frac{2640}{352}\text{ m/secs} \cr & = \frac{2640}{352}\times\frac{18}{5}\text{ kmph} \cr & = 27 \text{ kmph} }$$