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)?
A. 24 kmph B. 27 kmph C. 30 kmph D. 36 kmph Answer: Option BShow Answer
Solution(By Apex Team)
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
$\begin{array}{l}=\frac{2640\mathrm{~m}}{5\mathrm{~\min}.52\mathrm{\secs}}\\
=\frac{2640}{352}\mathrm{~m}/\mathrm{\sec s}\\
=\frac{2640}{352}\times\frac{18}{5}\mathrm{kmph}\\
=27\mathrm{kmph}\end{array}$
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