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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 B
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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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