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
Solution(By Examveda 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
$$\eqalign{ & = \frac{2640\text{ m}}{\text{5 min. 52 secs.}} \cr & = \frac{2640}{352}\text{ m/secs} \cr & = \frac{2640}{352}\times\frac{18}{5}\text{ kmph} \cr & = 27 \text{ kmph} }$$
This Question Belongs to Arithmetic Ability >> Speed Time And Distance
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Comments ( 4 )
Related Questions on Speed Time and Distance
A. 48 min.
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D. 62 min.
E. 66 min.
A. 262.4 km
B. 260 km
C. 283.33 km
D. 275 km
E. None of these
A. 4 hours
B. 4 hours 30 min.
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We can do the math by considering relative speed.
Time difference is [6*60-(5*60+52)]sec=8 sec
Covered distance by sound is (8×330)m
Time taken by men is (352) sec
So, speed of men is (8×330)/352 m/s=(15/2)×(18/5) kmph=27 kmph
Answer 27 kmph
I mean why relative speed not taken into consideration
What if the person is not approaching and running away in the direction bullet fired after 1st bullet and heard the 2nd bullet sound after 6min 8 sec. Will the answer change. And when 2nd bullet fired, both man and sound in motion, so why relative is not taken into consideration