A jogger running at 9 kmph along the side of running track is 250 meters ahead of the engine of a  110 meters long train running at 63 kmph in the same direction. In how much time will the train pass the jogger ?

# A jogger running at 9 kmph along the side of running track is 250 meters ahead of the engine of a  110 meters long train running at 63 kmph in the same direction. In how much time will the train pass the jogger ?

1. A

26 sec

2. B

24 sec

3. C

22 sec

4. D

28 sec

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### Solution:

$\text{Relative\hspace{0.17em}speed}={S}_{1}-{S}_{2}=\left(63-9\right)=54\text{\hspace{0.17em}}kmph$

$1\text{\hspace{0.17em}\hspace{0.17em}}km=1000\text{\hspace{0.17em}}m,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\text{\hspace{0.17em}\hspace{0.17em}}hr=3600\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{sec}$

$T=S×D$

$=54×\frac{1000}{3600}=15\text{\hspace{0.17em}}m/\mathrm{sec}$

$Time=\frac{360}{15}=24\text{\hspace{0.17em}}\mathrm{sec}$  +91

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