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Jawad001
Relative speed =(50-30) = 20 kmph
= 20(5/18) msec

Distance = 20*(5/18) * 18 meter
= 20*5
= 100 meter(B)

where do you get the 5 in 20(5/18) from?
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The faster train goes 50 - 30 = 20 kmph faster than the slower train. In other words, if the slower train stands still (i.e. speed =0), the faster train travels at 20 kmph.

In 3600 seconds the faster train travels 20,000 meters

In 18 seconds the faster train travels (20,000 X 18)/ 3600 = 100 meters.

Option: B.
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Stonely
we are talking about relative speed 50-30=20kph
20000m/h*1h/60min*1min/60sec
200/36=5.555555m/s
5.55555 m/s*18 s=99.999999 meters

the guy timing must have had a shaky finger and miss timed just a little ;)
answer is B
-------------------------------------------

But why are we calculating the relative speed when we just want to find out the length of the faster train wrt the man?

It can simply be: (50*1000*18)/(3600)?
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