Aptitude - Problems on Trains - Discussion

@All.

Refer Rajkumar explanation, he explained very well.

Thanks @Rajkumar.

Please explain it, easily. I can't able to do it.

Why considered only 36 in the above step?

I don't understand last 3 lines. Can anyone explain last 3 lines, that how much get answer 50?

Faster train 46 km/hr, and small train 36 km/hr then, 46 - 36 = 10 km/hr.
Then, 10 * 5/18 = 25/9.

It crosses 36 seconds then 36 * 9/25 = 12.96m then how it becomes 50 m?

I am confused. please explain

I have a doubt, hear if two trains are same length and same direction in that faster train speed 10km/h more then slower train in this case the faster train speed only 10km/h only and it crossed in slower train in 36 seconds that means faster train speed more then (25/9) m/s.

And they mentioned faster train takes 36sec to cross the slower train that means.

According to formula length = speed * time.

(25/9) * 36 = 200m.

The actual length of the train = 200m.

I think none of the above option is the answer.

After relative speed one have '0' and other '10kmph'.

Then,
-----------> <=1st train with 46kmph relative speed is '10 kmph.
-----------> <=2nd train with 36kmph relative speed is '0' kmph.

After passing,
-----------> <=with relative speed '10' kmph.
-----------> <=with relative speed '0' kmph.

So only one train length speed to be calculated, then why it take "2x"?

Please explain.

(X1 + x2/u - v) = 36sec. Formula x1, x2 = x.
(2x/ 46 - 36) = 36sec.
(2x/10) = 36sec.
(10 * 5/18) = (25/9).
(2x/36sec) = (25/9).
2x = 100m.
X = 50m.

Well explained. Thanks.

In other words, we can say that in this question slowest train is at rest and fast train in moving with speed of 10 km/hr and to pass slow train fast train only needs to pass it's length because both trains start at same point so the distance covered by train in 36 sec is,

10 * (5/18) * 36 = 100m.

So the correct answer is 100 not 50m.