Aptitude - Problems on Trains - Discussion

T = (L1 + L2)/(S1 + S2),
T * (S1 - S2) = L1 + L2,
36 * 10 * 5/18 = L1 + L2,
100 = L1 + L2.
Given that L1 = L2.
Then,
2L1 = 100.
L1 = 50.

@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.