Aptitude - Problems on Trains - Discussion

Total distance covered.
= 7/2 + 1/4.
=15/4 miles.

Can anyone explain me, how to convert it?

How come 15/4 could you explain?

So, if I change the speed to 90mph, then according to the way this is solved, the time would increase. Now that does not sound right to me. Can anyone explain this, please ?

I think here 75 is kmph not mph.

Can anyone explain about 7/2 how did we get it?

Here, it is asked for the time the front enters the rare emerges. So we will have to Add 1/4 more to the length of the Train Right.

3.75 miles.
3.75/75=.05,
.05 * 60 = 30min,

miles = meters.

The total distance should be,
(7/2 +2(1/4)) = 4miles,
So the required time will be 3.2 second.

75mph.

Where mph is miles per hour. Then only the answer is correct. If it is meters per hour then the given answer is wrong.

Here, consider only the head of the train. If the train is entering the tunnel, the head is at the entrance of the tunnel (obviously). After the train's head is reached at the end of the tunnel, it has covered a distance of the length of the tunnel. But we want the tail of the train to be at the end of the tunnel.

So, the train will have to move the distance equal to it's length. Now, it's clear that total distance=tunnel length+train length.