Aptitude - Problems on Trains - Discussion

The answer is C.

Distance=speed * time.

3.1/2+1/4=75 m/h *time,
3.5+.25=75m/h*time,
3.75=75m/h *time,
3.75/75m/h=time,
Time=.05*60=3min.

Given data is D1=3(1/2), D2=1/4 S=75m/s.
respective length is (7/2+1/4)=15/4,
S=D/T.
75=(15/4)/T,
T=(15/4)/(1/75),
=15/4*75,
=1/20.

SO ITS IN SEC.
CONVERT INTO HRS.
= (1/20)*60.
= 3 MINS.

7/2 + 1/4 = 15/4 how it comes? Please explain someone.

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.

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.

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

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

miles = meters.

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.

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