Aptitude - Problems on Trains - Discussion

How 15/4*75 hrs?

In the question as length of the tunnel is mentioned 3 miles and length of train 1 miles, how 7/2 and 1/4 is taken respectively?

We must add two times the distance of the train because we are just calculating the time needed to cross the tunnel if we just add one time the length of the train. So the answer must be 3.2min.

It doesn't need to explain the question that if we add only one time.

@Ashu.

They ask only for the time to cross the tunnel...
The train is driven by the engine, which is in the front part of the train.
;So we need to calculate the distance as;
7/2 mile + 1/4 mile = 3.75 mile.

Why put 7/2 In total distance covered?

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.

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

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.

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.