Aptitude - Problems on Trains - Discussion

Guys 1mile = 1.6km.
For train speed:
So 75mph = 75*1.6 = 120kmph.

For tunnel length and train length:
7/2miles*1.6=5.6km and 1/4miles*1.6=0.4km.
So total distance=5.6+0.4=6km.

Now to calculate the time we have the formulae is t=d/s.
Lets convert all data into metres.
So 6km*1000=6000m.

Speed=120*5/18 m/sec=100/3 m/sec.
So t=6000/100/3 which is equal to 18000/100=180seconds.

So to convert seconds into minutes is(1min=60sec) 180/60=3min.
Try to understand because it's too long.

Why in this sum we didn't change speed kmph into m/sec?

Actually, whenever a train crosses platform we are considering the length of train+length of the platform, similarly, in this case, we consider the length of tunnel+length of train i.e7/2+1/4.

Distant: 7/2+1/4+1/4 miles = 4 miles.
Speed: 5/4 miles/minute.
Time duration should be 3.2 minutes, I think.

(7/2+1/4) = (28+2)/4 = 30/4 = 15/2 , but how you will tell its as 15/4?

Can you explain it clearly?

@Palla Nagesh.

7/2+1/4 = 14 + 1/4.
Take LCM for 2 & 4 then final lcm is 4.
Because we write 4 at denominator then 2*2=4 2*7=14 4*1=4

And finally, we get the ans 15/2.

I think only the length of the tunnel must be considered because only the time taken for the front part to re-emerge is asked.

Correct me, if I am wrong.

75 mph is nothing, it's 75 miles per hour.

@All.

Given, A train travelling at a speed of 75 mph enters a tunnel 3 1/2 miles long.
Here, the speed is given in miles per hour it is not meter per hour and the distance is given wrong it is 7 1/2.
Now, calculate the answer will come as 0.05 hours to convert hours to min multiplied by 60.
So, 0.05 x 60 = 3 min.