Let the speed of 1st train be v.

Let the speed of 2nd train be u.

Let their combined length (distance) be d.

When going in opposite directions with original speeds, 

=> (v+u) = d/3

=> d= 3v + 3u..........(1)

When speed of one train is increased by 50 %, v ---> 1.5v

=> (1.5v+u)=d/2.5

=> d = 3.75v + 2.5 u............(2)

=>(2)-(1)

=>0.75v - 0.5u = 0

=>1.5v = u

time = distance/speed

When going in same direction, time= d/(u-v) 

Since d = 3v + 3u = 3v + 3*1.5v= 7.5 v

u - v= 0.5v

=> time = 7.5v/0.5v = 15 seconds

Please approve my answer if I was of any help.

Ok, I didnt see that.

The method is still the same, but in examination you would have to add a few lines to the solution.

In the first case, from ground frame,

The first train is moving to the RIGHT with velocity v.

The second train is moving to the LEFT with velocity u.

So, without using direction, we can say if 1st train is moving with velocity v, the second is moving with velocity (-u)<--- NOTICE the NEGATIVE sign.

So velocity of 1st train with respect to 2nd (relative velocity) will be v-(-u)=v+u.

Now you apply the same formula.

In the last case, while moving in the same direction, the velocities will be +v and +u, so relative velocity will be v-u. Apply the method in my first reply to get the answer.

Please approve!!