2 identical trains take 3 sec to pass 1 another when going in the opposite direction but only 2.5 sec if speed of 1 is increased by 50%.The time one would take to pass the other when going in the same direction at their original speed is?

let train A is moving along positie x axis with vel = +V

train B along negative x axis = -U

vel of train A with respect B is = V-(-U) = V+U

let S be distance

S=(V+U)3

S=(V + U + U/2)2.5

=> S= (V+3/2U)2.5

get v and u in terms of S

then apply S=(V-U)t    (now V along and u in same direction v=V-U)

and find t

may be answer 30/4 secs

working in ref frame attached to 2,

velocity of 1 wrt 2= v₁+v₂

let length of train be x metres each.

therefore time taken to cross each other= 2x/v₁₊v₂=3

=> 2x= 3v₁+3v_{2----------------------------------------------(1)}

Case 2:

velocity of 1= v₁+50v₁/100=3v₁/2

now, time taken to cross each other=      2x                                                             -------------- = 2.5

                                                             3v₁/2- v₂

=> 2x= 2.5v₂+ 7.5v₁/2---------------(2)

equating (1) and (2)

3v₁+3v₂= 2.5v₂+7.5v₁/2

=> 1.5v₁/2= v_{2--------------------------------(3)}

now if they move in same direction:

velocity of 1 wrt 2= v₁-v₂

time taken to cross each other= 2x/v₁-v₂

                                              = 2x/v₁-0.75v₁ (from 3)

                                              = 2x/0.25v₁

                                             = 3v₁+3 v₂/ 0.25v₁ (from 1)

                                              = 3v₁+ 0.75v₁/0.25v₁ (from 3)

                                              = 3.75v₁/0.25v₁

                                              = 15 sec

all the best!