Let the length of the 1st train = L mts

Speed of 1st train = 48 kmph

Now the length of the 2nd train = L/2 mts

Speed of 2nd train = 42 kmph

Let the length of the bridge = D mts

Distance = L + L/2 = 3L/2

Relative speed = 48 + 42 = 90 kmph = 90 x 5/18 = 25 m/s(opposite)

Time = 12 sec

=> 3L/2x25 = 12

=> L = 200 mts

Now it covers the bridge in 45 sec

=> distance = D + 200

Time = 45 sec

Speed = 48 x5/18 = 40/3 m/s

=> D + 200/(40/3) = 45

=> D = 600 - 200 = 400 mts

Hence, the length of the bridge = 400 mts.

Let 'd' be the distance and 's' be the speed and 't' be the time
d=sxt
45 mins = 3/4 hr and 48 mins = 4/5 hr
As distance is same in both cases;
s(3/4) = (s-5)(4/5)
3s/4 = (4s-20)/5
15s = 16s-80
s = 80 km.
=> d = 80 x 3/4 = 60 kms.

Speed =[ 72 x (5/18) ]m/sec= 20 m/sec.

Time = 26 sec.

Let the length of the train be x metres.

Then,[ (x+250)/26 ]= 20

=> x + 250 = 520

=> x = 270.

Let the speed of the faster train be 'S' kmph

Then speed of the slower train will be '(S-5)' kmph

Time taken by faster train = 350/S hrs

Time taken by slower train = 350/(S-5) hrs

  350 s - 5 - 350 s = 2 h r s 20 m i n = 2 1 3 = 7 3

=> S = 30 km/hr.

Given speed = 63 km/hr =  63 × 5 18 = 35 2 m/s

Let the length of the bridge = x mts

Given time taken to cover the distance of (170 + x)mts is 30 sec.

We know speed =  d i s tan c e t i m em/s

? 35 2 = 170 + x 30

 -->  x = 355 mts.

Distance = 70 x 1 ½ = 105 km

Relative Speed = 85 ? 70 = 15

Time = 105/15 = 7 hrs

4:30 + 7 hrs = 11.30 p.m.

We know that    S p e e d = d i s tan c e t i m e 

distance= speed * time

?  d = 40 × 5 18 × 9

   d= 100 mts.