Let the length of the first train be x metres.

Then, the length of the second train is\(\left(\frac{x}{2}\right)metres.\)

Relative speed = (48 + 42) kmph =\(\left(90\times\frac{5}{18}\right)m/sec = 25m/sec\)

\(\frac{[x+(\frac{x}{2})]}{25}=12 or \frac{3x}{2}=300 . or . x= 200\)

Therefore Length of first train = 200 m.

Let the length of platform be y metres.

Speed of the first train = \(\left(48\times\frac{5}{18}\right)m/sec = \frac{40}{3}m/sec\)

\(\therefore\left(200+y\right)\times\frac{3}{40}= 45\)

 600 + 3y = 1800

 y = 400 m.