(b) 700 m
Let the length of the first train be x metres.
Then, length of the second train = \(\frac{x}{2}\) metres
Relative speed = (36 + 54) km/hr = 90 km/h
= \(\big(90\times\frac5{18}\big)\) m/s = 25 m/s
∴ \(\frac{x+\frac{x}2}{25}=12\) ⇒ \(\frac{3x}{2} = 300 \) ⇒ \(x\) = 200.
∴ Length of the first train = 200 m.
Let the length of the platform be y metres.
Speed of the first train = \(\big(36\times\frac5{18}\big)\) m/s = 10 m/s
∴ (200 + y) x \(\frac1{10}\) = 90
⇒ 200 + y = 900 ⇒ y = 700 m.