a) 400 m
b) 450 m
c) 560 m
d) 600 m

Answer: a
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then,}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{is}}\, {\frac{x}{2}} \,{\text{metres}} \cr & {\text{Relative}}\,{\text{speed}} = \left( {48 + 42} \right)\,{\text{kmph}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {90 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25\,{\text{m/sec}} \cr & \frac{{ {x + \left( {x/2} \right)} }}{{25}} = 12 \cr & \frac{{3x}}{2} = 300 \cr & x = 200 \cr & {\text{Length}}\,{\text{of}}\,{\text{first}}\,{\text{train}} = 200\,{\text{m}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{platform}}\,{\text{be}}\,y\,{\text{metres}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr & = \left( {48 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \frac{{40}}{3}\,{\text{m/sec}} \cr & \therefore \left( {200 + y} \right) \times \frac{3}{{40}} = 45 \cr & \Rightarrow 600 + 3y = 1800 \cr & \Rightarrow y = 400\,{\text{m}} \cr} $$.