a) 200 m
b) 300 m
c) 350 m
d) 400 m

Answer: d
Explanation: Let the length of the train traveling at 48 kmph be 2x meters.
And length of the platform is y meters.
$$\eqalign{ & {\text{Relative speed of train}} \cr & = \left( {48 + 42} \right)\,{\text{kmph}} \cr & = {\frac{{90 \times 5}}{{18}}} \cr & = 25\,m/\sec \cr & {\text{And}}\,{\text{48}}\,{\text{kmph}} \cr & = \frac{{48 \times 5}}{{18}} \cr & = \frac{{40}}{3}\,m/\sec \cr & \cr & {\text{According to the question}}, \cr & \frac{{ {2x + x} }}{{25}} = 12; \cr & \,3x = 12 \times 25 = 300 \cr & \,x = \frac{{300}}{3} = 100m \cr & {\text{Then, length of the train}} \cr & = 2x \cr & = 100 \times 2 = 200m \cr & \frac{{200 + y}}{{ {\frac{{40}}{3}} }} = 45 \cr & 600 + 3y = 40 \times 45 \cr & {\text{Or}},\,3y = 1800 – 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200 \cr & {\text{Or}},\,y = \frac{{1200}}{3} = 400m \cr & {\text{Length of the platform}} \cr & = 400m \cr} $$