A) 400 m
B) 450 m
C) 560 m
D) 600 m
Correct Answer: A
$ \therefore $ Relative speed $ =(48+42)km/h $
$ =( 90\times \frac{5}{18} )=25,m/s $
According to the question,
$ \frac{( x+\frac{x}{2} )}{25}=12 $
$ \Rightarrow $ $ \frac{3x}{2}=300 $
$ \Rightarrow $ $ x=200m $
$ \therefore $ Length of first train $ =200m $ Let the length of platform be y m. Speed of the first train $ =( 48\times \frac{5}{18} )m/s=\frac{40}{3}m/s $ $ Time=\frac{Distance}{Speed} $
$ \therefore $ $ (200+y)\times \frac{3}{40}=45 $
$ \Rightarrow $ $ 600+3y=1800 $
$ \therefore $ $ y=400m $