Solution (By Examveda Team)
$$\eqalign{
& {\text{Let the length of train be x m}} \cr
& {\text{When a train crosses a light }} \cr
& {\text{post in 3 second the distance covered}} \cr
& {\text{ = length of train }} \cr
& \Rightarrow {\text{speed of train = }}\frac{x}{3} \cr
& {\text{Distance covered in crossing a}} \cr
& {\text{900 meter platfrom in 30 seconds}} \cr
& {\text{ = Length of platfrom + length of train}} \cr
& {\text{Speed of train = }}\frac{{x + 900}}{30} \cr
& \Rightarrow \frac{x}{3} = \frac{{x + 900}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr
& \Rightarrow \frac{x}{1} = \frac{{x + 900}}{{10}} \cr
& \Rightarrow 10x = x + 900 \cr
& \Rightarrow 10x - x = 900 \cr
& \Rightarrow 9x = 900 \cr
& \Rightarrow x = \frac{{900}}{9} = 100{\text{m}} \cr
& {\text{When the length of the platform be 800m,}} \cr
& {\text{then time T be taken by train to cross 800m}} \cr
& {\text{long platfrom}} \cr
& \frac{x}{3} = \frac{{x + 800}}{T} \cr
& \Rightarrow Tx = 3x + 2400 \cr
& \Rightarrow 100T = 300 + 2400 \cr
& \Rightarrow 100T = 2700 \cr
& \Rightarrow T = \frac{{2700}}{{100}} = 27{\text{ seconds}} \cr} $$
Join The Discussion