a) 32 seconds
b) 31 seconds
c) 33 seconds
d) 30 seconds

Answer: c
Explanation:
$$\eqalign{ & {\text{Let the length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 9 second the distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{9} \cr & {\text{Distance covered in crossing a}} \cr & {\text{700 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 700}}{30} \cr & \Rightarrow \frac{x}{9} = \frac{{x + 700}}{{30}}\left[ { {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \frac{x}{3} = \frac{{x + 700}}{{10}} \cr & 10x = 3x + 2100 \cr & 10x – 3x = 2100 \cr & 7x = 2100 \cr & x = \frac{{2100}}{7} = 300{\text{m}} \cr & {\text{When the length of the platform be 800m,}} \cr & {\text{then time T be taken by train to cross 800m}} \cr & {\text{long platform}} \cr & \frac{x}{9} = \frac{{x + 800}}{T} \cr & Tx = 9x + 7200 \cr & 300T = 2700 + 7200 \cr & 300T = 9900 \cr & T = \frac{{9900}}{{300}} = 33{\text{ seconds}} \cr} $$