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

Answer: Option B. -> 36 kmph
$$\eqalign{
& {\text{Distance = Speed }} \times {\text{Time}} \cr
& {\text{Distance covered by train with the}} \cr
& {\text{speed of 30 kmph in 12 minutes is }} \cr
& {\text{ = 30}} \times \frac{{12}}{{60}} = 6{\text{km}} \cr
& {\text{Distance covered by the same train}} \cr
& {\text{with the speed of 45 kmph in 8 minutes is }} \cr
& {\text{ = 45}} \times \frac{8}{{60}} = 6{\text{km}} \cr
& {\text{Average speed}} \cr
& {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{total time}}}}. \cr
& \Rightarrow \frac{{\left( {6 + 6} \right){\text{km}}}}{{\left( {12 + 8} \right)\min }} = \frac{{12}}{{20}} \times 60 \cr
& {\text{ = 36 kmph}} \cr} $$

Answer: Option A. -> 230 m
$$\eqalign{
& {\text{Relative}}\,{\text{speed}} \cr
& = \left( {120 + 80} \right)\,{\text{km/hr}} \cr
& = {200 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr
& = {\frac{{500}}{9}} \,{\text{m/sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{other}}\,{\text{train}}\,{\text{be}}\,{\text{x}}\,{\text{metres}}{\text{.}} \cr
& {\text{Then,}}\,\frac{{x + 270}}{9} = \frac{{500}}{9} \cr
& \Rightarrow x + 270 = 500 \cr
& \Rightarrow x = 230 \cr} $$

Answer: Option B. -> 6 sec
$$\eqalign{
& {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr
& = \left( {60 + 6} \right)\,{\text{km/hr}} \cr
& = 66\,{\text{km/hr}} \cr
& = {66 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr
& = {\frac{{55}}{3}} \,{\text{m/sec}} \cr
& \therefore {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr
& = {110 \times \frac{3}{{55}}} {\text{sec}} = 6\,{\text{sec}} \cr} $$

Answer: Option D. -> 10.8
$$\eqalign{
& {\text{Relative}}\,{\text{speed}} = \left( {60 + 40} \right)\,{\text{km/hr}} \cr
& = {100 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr
& = {\frac{{250}}{9}} \,{\text{m/sec}}. \cr
& {\text{Distance}}\,{\text{covered}}\,{\text{in}}\,{\text{crossing}}\,{\text{each}}\,{\text{other}} \cr
& = \left( {140 + 160} \right)m = 300\,m \cr
& {\text{Required}}\,{\text{time}} \cr
& = {300 \times \frac{9}{{250}}} \,{\text{sec}} \cr
& = \frac{{54}}{5}\,{\text{sec}} \cr
& = 10.8\,{\text{sec}} \cr} $$

Answer: Option C. -> 60 km/hr
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr
& {\text{Then,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}} = 2x\,{\text{m/sec}} \cr
& {\text{Relative}}\,{\text{speed}} = \,\left( {x + 2x} \right)\,{\text{m/sec}} = 3x\,{\text{m/sec}} \cr
& \therefore \frac{{ {100 + 100} }}{8} = 3x \cr
& \Rightarrow 24x = 200 \cr
& \Rightarrow x = \frac{{25}}{3} \cr
& {\text{So,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}}\, = \frac{{50}}{3}\,{\text{m/sec}} \cr
& = {\frac{{50}}{3} \times \frac{{18}}{5}} \,{\text{km/hr}} \cr
& = 60\,{\text{km/hr}} \cr} $$

Answer: Option D. -> 270 m
$$\eqalign{
& {\text{Speed}} = {72 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\,{\text{m/sec}} \cr
& {\text{Time}} = 26\,{\text{sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}{\text{.}} \cr
& {\text{Then}},\,\frac{{x + 250}}{{26}} = 20 \cr
& \Rightarrow x + 250 = 520 \cr
& \Rightarrow x = 270 \cr} $$

Answer: Option B. -> 24 sec
$$\eqalign{
& {\text{Relative}}\,{\text{speed}} \cr
& = \left( {45 + 30} \right)\,{\text{km/hr}} \cr
& = \left(75 \times \frac{5}{18} \right)\, \text{m/sec} \cr
& = {\frac{{125}}{6}} \,{\text{m/sec}} \cr
& {\text{We}}\,{\text{have}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{time}}\,{\text{taken}}\,{\text{by}}\,{\text{the}} \cr
& {\text{slower}}\,{\text{train}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{DRIVER}}\,{\text{of}}\, \cr
& {\text{The}}\,{\text{faster}}\,{\text{train}}\,{\text{and}}\,{\text{not}}\,{\text{the}}\,{\text{complete}}\,{\text{train}}{\text{.}} \cr
& \cr
& {\text{So,}}\,{\text{distance}}\,{\text{covered = Length}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}. \cr
& {\text{Therefore,}}\,{\text{Distance}}\,{\text{covered = 500}}\,{\text{m}}. \cr
& \therefore {\text{Required}}\,{\text{time}} \cr
& = {500 \times \frac{6}{{125}}} \cr
& = 24\,\sec \cr} $$

Answer: Option C. -> 36
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{each}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}}. \cr
& {\text{Then,}}\,{\text{relative}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{two}}\,{\text{trains}} = 2x\,{\text{m/sec}} \cr
& {\text{So}},\,2x = \frac{{ {120 + 120} }}{{12}} \cr
& \Rightarrow 2x = 20 \cr
& \Rightarrow x = 10 \cr
& \therefore {\text{Speed}}\,{\text{of}}\,{\text{each}}\,{\text{train}} = 10\,{\text{m/sec}} \cr
& = {10 \times \frac{{18}}{5}} \,{\text{km/hr}} \cr
& = 36\,{\text{km/hr}} \cr} $$

Answer: Option B. -> 30
$$\eqalign{
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr
& = \left( {63 - 3} \right)\,{\text{km/hr}} \cr
& = 60\,{\text{km/hr}} \cr
& = \left( {60 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& = \frac{50}{3}\, \text{m/sec} \cr
& \therefore {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr
& = \left( {500 \times \frac{3}{{50}}} \right)\,\sec \cr
& = 30\,\sec \cr} $$