Answer: Option C. -> 36 sec
$$\eqalign{
& {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{jogger}} \cr
& = \left( {45 - 9} \right)\,{\text{km/hr}} \cr
& = 36\,{\text{km/hr}} \cr
& {36 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr
& = 10\,{\text{m/sec}} \cr
& {\text{Distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr
& = \left( {240 + 120} \right)\,m \cr
& = 360\,m \cr
& \therefore {\text{Time}}\,{\text{taken}} \cr
& = {\frac{{360}}{{10}}} \,{\text{sec}} \cr
& = 36\,{\text{sec}} \cr} $$

Answer: Option B. -> 3 min
$$\eqalign{
& {\text{Total}}\,{\text{distance}}\,{\text{covered}} \cr
& = \left( {\frac{7}{2} + \frac{1}{4}} \right)\,{\text{miles}} \cr
& = \frac{{15}}{4}\,{\text{miles}} \cr
& \therefore {\text{Time}}\,{\text{taken}} \cr
& = \left( {\frac{{15}}{{4 \times 75}}} \right)\,{\text{hrs}} \cr
& = \frac{1}{{20}}\,{\text{hrs}} \cr
& = \left( {\frac{1}{{20}} \times 60} \right)\,\min \cr
& = 3\,\min \cr} $$

Answer: Option B. -> 150 m
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr
& \,{\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{by}}\,y\,{\text{m/sec}} \cr
& Then,\,\frac{x}{y} = 15\,\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \frac{x}{{15}} \cr
& \therefore \frac{{x + 100}}{{25}} = \frac{x}{{15}} \cr
& \Rightarrow 15\left( {x + 100} \right) = 25x \cr
& \Rightarrow 15x + 1500 = 25x \cr
& \Rightarrow 1500 = 10x \cr
& \Rightarrow x = 150m \cr} $$

Answer: Option D. -> 79.2 km/hr
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr
& \,{\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{by}}\,y\,{\text{m/sec}} \cr
& {\text{Then}},\,\frac{x}{y} = 8\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 8y \cr
& {\text{Now}},\,\frac{{x + 264}}{{20}} = y \cr
& \Rightarrow 8y + 264 = 20y \cr
& \Rightarrow y = 22 \cr
& \therefore {\text{Speed}} = 22\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {22 \times \frac{{18}}{5}} \,{\text{km/hr}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 79.2\,{\text{km/hr}} \cr} $$

Answer: Option C. -> 500
$$\eqalign{
& {\text{Speed}} = \left( {78 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\frac{{65}}{3}} \,{\text{m/sec}} \cr
& {\text{Time = }}\,{\text{1}}\,{\text{minute = 60}}\,{\text{second}}. \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{tunnel}}\,{\text{be}}\,x\,{\text{metres}}. \cr
& {\text{Then}},\, {\frac{{800 + x}}{{60}}} = \frac{{65}}{3} \cr
& \Rightarrow 3\left( {800 + x} \right) = 3900 \cr
& \Rightarrow x = 500 \cr} $$

Answer: Option B. -> 350 m
$$\eqalign{
& {\text{Speed}} = {\frac{{300}}{{18}}} \,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50}}{3}\,{\text{m/sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{platform}}\,{\text{be}}\,x\,{\text{metres}}{\text{.}} \cr
& {\text{Then}}, {\frac{{x + 300}}{{39}}} = \frac{{50}}{3} \cr
& \Rightarrow 3\left( {x + 300} \right) = 1950 \cr
& \Rightarrow x = 350\,m. \cr} $$

Answer: Option B. -> 4 : 3
$$\eqalign{
& {\text{Let}}\,{\text{us}}\,{\text{name}}\,{\text{the}}\,{\text{trains}}\,{\text{as}}\,{\text{A}}\,{\text{and}}\,{\text{B}}{\text{.}}\,{\text{Then}}, \cr
& \left( {{\text{A's}}\,{\text{speed}}} \right):\left( {{\text{B's}}\,{\text{speed}}} \right) \cr
& = \sqrt b :\sqrt a \cr
& = \sqrt {16} :\sqrt 9 \cr
& = 4:3\, \cr} $$

Answer: Option E. -> None of these
$$\eqalign{
& {\text{Let the length of the train be x metres}}{\text{.}} \cr
& {\text{Then, length of the platform = (2}}x{\text{) metres}}{\text{.}} \cr
& {\text{Speed of the train}} \cr
& {\text{ = }}\left( {90 \times \frac{5}{{18}}} \right)m/\sec \cr
& = 25m/sec \cr
& \therefore \frac{{x + 2x}}{{25}} = 36 \cr
& \Rightarrow 3x = 900 \cr
& \Rightarrow x = 300 \cr
& {\text{Hence, length of platform}} \cr
& {\text{ = }}2x = \left( {2 \times 300} \right){\text{m}} = 600{\text{m}} \cr} $$

Answer: Option C. -> 14.4 sec
$$\eqalign{
& {\text{Speed }} \cr
& {\text{ = }}\left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr
& {\text{ = }}\frac{{50}}{3}{\text{ m/sec}} \cr
& {\text{Total distance covered}} \cr
& {\text{ = (100 + 140) m = 240 m}} \cr
& \therefore {\text{Required time}} \cr
& {\text{ = }}\left( {240 \times \frac{3}{{50}}} \right){\text{sec}} \cr
& {\text{ = }}\frac{{72}}{5}{\text{sec}} \cr
& {\text{ = 14}}{\text{.4 sec}} \cr} $$

Answer: Option C. -> 79.2 km/hr
$$\eqalign{
& {\text{Speed}} \cr
& {\text{ = }}\left( {\frac{{132}}{6}} \right){\text{m/sec}} \cr
& {\text{ = }}\left( {22 \times \frac{{18}}{5}} \right){\text{km/sec}} \cr
& {\text{ = 79}}{\text{.2 km/hr}} \cr} $$