Answer: Option C. -> 250 meters
$$\eqalign{
& {\text{Speed}} \cr
& {\text{ = }}\left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr
& {\text{ = }}\frac{{50}}{3}{\text{m/sec}} \cr
& {\text{Time = 27 sec}}{\text{.}} \cr
& {\text{Let the length of the train be }}x{\text{ metres}}{\text{.}} \cr
& {\text{Then,}}\frac{{x + 200}}{{27}}{\text{ = }}\frac{{50}}{3}{\text{ }} \cr
& \Leftrightarrow x + 200 = \left( {\frac{{50}}{3} \times 27} \right) = 450 \cr
& \Leftrightarrow x = 450 - 200 = 250{\text{ metres}} \cr} $$

Answer: Option D. -> 54 kmph
Speed of first train = $$\frac{{150}}{{15}}$$ m/sec = 10 m/sec
Let the speed of second train be x m/sec
Relative speed = (10 + x) m/sec
∴ $$\frac{{300}}{{10 + {\text{x}}}}$$  = 12
⇒ 300 = 120 + 12x
⇒ 12x = 180
⇒ x = $$\frac{{180}}{{12}}$$ = 15 m/sec
Hence, speed of other train
= $$\left( {15 \times \frac{{18}}{5}} \right)$$  kmph
= 54 kmph

Answer: Option B. -> 2 km/hr
Speed of the train relative to first man
$$\eqalign{
& = \frac{{75}}{{7.5}}{\text{m/sec}} = 10\,{\text{m/sec}} \cr
& = \left( {10 \times \frac{{18}}{5}} \right){\text{km/hr}} = 36\,{\text{km/hr}} \cr} $$
Let the speed of the train be x km/hr.
Then, relative speed = (x - 6) km/hr
∴ x - 6 = 36
⇒ x = 42 km/hr
Speed of the train relative to second man
$$\eqalign{
& {\text{ = }}\frac{{75}}{{6\frac{3}{4}}}\,{\text{m/sec}} \cr
& = \left( {75 \times \frac{4}{{27}}} \right){\text{m/sec}} \cr
& = \frac{{100}}{9}{\text{m/sec}} \cr
& = \left( {\frac{{100}}{9} \times \frac{{18}}{5}} \right){\text{km}} \cr
& = 40\,{\text{km/hr}} \cr} $$
Let the speed of the second man be y kmph.
Then, relative speed = (42 - y) kmph
∴ 42 - y = 40
⇒ y = 2 km/hr

Answer: Option D. -> 36 km/hr
Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec
So, 2x = $$\frac{{120 + 120}}{{12}}$$
⇒ 2x = 20
⇒ x = 10
∴ Speed of each train = 10 m/sec
= $$\left( {10 \times \frac{{18}}{5}} \right)$$  km/hr
= 36 km/hr

Answer: Option B. -> $$3\frac{3}{7}$$ seconds
Let the length of each train be x meters
Then, speed of first train = $$\frac{{\text{x}}}{3}$$ m/sec
Speed of second train = $$\frac{{\text{x}}}{4}$$ m/sec
∴ Required time
$$\eqalign{
& = \left[ {\frac{{{\text{x}} + {\text{x}}}}{{\left( {\frac{{\text{x}}}{3} + \frac{{\text{x}}}{4}} \right)}}} \right]{\text{sec}} \cr
& = \left[ {\frac{{2{\text{x}}}}{{\left( {\frac{{7{\text{x}}}}{{12}}} \right)}}} \right]{\text{sec}} \cr
& = \left( {2 \times \frac{{12}}{7}} \right){\text{sec}} \cr
& = \frac{{24}}{7}{\text{sec}} \cr
& = 3\frac{3}{7}{\text{sec}} \cr} $$

Answer: Option B. -> 42 m/sec
Let the speeds of the faster and slower trains be x m/sec and y m/sec respectively.
Then, $$\frac{{240}}{{{\text{x}} - {\text{y}}}}$$  = 60
⇒ x - y = 4 . . . . . . . . (i)
And, $$\frac{{240}}{{{\text{x}} + {\text{y}}}}$$  = 3
⇒ x + y = 80 . . . . . . . . (ii)
Adding (i) and (ii), we get
2x = 84
⇒ x = 42
Putting x = 42 in (i), we get: y = 38
Hence, speed of faster train = 42 m/sec

Answer: Option C. -> 80.64
$$\eqalign{
& {\text{Length of train}} \cr
& {\text{ = 280 m }} \cr
& {\text{Length of platform}} \cr
& {\text{ = (3}} \times {\text{280) m = 840m}} \cr
& \therefore {\text{Speed of train}} \cr
& {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr
& = \frac{{1120}}{{50}}m/\sec \cr
& = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr
& = 80.64\,km/hr \cr} $$

Answer: Option C. -> 40
$$\eqalign{
& {\text{Speed = }}\left( {\frac{{150 + 300}}{{40.5}}} \right)m/\sec \cr
& = \left( {\frac{{450}}{{40.5}} \times \frac{{18}}{5}} \right)km/hr \cr
& = 40km/hr \cr} $$

Answer: Option D. -> 2 min 24 sec
$$\eqalign{
& {\text{Relative speed}} \cr
& {\text{ = (72}} - {\text{60) km/hr}} \cr
& {\text{ = 12 km/hr}} \cr
& = \left( {12 \times \frac{5}{{18}}} \right)m/\sec \cr
& = \left( {\frac{{10}}{3}} \right)m/\sec \cr
& {\text{Total distance covered}} \cr
& {\text{ = Sum of lengths of trains}} \cr
& {\text{ = (240 + 240) m}} \cr
& {\text{ = 480 m}} \cr
& {\text{Time taken}} \cr
& {\text{ = }}\left( {480 \times \frac{3}{{10}}} \right)\sec \cr
& = 144\sec \cr
& = 2\min \,24sec \cr} $$

Answer: Option C. -> 48
$$\eqalign{
& {\text{Relative speed}} \cr
& {\text{ = (60 + 90) km/hr}} \cr
& {\text{ = }}\left( {150 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr
& {\text{ = }}\left( {\frac{{125}}{3}} \right){\text{m/sec}} \cr
& {\text{Distance coverd}} \cr
& {\text{ = (1}}{\text{.10 + 0}}{\text{.9)km}} \cr
& {\text{ = 2 km}} \cr
& {\text{ = 2000 m}}{\text{}} \cr
& {\text{Required time}} \cr
& {\text{ = }}\left( {2000 \times \frac{3}{{125}}} \right)\sec \cr
& = 48{\text{ sec}}\cr} $$