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

$$\eqalign{ & {\text{Speed of train}} \cr & {\text{ = }}\frac{{700 + 500}}{{10}} \cr & {\text{ = 120 ft/second}} \cr} $$

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

$$\eqalign{ & {\text{Let length of train }}x{\text{ m }} \cr & {\text{Speed of train }} \cr & {\text{ = }}\frac{{\left( {{\text{Length of train + length of bridge }}} \right)}}{{{\text{Time taken in crossing}}}}{\text{ }} \cr & {\text{According to information we get}} \cr & \Rightarrow \frac{{x + 500}}{{100}} = \frac{{x + 250}}{{60}} \cr & \Rightarrow 60\left( {x + 500} \right) = 100\left( {x + 250} \right) \cr & \Rightarrow 3\left( {x + 500} \right) = 5\left( {x + 250} \right) \cr & \Rightarrow 5x + 1250 = 3x + 1500 \cr & \Rightarrow 5x - 3x = 1500 - 1250 \cr & \Rightarrow 2x = 250 \cr & \Rightarrow x = \frac{{250}}{2} = 125\,{\text{m}} \cr} $$

$$\eqalign{ & {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{120}}{{10}}} \right){\text{ m/sec}} \cr & {\text{ = 12 m/sec}} \cr & {\text{Speed of second train}} \cr & {\text{ = }}\left( {\frac{{120}}{{15}}} \right){\text{ m/sec}} \cr & {\text{ = 8 m/sec}} \cr & {\text{Relative speed}} \cr & {\text{ = (12 + 8)m/sec}} \cr & {\text{ = 20 m/sec}} \cr & \refore {\text{Required time}} \cr & {\text{ = }}\frac{{\left( {120 + 120} \right)}}{{20}}\,\sec \cr & = 12\,\sec \cr} $$

Let length of train be x meters and its speed be y m/sec.
Then, $$\frac{{\text{x}}}{{\text{y}}}$$ = 20
⇒ y = $$\frac{{\text{x}}}{{20}}$$
∴ $$\frac{{{\text{x}} + 100}}{{30}}$$  = $$\frac{{\text{x}}}{{20}}$$
⇒ 30x = 20x + 2000
⇒ 10x = 2000
⇒ x = 200 meters

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

Let speeds of 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

$$\eqalign{ & {\text{Speed = 132 km/hr }} \cr & {\text{ = 132}} \times \frac{5}{{18}}{\text{m/sec}} \cr & {\text{ = }}\frac{{110}}{3}m/\sec \cr & T = \frac{D}{S} \cr & \,\,\,\,\,\, = \frac{{110 + 165}}{{\frac{{100}}{3}}} \cr & \,\,\,\,\,\, = \frac{{3\left( {275} \right)}}{{110}} \cr & \,\,\,\,\,\, = 7.5\sec \cr} $$

$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr & = {\frac{{125}}{{10}}} {\text{ m/sec}} \cr & = {\frac{{25}}{2}} {\text{ m/sec}} \cr & = {\frac{{25}}{2} \times \frac{{18}}{5}} {\text{ km/hr}} \cr & = 45\,{\text{km/hr}} \cr & {\text{Let}}\,{\text{}}\,{\text{speed}}\,{\text{of}}\,{\text{}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr & \text{Then, relative speed} = \left( {x - 5} \right)\,{\text{km/hr}} \cr & \refore x - 5 = 45 \cr & \Rightarrow x = 50\,{\text{km/hr}} \cr} $$