$$\eqalign{ & {\text{Speed}} = {\frac{{240}}{{24}}} \,{\text{m/sec}} = 10\,{\text{m/sec}} \cr & \refore {\text{Required}}\,{\text{time}} \cr & {\text{ = }}\, {\frac{{240 + 650}}{{10}}} \,{\text{sec}}. \cr & = 89\,sec. \cr} $$

Speed of 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 speed of train be x km/hr.
Then, relative speed = (x - 6) km/hr
∴ x - 6 = 36
⇒ x = 42 km/hr
Speed of 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 speed of second man be y kmph.
Then, relative speed = (42 - y) kmph
∴ 42 - y = 40
⇒ y = 2 km/hr

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

Speed = $$\left( {90 \times \frac{5}{{18}}} \right)$$   m/sec = 25 m/sec
Total distance covered
= (120 + 230) m
= 350 m
∴ Required time
= $$\frac{{350}}{{25}}$$ seconds
= 14 seconds

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (80 + 55)km/hr}} \cr & {\text{ = 135 km/hr}} \cr & {\text{ = }}\left( {135 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{75}}{2}} \right)m/\sec \cr & {\text{Distance covered}} \cr & {\text{ = (120 + 90 + 90)m}} \cr & {\text{ = 300m}} \cr & {\text{Required time}} \cr & {\text{ = }}\left( {300 \times \frac{2}{{75}}} \right)\sec \cr & = 8\sec \cr} $$

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

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

Length of train
$$ = \frac{{{\text{Length}}\,{\text{of}}\,{\text{}}\,{\text{platform}}}}{{{\text{Difference}}\,{\text{in time}}}}$$     × (Time taken to cross a lamp post)
$$\eqalign{ & = \frac{{390}}{{28 - 7}} \times 7 \cr & = \frac{{390}}{{21}} \times 7 \cr & = \frac{{390}}{3} \cr & = 130\,{\text{m}} \cr} $$

$$\eqalign{ & 2\,kmph = \left( {2 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{9}\,{\text{m/sec}} \cr & 4\,kmph = \left( {4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{9}\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{}}\,{\text{length}}\,{\text{of}}\,{\text{}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}\, \cr & {\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{be}}\,y\,{\text{m/sec}} \cr & {\text{Then}},\, {\frac{x}{{y - \frac{5}{9}}}} = 9\,{\text{and}}\, {\frac{x}{{y - \frac{{10}}{9}}}} = 10 \cr & \refore 9y - 5 = x\,{\text{and}}\,10\left( {9y - 10} \right) = 9x \cr & \Rightarrow 9y - x = 5\,{\text{and}}\,90y - 9x = 100 \cr & {\text{On}}\,{\text{solving,}}\,{\text{we}}\,{\text{get}}:\,x = 50 \cr & \refore {\text{Length}}\,{\text{of}}\,{\text{}}\,{\text{train}}\,{\text{is}}\,50\,m \cr} $$

$$\eqalign{ & {\text{Time taken by train to cross a pole}} \cr & {\text{ = 9 sec}} \cr & {\text{Distance covered in crossing a pole}} \cr & {\text{ = length of train}} \cr & {\text{Speed of train}} \cr & {\text{ = 48 km/h}} \cr & = \left( {\frac{{48 \times 5}}{{18}}} \right)m/\sec \cr & = \frac{{40}}{3}m/\sec \cr & \refore {\text{Length of train}} \cr & {\text{ = Speed }} \times {\text{Time}} \cr & {\text{ = }}\frac{{40}}{3} \times 9 \cr & {\text{ = 120 m}} \cr} $$