$$\eqalign{ & {\text{Speed}} = \left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} = {\frac{{50}}{3}} {\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{}}\,{\text{train}} = \left( {{\text{Speed}} \times {\text{Time}}} \right) \cr & \refore {\text{Length}}\,{\text{of}}\,{\text{}}\,{\text{train}} \cr & = \left( {\frac{{50}}{3} \times 9} \right)m = 150m \cr} $$

Let length of each train be x meters and let speed of each of m by y m/sec
Then, $$\frac{{{\text{2x}}}}{{2{\text{y}}}}$$ = 120
⇒ $$\frac{{{\text{x}}}}{{{\text{y}}}}$$ = 120 . . . . . . . (i)
New length of train A $$ = \left( {\frac{{16}}{{12}}{\text{x}}} \right){\text{m}} = \left( {\frac{{4{\text{x}}}}{3}} \right){\text{m}}$$
∴ Time taken by trains to cross each or
$$\eqalign{ & = \left( {\frac{{{\text{x}} + \frac{{4{\text{x}}}}{3}}}{{2{\text{y}}}}} \right){\text{sec}} \cr & = \frac{{7{\text{x}}}}{{6{\text{y}}}} \cr & = \frac{7}{6} \times \frac{{\text{x}}}{{\text{y}}} \cr & = \left( {\frac{7}{6} \times 120} \right){\text{sec}} \cr & = 140\,{\text{sec}} \cr} $$
Hence, difference in times taken
= (140 - 120) sec
= 20 sec

$$\eqalign{ & {\text{Let length of train be x metres}}{\text{.}} \cr & {\text{Then, length of platform = (2}}x{\text{) metres}}{\text{.}} \cr & {\text{Speed of train}} \cr & {\text{ = }}\left( {90 \times \frac{5}{{18}}} \right)m/\sec \cr & = 25m/sec \cr & \refore \frac{{x + 2x}}{{25}} = 36 \cr & \Rightarrow 3x = 900 \cr & \Rightarrow x = 300 \cr & {\text{Hence, length of platform}} \cr & {\text{ = }}2x = \left( {2 \times 300} \right){\text{m}} = 600{\text{m}} \cr} $$

Speed = $$\left( {54 \times \frac{5}{{18}}} \right)$$  m/sec = 15 m/sec
Length of train = (15 × 20) m = 300 m
Let length of platform be x meters
Then, $$\frac{{{\text{x}} + 300}}{{36}}$$  = 15
⇒ x + 300 = 540
⇒ x = 240 meters

$$\eqalign{ & {\text{Relative speed of trains }} \cr & {\text{ = }}\left( {\frac{{100 + 200}}{{2 \times 60}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{5}{2}} \right){\text{m/sec}} \cr & {\text{Speed of train B}} \cr & {\text{ = 120 kmph}} \cr & = \left( {120 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{{100}}{3}} \right){\text{m/sec}} \cr & {\text{Let speed of second train be }}x{\text{ m/sec}} \cr & {\text{Then, }} \frac{{100}}{3} - x = \frac{5}{2} \cr & \Rightarrow x = \left( {\frac{{100}}{3} - \frac{5}{2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{185}}{6}} \right){\text{m/sec}} \cr & \refore {\text{Speed of second train}} \cr & {\text{ = }}\left( {\frac{{185}}{6} \times \frac{{18}}{5}} \right){\text{ kmph}} \cr & {\text{ = 111 kmph}} \cr} $$

$$\eqalign{ & {\text{Time taken by trains to cross each }} \cr & {\text{or in opposite direction}} \cr & {\text{ = }}\frac{{{l_1} + {l_2}}}{{{\text{relative speed in opposite direction}}}} \cr & {\text{ = }}\frac{{\left( {180 + 120} \right)}}{{\left( {65 + 55} \right)}} \cr & {\text{ = }}\frac{{300}}{{120 \times \frac{5}{{18}}}} \cr & {\text{ = 9 seconds}} \cr} $$

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

$$\eqalign{ & {\text{In se type of questions use given}} \cr & {\text{below formula to save your valuable time}} \cr & \frac{{{{\text{S}}_1}}}{{{{\text{S}}_2}}}{\text{ = }}\sqrt {\frac{{{{\text{T}}_2}}}{{{{\text{T}}_1}}}} {\text{ }} \cr & {\text{Where }}{{\text{S}}_1}{\text{,}}{{\text{S}}_2}{\text{ and }}{{\text{T}}_1}{\text{, }}{{\text{T}}_2}{\text{ are respective}} \cr & {\text{speeds and times of objects}} \cr & \Rightarrow \frac{{45}}{{{{\text{S}}_2}}} = \sqrt {3\frac{1}{3} \div 4\frac{4}{5}} \cr & {\text{ = }}{{\text{S}}_2}{\text{ = 45}} \times \frac{6}{5}{\text{ = 54 km/hr}} \cr & \refore {\text{Required speed = 54 km/hr}} \cr} $$

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

Let lengths of train and bridge be x meters and y meters respectively.
Speed of first train
= 90 km/hr
= $$\left( {90 \times \frac{5}{{18}}} \right)$$  m/sec
= 25 m/sec
Speed of second train = 45 km/hr
= $$\left( {45 \times \frac{5}{{18}}} \right)$$  m/sec
= $$\frac{{25}}{2}$$ m/sec
Then, $$\frac{{{\text{x}} + {\text{y}}}}{{36}}$$ = 25
⇒ x + y = 900
∴ Required time
$$\eqalign{ & = \left[ {\frac{{\left( {{\text{x}} - 100} \right) + {\text{y}}}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr & = \left[ {\frac{{\left( {{\text{x}} + {\text{y}}} \right) - 100}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr & = \left( {800 \times \frac{2}{{25}}} \right){\text{sec}} \cr & = 64\,{\text{sec}} \cr} $$