Answer: Option B. -> 6 sec
$$\eqalign{
& {\text{Speed of train relative to man}} \cr
& {\text{ = }}\left( {60 + 6} \right){\text{km/hr}} \cr
& = 66\,{\text{km/hr}} \cr
& = \left( {66 \times \frac{5}{{18}}} \right)m/\sec \cr
& = \left( {\frac{{55}}{3}} \right)m/\sec \cr
& \therefore {\text{Time taken to pass the man}} \cr
& = \left( {110 \times \frac{3}{{55}}} \right)\sec \cr
& = 6\,\sec \cr} $$

Answer: Option C. -> 83.4 kmph
$$\eqalign{
& {\text{Speed of the train relative to man}} \cr
& {\text{ = }}\left( {\frac{{240}}{{10}}} \right){\text{m/sec}} \cr
& {\text{ = 24 m/sec}} \cr
& {\text{ = }}\left( {24 \times \frac{{18}}{5}} \right){\text{ km/sec}} \cr
& {\text{ = }}\frac{{432}}{5}{\text{km/hr}} \cr
& {\text{Let the speed of the train be x kmph}}{\text{.}} \cr
& {\text{Then relative speed = }}\left( {x + 3} \right){\text{kmph}} \cr
& \therefore x{\text{ + 3 = }}\frac{{432}}{5} \cr
& \Rightarrow x = \frac{{432}}{5} - 3 \cr
& \Rightarrow x = \frac{{417}}{5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 83.4\,{\text{kmph}} \cr} $$

Answer: Option B. -> $$1\frac{1}{7}$$ hr
$$\eqalign{
& {\text{Relative speed}} \cr
& {\text{ = (75 + 100)km/hr}} \cr
& {\text{ = 175 km/hr}} \cr
& {\text{Time taken to cover 175 km}} \cr
& {\text{at relative speed = 1 hr}} \cr
& \therefore {\text{T = Time taken to cover 200 km}} \cr
& {\text{ = }}\left( {\frac{1}{{175}} \times 200} \right)\, \text{hr} \cr
& = \frac{8}{7}\, \text{hr} \cr
& = 1\frac{1}{7}\, \text{hr} \cr} $$

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

Answer: Option A. -> 50 m
$$\eqalign{
& {\text{Let the length of each train be }}x{\text{ metres}} \cr
& {\text{Then distance covered}} \cr
& {\text{ = 2x metres}} \cr
& {\text{Relative speed}} \cr
& {\text{ = (46}} - {\text{36)km/hr}} \cr
& {\text{ = }}\left( {10 \times \frac{5}{{18}}} \right)m/\sec \cr
& = \left( {\frac{{25}}{9}} \right)m/\sec \cr
& \therefore \frac{{2x}}{{36}} = \frac{{25}}{9} \Leftrightarrow 2x = 100 \Leftrightarrow x = 50 \cr} $$

Answer: Option B. -> 12
$$\eqalign{
& {\text{Speed of the train}} \cr
& {\text{ = }}\left( {\frac{{120}}{{10}}} \right){\text{ m/sec}} \cr
& {\text{ = 12 m/sec}} \cr
& {\text{Speed of the second train}} \cr
& {\text{ = }}\left( {\frac{{120}}{{15}}} \right){\text{ m/sec}} \cr
& {\text{ = 8 m/sec}} \cr
& {\text{Relative speed}} \cr
& {\text{ = (12 + 8)m/sec}} \cr
& {\text{ = 20 m/sec}} \cr
& \therefore {\text{Required time}} \cr
& {\text{ = }}\frac{{\left( {120 + 120} \right)}}{{20}}\,\sec \cr
& = 12\,\sec \cr} $$

Answer: Option A. -> 20 sec
Let the length of each train be x meters and let the speed of each of them 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 other
$$\eqalign{
& = \left( {\frac{{{\text{x}} + \frac{{4{\text{x}}}}{3}}}{{2{\text{y}}}}} \right){\text{sec}} \cr
& = \frac{{7{\text{x}}}}{{6{\text{y}}}} \cr
& = \frac{7}{6} \times \frac{{\text{x}}}{{\text{y}}} \cr
& = \left( {\frac{7}{6} \times 120} \right){\text{sec}} \cr
& = 140\,{\text{sec}} \cr} $$
Hence, difference in times taken
= (140 - 120) sec
= 20 sec

Answer: Option B. -> 444 km
$$\eqalign{
& {\text{At the time of meeting ,}} \cr
& {\text{let the distance travelled by the}} \cr
& {\text{first train be }}x{\text{ km}}{\text{.}} \cr
& {\text{Then distance travelled by the }} \cr
& {\text{second train is (}}x{\text{ + 60) km}} \cr
& \therefore \frac{x}{{16}} = \frac{{x + 60}}{{21}} \cr
& \Rightarrow 21x = 16x + 960 \cr
& \Rightarrow 5x = 960 \Rightarrow x = 192 \cr
& {\text{Hence,}} \cr
& {\text{distance between two stations}} \cr
& {\text{ = (192 + 192 + 60) km}} \cr
& {\text{ = 444 km}}{\text{.}} \cr} $$

Answer: Option B. -> 24 km/hr
$$\eqalign{
& {\text{Ratio of speeds}} \cr
& {\text{ = }}\sqrt 4 :\sqrt {6\frac{1}{4}} \cr
& = \sqrt 4 :\sqrt {\frac{{25}}{4}} \cr
& = 2:\frac{5}{2} \cr
& = 4:5 \cr }$$
Let the speeds of the two trains be 4x and 5x km/hr respectively
Then time taken by trains to meet each other
$$\eqalign{
& {\text{ = }}\left( {\frac{{270}}{{4x + 5x}}} \right){\text{hr}} \cr
& {\text{ = }}\left( {\frac{{270}}{{9x}}} \right){\text{hr = }}\left( {\frac{{30}}{x}} \right){\text{hr}} \cr
& {\text{Time taken by slower train to travel}} \cr
& {\text{ 270 km = }}\left( {\frac{{270}}{{4x}}} \right){\text{hr}} \cr
& \therefore \frac{{270}}{{4x}} = \frac{{30}}{x} + 6\frac{1}{4} \cr
& \Rightarrow \frac{{270}}{{4x}} - \frac{{30}}{x} = \frac{{25}}{4} \cr
& \Rightarrow \frac{{150}}{{4x}} = \frac{{25}}{4} \cr
& \Rightarrow 100x = 600 \cr
& \Rightarrow x = 6 \cr
& {\text{Hence speed of slower train}} \cr
& {\text{ = 4}}x \cr
& = \,24\,{\text{km/hr}} \cr} $$

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