Answer: Option C. -> $$15\frac{1}{2}$$ second
$$\eqalign{
& {\text{Length of the train}} \cr
& {\text{ = Speed }} \times {\text{time}} \cr
& {\text{ = 36 km/hr}} \times {\text{10 sec}} \cr
& {\text{ = 36}} \times \frac{5}{{18}}{\text{m/s}} \times 10\sec \cr
& = 100{\text{ metres}} \cr
& {\text{Therefore, }} \cr
& {\text{Time taken by train to cross a plateform}} \cr
& {\text{ of 55 metre long in time}} \cr
& {\text{ = }}\frac{{\left( {100 + 55} \right)}}{{36 \times \frac{5}{{18}}}} \cr
& = \frac{{155}}{{10}} \cr
& {\text{Time}} = 15\frac{1}{2}\,\sec \cr} $$

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

Answer: Option A. -> 2 hr
$$\eqalign{
& {\text{Relative speed}} \cr
& {\text{ = 60 + 50}} \cr
& {\text{ = 110 km/h}} \cr
& {\text{Time taken}} \cr
& {\text{ = }}\frac{{220}}{{110}} \cr
& {\text{ = 2 hr}} \cr} $$

Answer: Option A. -> 875 km
$$\eqalign{
& {\text{Let the trains meet after t hours}} \cr
& {\text{Speed of train A}} \cr
& {\text{ = 75 km/hr}} \cr
& {\text{Speed of train B}} \cr
& {\text{ = 50 km/hr}} \cr
& {\text{Distance covered by train A}} \cr
& {\text{ = 75}} \times {\text{t = 75t}} \cr
& {\text{Distance covered by train B}} \cr
& {\text{ = 50}} \times {\text{t = 50t}} \cr
& {\text{Distance}}\,{\text{ = Speed }} \times {\text{Time}} \cr
& {\text{According to question}} \cr
& 75{\text{t}} - 50{\text{t}} = 175 \cr
& \Rightarrow 25{\text{t}} = 175 \cr
& \Rightarrow {\text{t}} = \frac{{175}}{{25}} = 7\,{\text{hour}} \cr
& \therefore {\text{Distance between A and B }} \cr
& {\text{ = 75t}} + 50{\text{t}} = 125{\text{t}} \cr
& = 125 \times 7 = 875\,{\text{km}} \cr} $$

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

Answer: Option A. -> 45 km/h
$$\eqalign{
& {\text{Relative speed of man & train}} \cr
& {\text{ = }}\frac{{100 \times 5}}{{36}} \times \frac{{18}}{5} \cr
& {\text{ = 50km/hr}} \cr
& \therefore {\text{speed of train}} \cr
& {\text{ = 50}} - {\text{5}} \cr
& {\text{ = 45 km/hr}} \cr} $$

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

Answer: Option B. -> 5 : 4
The speed of train A is 50km/hr and A starts its journey at 7 AM and reaches C at 10 AM. Total Travel time = 3hr
∴ Distance cover by A in 3hr = 50 × 3 = 150KM
Similarly, the speed of train B is 60km/hr and B starts its journey at 8 AM and reaches C at 10 AM. Total Travel time = 2hr
∴ Distance cover by B in 2hr = 60 × 2 = 120KM
The ratio of the distance between AC : BC
= 150 : 120
= 5 : 4

Answer: Option D. -> 10
$$\eqalign{
& {\text{Speed of train A}} \cr
& {\text{ = 63 kmph}} \cr
& {\text{ = }}\left( {\frac{{63 \times 5}}{{18}}} \right){\text{m/sec}} \cr
& {\text{ = 17}}{\text{.5 m/sec}} \cr
& {\text{Speed of train B}} \cr
& {\text{ = 54 kmph}} \cr
& {\text{ = }}\left( {\frac{{54 \times 5}}{{18}}} \right){\text{m/sec = 15 m/sec}} \cr
& {\text{If the length of train A be }}x{\text{ metre,}} \cr
& {\text{then}} \cr
& {\text{Speed of train A}} \cr
& {\text{ = }}\frac{{{\text{Length of train + length of platform}}}}{{{\text{Time taken in crossing}}}}{\text{ }} \cr
& \Rightarrow 17.5 = \frac{{x + 199.5}}{{21}} \cr
& \Rightarrow 17.5 \times 21 = x + 199.5 \cr
& \Rightarrow 367.5 = x + 199.5 \cr
& \Rightarrow x = 367.5 - 199.5 \cr
& \Rightarrow 168\,{\text{metres}} \cr
& {\text{Relative speed}} \cr
& {\text{ = ( Speed train A + Speed train B)}} \cr
& {\text{ = (17}}{\text{.5 + 15) m/sec}} \cr
& {\text{ = 32}}{\text{.5 m/sec}} \cr
& {\text{Required time}} \cr
& {\text{ = }}\frac{{{\text{ Length of train A + Length of train B}}}}{{{\text{Relative speed }}}} \cr
& = \left( {\frac{{168 + 157}}{{32.5}}} \right){\text{seconds}} \cr
& = 10\,{\text{seconds}} \cr} $$

Answer: Option A. -> 70 km
$$\eqalign{
& {\text{Average speed of train}} \cr
& {\text{ = 40 km/hr}} \cr
& {\text{Reach at its destination at on time }} \cr
& {\text{New average speed of train}} \cr
& {\text{ = 35 km/h}} \cr
& {\text{Time = 15 minutes}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\frac{{15}}{{60}}{\text{hours }} \cr
& {\text{Then distance travelled}} \cr
& {\text{ = }}\frac{{40 \times 35}}{{40 - 35}}{\text{ }} \times \frac{{15}}{{60}} \cr
& {\text{ = }}\frac{{40 \times 35}}{5}{\text{ }} \times \frac{{15}}{{60}} \cr
& {\text{ = 70}}\,{\text{km}} \cr} $$