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

Answer: Option B. -> 250 meters
Speed = $$\left( {30 \times \frac{5}{{18}}} \right)$$  m/sec = $$\frac{{25}}{3}$$ m/sec
Time = 36 second
Let the length of the bridge be x meters.
Then, $$\frac{{50 + {\text{x}}}}{{36}}$$   = $$\frac{{25}}{3}$$
⇒ 3(50 + x) = 900
⇒ 50 + x = 300
⇒ x = 250 meters

Answer: Option D. -> 14 seconds
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

Answer: Option A. -> 3501
Distance covered by the train in 5 hours
= (42 × 5) km
= 210 km
= 210000 m
∴ Number of pillars counted by the man
= $$\left( {\frac{{210000}}{{60}} + 1} \right)$$
= 3500 + 1
= 3501

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

Answer: Option C. -> 20 minutes
Let the length of the train be x metres.
Time taken to cover x meters = 5 min
= (5 × 60) sec
= 300 sec
Speed of the train = $$\frac{{\text{x}}}{{300}}$$ m/sec
Length of the second train = 2x meters
Length of the platform = 2x meters
∴ Required time
$$\eqalign{
& = \left[ {\frac{{2{\text{x}} + 2{\text{x}}}}{{\left( {\frac{{\text{x}}}{{300}}} \right)}}} \right]{\text{sec}} \cr
& = \left( {\frac{{4{\text{x}} \times 300}}{{\text{x}}}} \right){\text{sec}} \cr
& = 1200\,{\text{sec}} \cr
& = \frac{{1200}}{{60}}\,{\text{min}} \cr
& = 20\,{\text{minutes}} \cr} $$

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

Answer: Option B. -> (a + b + 1) m/s
$$\eqalign{
& {\text{Let the length of the train be }}x{\text{ metres}} \cr
& {\text{and its speed be }}y{\text{ m/s}} \cr
& {\text{Then,}} \cr
& {\text{ }}\frac{x}{{y - a}}{\text{ = b}}\,\,{\text{and}}\, \cr
& \,\frac{x}{{y - \left( {a + 1} \right)}} = \left( {b + 1} \right) \cr
& \Leftrightarrow {\text{ }}x{\text{ = }}b\left( {y - a} \right){\text{ and}} \cr
& \,\,\,\,\,\,\,\,\,\,{\text{ }}x = \left( {b + 1} \right)\left( {y - a - 1} \right) \cr
& \Leftrightarrow b\left( {y - a} \right) = \left( {b + 1} \right)\left( {y - a - 1} \right) \cr
& \Leftrightarrow by - ba = by - ba - b + y - a - 1 \cr
& \Leftrightarrow y = \left( {a + b + 1} \right) \cr} $$

Answer: Option D. -> 45 km/hr
$$\eqalign{
& {\text{Distance travelled in 14 sec}} \cr
& {\text{ = 50 + }}l \cr
& {\text{Distance travelled in 10 sec}} \cr
& {\text{ = }}l \cr
& {\text{So speed of train}} \cr
& {\text{ = }}\frac{{50}}{{14 - 10}}{\text{m/sec}} \cr
& {\text{ = }}\frac{{50}}{4} \times \frac{{18}}{5}{\text{km/hr}} \cr
& {\text{ = 45 km/hr}} \cr} $$

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