$$\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {60 + 90} \right)\,{\text{km/hr}} \cr & = {150 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{125}}{3}} \,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{covered}} \cr & = \left( {1.10 + 0.9} \right)\,km \cr & = 2\,km \cr & = \,2000\,m \cr & {\text{Required}}\,{\text{time}} \cr & = {2000 \times \frac{3}{{125}}} \,{\text{sec}} \cr & = 48\,sec \cr} $$

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

$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{}}\,{\text{first}}\,{\text{train}} \cr & = {\frac{{120}}{{10}}} \,{\text{m/sec}} \cr & = 12\,{\text{m/sec}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{}}\,{\text{second}}\,{\text{train}} \cr & {\frac{{120}}{{15}}} \,{\text{m/sec}} \cr & = 8\,{\text{m/sec}} \cr & {\text{Relative}}\,{\text{speed}} = {12 + 8} = 20\,{\text{m/sec}} \cr & \refore {\text{Required}}\,{\text{time}} \cr & = {\frac{{ {120 + 120} }}{{20}}} \,{\text{ sec}} \cr & = 12\,{\text{sec}} \cr} $$

$$\eqalign{ & {\text{Relative speed of man & train}} \cr & {\text{ = }}\frac{{100 \times 5}}{{36}} \times \frac{{18}}{5} \cr & {\text{ = 50km/hr}} \cr & \refore {\text{speed of train}} \cr & {\text{ = 50}} - {\text{5}} \cr & {\text{ = 45 km/hr}} \cr} $$

$$\eqalign{ & {\text{Speed}} = \left( {78 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\frac{{65}}{3}} \,{\text{m/sec}} \cr & {\text{Time = }}\,{\text{1}}\,{\text{minute = 60}}\,{\text{second}}. \cr & {\text{Let}}\,{\text{}}\,{\text{length}}\,{\text{of}}\,{\text{}}\,{\text{tunnel}}\,{\text{be}}\,x\,{\text{metres}}. \cr & {\text{Then}},\, {\frac{{800 + x}}{{60}}} = \frac{{65}}{3} \cr & \Rightarrow 3\left( {800 + x} \right) = 3900 \cr & \Rightarrow x = 500 \cr} $$

$$\eqalign{ & {\text{Let length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 3 second distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{3} \cr & {\text{Distance covered in crossing a}} \cr & {\text{900 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 900}}{30} \cr & \Rightarrow \frac{x}{3} = \frac{{x + 900}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \Rightarrow \frac{x}{1} = \frac{{x + 900}}{{10}} \cr & \Rightarrow 10x = x + 900 \cr & \Rightarrow 10x - x = 900 \cr & \Rightarrow 9x = 900 \cr & \Rightarrow x = \frac{{900}}{9} = 100{\text{m}} \cr & {\text{When length of platform be 800m,}} \cr & {\text{n time T be taken by train to cross 800m}} \cr & {\text{long platfrom}} \cr & \frac{x}{3} = \frac{{x + 800}}{T} \cr & \Rightarrow Tx = 3x + 2400 \cr & \Rightarrow 100T = 300 + 2400 \cr & \Rightarrow 100T = 2700 \cr & \Rightarrow T = \frac{{2700}}{{100}} = 27{\text{ seconds}} \cr} $$

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

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

$$\eqalign{ & {\text{Let}}\,{\text{us}}\,{\text{name}}\,{\text{}}\,{\text{trains}}\,{\text{as}}\,{\text{A}}\,{\text{and}}\,{\text{B}}{\text{.}}\,{\text{Then}}, \cr & \left( {{\text{A's}}\,{\text{speed}}} \right):\left( {{\text{B's}}\,{\text{speed}}} \right) \cr & = \sqrt b :\sqrt a \cr & = \sqrt {16} :\sqrt 9 \cr & = 4:3\, \cr} $$

Let speed of each train be x m/sec.
Then, relative speed of two trains = 2x m/sec
So, 2x = $$\frac{{120 + 120}}{{12}}$$
⇒ 2x = 20
⇒ x = 10
∴ Speed of each train = 10 m/sec
= $$\left( {10 \times \frac{{18}}{5}} \right)$$  km/hr
= 36 km/hr