Answer: Option D. -> 120 ft/second
$$\eqalign{
& {\text{Speed of the train}} \cr
& {\text{ = }}\frac{{700 + 500}}{{10}} \cr
& {\text{ = 120 ft/second}} \cr} $$

Answer: Option B. -> 52 km/hr, 26 km/hr
$$\eqalign{
& {\text{Let the speed of first train be }} \cr
& {{\text{S}}_1}{\text{ km/hr and speed of second train}} \cr
& {\text{is }}{{\text{S}}_2}{\text{km/hr }} \cr
& {\text{As we know,}} \cr
& {\text{Time }} \cr
& {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{relative speed in same/opposite direction}}}} \cr
& {\text{In the same direction}} \cr
& \Rightarrow {\text{27 sec = }}\frac{{\left( {100 + 95} \right)}}{{\left( {{\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2}} \right) \times \frac{5}{{18}}}} \cr
& \Rightarrow 27 = \frac{{195 \times 18}}{{\left( {{\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2}} \right) \times 5}} \cr
& \Rightarrow {\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2} = 26.......................(i) \cr
& {\text{In the opposite direction,}} \cr
& \Rightarrow 9 = \frac{{\left( {100 + 95} \right)}}{{\left( {{\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2}} \right) \times \frac{5}{{18}}}} \cr
& \Rightarrow 9 = \frac{{195 \times 18}}{{\left( {{\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2}} \right) \times 5}} \cr
& \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2} = 39 \times 2 \cr
& \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2} = 78 \cr
& {\text{From equation (i) and (ii)}} \cr
& \Rightarrow {\text{ }}{{\text{S}}_1} - {\text{ }}{{\text{S}}_2} = 26 \cr
& \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ + }}{{\text{S}}_2} = 78 \cr
& \Rightarrow {\text{ }}{{\text{S}}_1} = \frac{{26 + 78}}{2} \cr
& \Rightarrow {\text{ }}{{\text{S}}_1} = \frac{{104}}{2} \cr
& \Rightarrow {\text{ }}{{\text{S}}_1}{\text{ = 52 km/hr and }} \cr
& \,\,\,\,\,\,\,\,\,\,{{\text{S}}_2}{\text{ = 26 km/hr}} \cr} $$

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

Answer: Option B. -> 125 m
$$\eqalign{
& {\text{Let the length of train }}x{\text{ m }} \cr
& {\text{Speed of train }} \cr
& {\text{ = }}\frac{{\left( {{\text{Length of train + length of bridge }}} \right)}}{{{\text{Time taken in crossing}}}}{\text{ }} \cr
& {\text{According to information we get}} \cr
& \Rightarrow \frac{{x + 500}}{{100}} = \frac{{x + 250}}{{60}} \cr
& \Rightarrow 60\left( {x + 500} \right) = 100\left( {x + 250} \right) \cr
& \Rightarrow 3\left( {x + 500} \right) = 5\left( {x + 250} \right) \cr
& \Rightarrow 5x + 1250 = 3x + 1500 \cr
& \Rightarrow 5x - 3x = 1500 - 1250 \cr
& \Rightarrow 2x = 250 \cr
& \Rightarrow x = \frac{{250}}{2} = 125\,{\text{m}} \cr} $$

Answer: Option A. -> 3 km/h
$$\eqalign{
& {\text{Given , }} \cr
& {\text{Train cover 3584 kms in 2 days 8 hours}} \cr
& \left( {2\,{\text{days 8 hours = }}\frac{7}{3}{\text{ days}}} \right) \cr
& {\text{Average speed = }} {\frac{{3584}}{{ {\frac{{7}}{{3}}}}}} \cr
& {\text{ = 1536 km/day = }}\frac{{1536}}{{24}}{\text{ = 64 km/h}} \cr
& {\text{Distance covered in two days}} \cr
& {\text{ = 1440 + 1608 = 3048 km}} \cr
& {\text{Remaining distance for third day}} \cr
& {\text{ = 3584 }} - {\text{3048 = 536 km}} \cr
& {\text{Third day 536 km is covered in }} \cr
& {\text{8 hour with speed of}} \cr
& {\text{ = }}\frac{{536}}{8} = 67{\text{ km/h }} \cr
& {\text{( 3rd day total 536 km distance}} \cr
& {\text{ covered by 67 km/hr in 8 hr)}} \cr
& \therefore {\text{Difference of average speedm}} \cr
& {\text{ = 67}} - {\text{64 = 3 km/hr}} \cr} $$

Answer: Option C. -> 100 m
$$\eqalign{
& {\text{Let length of train }} \cr
& {\text{ = }}l\,{\text{metre}} \cr
& \Rightarrow {\text{Time }} \cr
& {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{relative speed in opposite direction}}}} \cr
& \Rightarrow 4\sec \, = \,\frac{{l + 0}}{{\left( {84 + 6} \right) \times \frac{5}{{18}}{\text{m/s}}}} \cr
& \Rightarrow 4\, = \frac{l}{{90 \times \frac{5}{{18}}}} \cr
& \Rightarrow \,l\, = \,100\,{\text{m}} \cr
& \therefore {\text{ length of the train = 100 m}} \cr} $$

Answer: Option B. -> 89 sec
$$\eqalign{
& {\text{Speed}} = {\frac{{240}}{{24}}} \,{\text{m/sec}} = 10\,{\text{m/sec}} \cr
& \therefore {\text{Required}}\,{\text{time}} \cr
& {\text{ = }}\, {\frac{{240 + 650}}{{10}}} \,{\text{sec}}. \cr
& = 89\,sec. \cr} $$

Answer: Option A. -> 40 sec
$$\eqalign{
& {\text{Formula}}\,{\text{for}}\,{\text{converting}}\,{\text{from}}\,{\text{km/hr}}\,{\text{to}}\,{\text{m/s:}} \cr
& X\,{\text{km/hr}} = {X \times \frac{5}{{18}}} \,{\text{m/s}} \cr
& {\text{Therefore,}}\,{\text{Speed}} \cr
& = {45 \times \frac{5}{{18}}} \,{\text{m/sec}} = \frac{{25}}{2}{\text{m/sec}} \cr
& {\text{Total}}\,{\text{distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr
& = \left( {360 + 140} \right)m = 500\,m \cr
& {\text{Formula}}\,{\text{for}}\,{\text{finding}}\,{\text{Time}} \cr
& = {\frac{{{\text{Distance}}}}{{{\text{Speed}}}}} \cr
& \therefore {\text{Required}}\,{\text{time}} \cr
& = \left( {\frac{{500 \times 2}}{{25}}} \right)\,\sec \cr
& = 40\,\sec . \cr} $$

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

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