Problems On Trains — Part 2 MCQs with Explanations

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

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

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

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

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

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

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

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

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

$$\eqalign{ & {\text{Speed of the train}} \cr & {\text{ = }}\frac{{700 + 500}}{{10}} \cr & {\text{ = 120 ft/second}} \cr} $$

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

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

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

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

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

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = 60 + 50}} \cr & {\text{ = 110 km/h}} \cr & {\text{Time taken}} \cr & {\text{ = }}\frac{{220}}{{110}} \cr & {\text{ = 2 hr}} \cr} $$

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

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

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

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

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

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

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

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

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

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

$$\eqalign{ & {\text{Let the length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 9 second the distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{9} \cr & {\text{Distance covered in crossing a}} \cr & {\text{700 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 700}}{30} \cr & \Rightarrow \frac{x}{9} = \frac{{x + 700}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \Rightarrow \frac{x}{3} = \frac{{x + 700}}{{10}} \cr & \Rightarrow 10x = 3x + 2100 \cr & \Rightarrow 10x - 3x = 2100 \cr & \Rightarrow 7x = 2100 \cr & \Rightarrow x = \frac{{2100}}{7} = 300{\text{m}} \cr & {\text{When the length of the platform be 800m,}} \cr & {\text{then time T be taken by train to cross 800m}} \cr & {\text{long platform}} \cr & \frac{x}{9} = \frac{{x + 800}}{T} \cr & \Rightarrow Tx = 9x + 7200 \cr & \Rightarrow 300T = 2700 + 7200 \cr & \Rightarrow 300T = 9900 \cr & \Rightarrow T = \frac{{9900}}{{300}} = 33{\text{ seconds}} \cr} $$

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

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

$$\eqalign{ & {\text{Distance = Speed }} \times {\text{Time}} \cr & {\text{Distance covered by train with the}} \cr & {\text{speed of 30 kmph in 12 minutes is }} \cr & {\text{ = 30}} \times \frac{{12}}{{60}} = 6{\text{km}} \cr & {\text{Distance covered by the same train}} \cr & {\text{with the speed of 45 kmph in 8 minutes is }} \cr & {\text{ = 45}} \times \frac{8}{{60}} = 6{\text{km}} \cr & {\text{Average speed}} \cr & {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{total time}}}}. \cr & \Rightarrow \frac{{\left( {6 + 6} \right){\text{km}}}}{{\left( {12 + 8} \right)\min }} = \frac{{12}}{{20}} \times 60 \cr & {\text{ = 36 kmph}} \cr} $$

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

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

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

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 $$\eqalign{ & = \left[ {\frac{{2{\text{x}} + 2{\text{x}}}}{{\left( {\frac{{\text{x}}}{{300}}} \right)}}} \right]{\text{sec}} \cr & = \left( {\frac{{4{\text{x}} \times 300}}{{\text{x}}}} \right){\text{sec}} \cr & = 1200\,{\text{sec}} \cr & = \frac{{1200}}{{60}}\,{\text{min}} \cr & = 20\,{\text{minutes}} \cr} $$

Let the length of the train be x meters and its speed be y m/sec. Then, $$\frac{{\text{x}}}{{\text{y}}}$$ = 20 ⇒ y = $$\frac{{\text{x}}}{{20}}$$ ∴ $$\frac{{{\text{x}} + 100}}{{30}}$$ = $$\frac{{\text{x}}}{{20}}$$ ⇒ 30x = 20x + 2000 ⇒ 10x = 2000 ⇒ x = 200 meters

Speed of train relative to man = (42 - 6) kmph = 36 kmph = $$\left( {36 \times \frac{5}{{18}}} \right)$$ m/sec = 10 m/sec ∴ Time taken to pass the man = $$\frac{{180}}{{10}}$$ sec = 18 sec

Relative speed = (45 - 40) km/hr = 5 km/hr = $$\left( {5 \times \frac{5}{{18}}} \right)$$ m/sec = $$\frac{{25}}{{18}}$$ m/sec Total distance covered = Sum of lengths of trains = (200 + 150) m = 350 m ∴ Time taken = $$\left( {350 \times \frac{{18}}{{25}}} \right)$$ sec = 252 seconds

Let the lengths of the train and the bridge be x meters and y meters respectively. Speed of the first train = 90 km/hr = $$\left( {90 \times \frac{5}{{18}}} \right)$$ m/sec = 25 m/sec Speed of the second train = 45 km/hr = $$\left( {45 \times \frac{5}{{18}}} \right)$$ m/sec = $$\frac{{25}}{2}$$ m/sec Then, $$\frac{{{\text{x}} + {\text{y}}}}{{36}}$$ = 25 ⇒ x + y = 900 ∴ Required time $$\eqalign{ & = \left[ {\frac{{\left( {{\text{x}} - 100} \right) + {\text{y}}}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr & = \left[ {\frac{{\left( {{\text{x}} + {\text{y}}} \right) - 100}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr & = \left( {800 \times \frac{2}{{25}}} \right){\text{sec}} \cr & = 64\,{\text{sec}} \cr} $$

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

Speed of the train relative to first man $$\eqalign{ & = \frac{{75}}{{7.5}}{\text{m/sec}} = 10\,{\text{m/sec}} \cr & = \left( {10 \times \frac{{18}}{5}} \right){\text{km/hr}} = 36\,{\text{km/hr}} \cr} $$ Let the speed of the train be x km/hr. Then, relative speed = (x - 6) km/hr ∴ x - 6 = 36 ⇒ x = 42 km/hr Speed of the train relative to second man $$\eqalign{ & {\text{ = }}\frac{{75}}{{6\frac{3}{4}}}\,{\text{m/sec}} \cr & = \left( {75 \times \frac{4}{{27}}} \right){\text{m/sec}} \cr & = \frac{{100}}{9}{\text{m/sec}} \cr & = \left( {\frac{{100}}{9} \times \frac{{18}}{5}} \right){\text{km}} \cr & = 40\,{\text{km/hr}} \cr} $$ Let the speed of the second man be y kmph. Then, relative speed = (42 - y) kmph ∴ 42 - y = 40 ⇒ y = 2 km/hr