Practice MCQ Questions and Answer on Problems on Trains

Solution: $$\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {60 + 40} \right)\,{\text{km/hr}} \cr & = {100 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{250}}{9}} \,{\text{m/sec}}. \cr & {\text{Distance}}\,{\text{covered}}\,{\text{in}}\,{\text{crossing}}\,{\text{each}}\,{\text{other}} \cr & = \left( {140 + 160} \right)m = 300\,m \cr & {\text{Required}}\,{\text{time}} \cr & = {300 \times \frac{9}{{250}}} \,{\text{sec}} \cr & = \frac{{54}}{5}\,{\text{sec}} \cr & = 10.8\,{\text{sec}} \cr} $$

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

Solution: $$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{jogger}} \cr & = \left( {45 - 9} \right)\,{\text{km/hr}} \cr & = 36\,{\text{km/hr}} \cr & {36 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = 10\,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {240 + 120} \right)\,m \cr & = 360\,m \cr & \therefore {\text{Time}}\,{\text{taken}} \cr & = {\frac{{360}}{{10}}} \,{\text{sec}} \cr & = 36\,{\text{sec}} \cr} $$

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

Solution: 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

Solution: $$\eqalign{ & {\text{Length of train}} \cr & {\text{ = 280 m }} \cr & {\text{Length of platform}} \cr & {\text{ = (3}} \times {\text{280) m = 840m}} \cr & \therefore {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr & = \frac{{1120}}{{50}}m/\sec \cr & = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr & = 80.64\,km/hr \cr} $$

Solution: $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speeds}}\,{\text{of}}\,{\text{the}}\,{\text{two}}\,{\text{trains}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{and}}\,y\,{\text{m/sec}}\,{\text{respectively}}. \cr & {\text{Then,}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} = 27x\,{\text{metres}}, \cr & {\text{and}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}} = 17y\,{\text{metres}}. \cr & \therefore \frac{{27x + 17y}}{{x + y}} = 23 \cr & \Rightarrow 27x + 17y = 23x + 23y \cr & \Rightarrow 4x = 6y \cr & \Rightarrow \frac{x}{y} = \frac{3}{2} \cr} $$

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

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

Solution: $$\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {40 - 20} \right)\,{\text{km/hr}} \cr & = \left( {20 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = {\frac{{50}}{9}} \,{\text{m/sec}} \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{faster}}\,{\text{train}} \cr & = \left( {\frac{{50}}{9} \times 5} \right)\,m \cr & = \frac{{250}}{9}\,m \cr & = 27\frac{7}{9}\,m \cr} $$