Practice MCQ Questions and Answer on Problems on Trains

Solution: Let the length of each train be x meters Then, speed of first train = $$\frac{{\text{x}}}{3}$$ m/sec Speed of second train = $$\frac{{\text{x}}}{4}$$ m/sec ∴ Required time $$\eqalign{ & = \left[ {\frac{{{\text{x}} + {\text{x}}}}{{\left( {\frac{{\text{x}}}{3} + \frac{{\text{x}}}{4}} \right)}}} \right]{\text{sec}} \cr & = \left[ {\frac{{2{\text{x}}}}{{\left( {\frac{{7{\text{x}}}}{{12}}} \right)}}} \right]{\text{sec}} \cr & = \left( {2 \times \frac{{12}}{7}} \right){\text{sec}} \cr & = \frac{{24}}{7}{\text{sec}} \cr & = 3\frac{3}{7}{\text{sec}} \cr} $$

Solution: 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 $$\eqalign{ & = \left[ {\frac{{\left( {{\text{x}} - 100} \right) + {\text{y}}}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr & = \left[ {\frac{{\left( {{\text{x}} + {\text{y}}} \right) - 100}}{{\frac{{25}}{2}}}} \right]{\text{sec}} \cr & = \left( {800 \times \frac{2}{{25}}} \right){\text{sec}} \cr & = 64\,{\text{sec}} \cr} $$

Solution: 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

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

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

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

Solution: $$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr & = \left( {63 - 3} \right)\,{\text{km/hr}} \cr & = 60\,{\text{km/hr}} \cr & = \left( {60 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \frac{50}{3}\, \text{m/sec} \cr & \therefore {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr & = \left( {500 \times \frac{3}{{50}}} \right)\,\sec \cr & = 30\,\sec \cr} $$

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

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

Solution: $$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr & {\text{Relative}}\,{\text{speed}}\, \cr & = \,\left( {x + 50} \right)\,{\text{km/hr}} \cr & = \left[ {\left( {x + 50} \right) \times \frac{5}{{18}}} \right]\,{\text{m/sec}} \cr & = {\frac{{250 + 5x}}{{18}}} \,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{covered}} \cr & = \left( {108 + 112} \right) = 220\,m \cr & \therefore \frac{{220}}{{ {\frac{{250 + 5x}}{{18}}} }} = 6 \cr & \Rightarrow 250 + 5x = 660 \cr & \Rightarrow x = 82\,{\text{km/hr}} \cr} $$