Problems On Trains - Aptitude MCQ Questions and Solutions with Explanations

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

$$\eqalign{ & {\text{Ratio of speeds}} \cr & {\text{ = }}\sqrt 4 :\sqrt {6\frac{1}{4}} \cr & = \sqrt 4 :\sqrt {\frac{{25}}{4}} \cr & = 2:\frac{5}{2} \cr & = 4:5 \cr }$$
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)
$$\eqalign{ & = \frac{{390}}{{28 - 7}} \times 7 \cr & = \frac{{390}}{{21}} \times 7 \cr & = \frac{{390}}{3} \cr & = 130\,{\text{m}} \cr} $$

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

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

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

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

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

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