Problems On Trains - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option D

$$\eqalign{ & {\text{Speed}} = \left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} = {\frac{{50}}{3}} {\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {{\text{Speed}} \times {\text{Time}}} \right) \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \left( {\frac{{50}}{3} \times 9} \right)m = 150m \cr} $$

Answer: Option B

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

Answer: Option C

$$\eqalign{ & {\text{Speed}} = {45 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\frac{{25}}{2}} \,{\text{m/sec}} \cr & {\text{Time}} = 30\,{\text{sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{bridge}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then}},\,\frac{{130 + x}}{{30}} = \frac{{25}}{2} \cr & \Rightarrow 2\left( {130 + x} \right) = 750 \cr & \Rightarrow x = 245\,m \cr} $$

Answer: Option B

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

Answer: Option B

$$\eqalign{ & {\text{Speed}} = {54 \times \frac{5}{{18}}} \,{\text{m/sec}} = 15\,{\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {15 \times 20} \right){\text{m}} = 300\,{\text{m}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{platform}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then}},\,\frac{{x + 300}}{{36}} = 15 \cr & \Rightarrow x + 300 = 540 \cr & \Rightarrow x = 240\,{\text{m}} \cr} $$