Problems On Trains — Part 1 MCQs with Explanations

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

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

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

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

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

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

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

$$\eqalign{ & {\text{Formula}}\,{\text{for}}\,{\text{converting}}\,{\text{from}}\,{\text{km/hr}}\,{\text{to}}\,{\text{m/s:}} \cr & X\,{\text{km/hr}} = {X \times \frac{5}{{18}}} \,{\text{m/s}} \cr & {\text{Therefore,}}\,{\text{Speed}} \cr & = {45 \times \frac{5}{{18}}} \,{\text{m/sec}} = \frac{{25}}{2}{\text{m/sec}} \cr & {\text{Total}}\,{\text{distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {360 + 140} \right)m = 500\,m \cr & {\text{Formula}}\,{\text{for}}\,{\text{finding}}\,{\text{Time}} \cr & = {\frac{{{\text{Distance}}}}{{{\text{Speed}}}}} \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = \left( {\frac{{500 \times 2}}{{25}}} \right)\,\sec \cr & = 40\,\sec . \cr} $$

$$\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {60 + 90} \right)\,{\text{km/hr}} \cr & = {150 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{125}}{3}} \,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{covered}} \cr & = \left( {1.10 + 0.9} \right)\,km \cr & = 2\,km \cr & = \,2000\,m \cr & {\text{Required}}\,{\text{time}} \cr & = {2000 \times \frac{3}{{125}}} \,{\text{sec}} \cr & = 48\,sec \cr} $$

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

$$\eqalign{ & {\text{Relative}}\,{\text{speed}} \cr & = \left( {120 + 80} \right)\,{\text{km/hr}} \cr & = {200 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{500}}{9}} \,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{other}}\,{\text{train}}\,{\text{be}}\,{\text{x}}\,{\text{metres}}{\text{.}} \cr & {\text{Then,}}\,\frac{{x + 270}}{9} = \frac{{500}}{9} \cr & \Rightarrow x + 270 = 500 \cr & \Rightarrow x = 230 \cr} $$

$$\eqalign{ & {\text{Speed}} = {72 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\,{\text{m/sec}} \cr & {\text{Time}} = 26\,{\text{sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}{\text{.}} \cr & {\text{Then}},\,\frac{{x + 250}}{{26}} = 20 \cr & \Rightarrow x + 250 = 520 \cr & \Rightarrow x = 270 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{Then,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}} = 2x\,{\text{m/sec}} \cr & {\text{Relative}}\,{\text{speed}} = \,\left( {x + 2x} \right)\,{\text{m/sec}} = 3x\,{\text{m/sec}} \cr & \therefore \frac{{ {100 + 100} }}{8} = 3x \cr & \Rightarrow 24x = 200 \cr & \Rightarrow x = \frac{{25}}{3} \cr & {\text{So,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}}\, = \frac{{50}}{3}\,{\text{m/sec}} \cr & = {\frac{{50}}{3} \times \frac{{18}}{5}} \,{\text{km/hr}} \cr & = 60\,{\text{km/hr}} \cr} $$

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

$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr & = \left( {60 + 6} \right)\,{\text{km/hr}} \cr & = 66\,{\text{km/hr}} \cr & = {66 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{55}}{3}} \,{\text{m/sec}} \cr & \therefore {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr & = {110 \times \frac{3}{{55}}} {\text{sec}} = 6\,{\text{sec}} \cr} $$

$$\eqalign{ & {\text{Total}}\,{\text{distance}}\,{\text{covered}} \cr & = \left( {\frac{7}{2} + \frac{1}{4}} \right)\,{\text{miles}} \cr & = \frac{{15}}{4}\,{\text{miles}} \cr & \therefore {\text{Time}}\,{\text{taken}} \cr & = \left( {\frac{{15}}{{4 \times 75}}} \right)\,{\text{hrs}} \cr & = \frac{1}{{20}}\,{\text{hrs}} \cr & = \left( {\frac{1}{{20}} \times 60} \right)\,\min \cr & = 3\,\min \cr} $$

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

$$\eqalign{ & {\text{Speed}} = {\frac{{300}}{{18}}} \,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50}}{3}\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{platform}}\,{\text{be}}\,x\,{\text{metres}}{\text{.}} \cr & {\text{Then}}, {\frac{{x + 300}}{{39}}} = \frac{{50}}{3} \cr & \Rightarrow 3\left( {x + 300} \right) = 1950 \cr & \Rightarrow x = 350\,m. \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr & \,{\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{by}}\,y\,{\text{m/sec}} \cr & Then,\,\frac{x}{y} = 15\,\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \frac{x}{{15}} \cr & \therefore \frac{{x + 100}}{{25}} = \frac{x}{{15}} \cr & \Rightarrow 15\left( {x + 100} \right) = 25x \cr & \Rightarrow 15x + 1500 = 25x \cr & \Rightarrow 1500 = 10x \cr & \Rightarrow x = 150m \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr & \,{\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{by}}\,y\,{\text{m/sec}} \cr & {\text{Then}},\,\frac{x}{y} = 8\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 8y \cr & {\text{Now}},\,\frac{{x + 264}}{{20}} = y \cr & \Rightarrow 8y + 264 = 20y \cr & \Rightarrow y = 22 \cr & \therefore {\text{Speed}} = 22\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {22 \times \frac{{18}}{5}} \,{\text{km/hr}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 79.2\,{\text{km/hr}} \cr} $$

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

$$\eqalign{ & {\text{Relative}}\,{\text{speed}} \cr & = \left( {45 + 30} \right)\,{\text{km/hr}} \cr & = \left(75 \times \frac{5}{18} \right)\, \text{m/sec} \cr & = {\frac{{125}}{6}} \,{\text{m/sec}} \cr & {\text{We}}\,{\text{have}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{time}}\,{\text{taken}}\,{\text{by}}\,{\text{the}} \cr & {\text{slower}}\,{\text{train}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{DRIVER}}\,{\text{of}}\, \cr & {\text{The}}\,{\text{faster}}\,{\text{train}}\,{\text{and}}\,{\text{not}}\,{\text{the}}\,{\text{complete}}\,{\text{train}}{\text{.}} \cr & \cr & {\text{So,}}\,{\text{distance}}\,{\text{covered = Length}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}. \cr & {\text{Therefore,}}\,{\text{Distance}}\,{\text{covered = 500}}\,{\text{m}}. \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = {500 \times \frac{6}{{125}}} \cr & = 24\,\sec \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{each}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}}. \cr & {\text{Then,}}\,{\text{relative}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{two}}\,{\text{trains}} = 2x\,{\text{m/sec}} \cr & {\text{So}},\,2x = \frac{{ {120 + 120} }}{{12}} \cr & \Rightarrow 2x = 20 \cr & \Rightarrow x = 10 \cr & \therefore {\text{Speed}}\,{\text{of}}\,{\text{each}}\,{\text{train}} = 10\,{\text{m/sec}} \cr & = {10 \times \frac{{18}}{5}} \,{\text{km/hr}} \cr & = 36\,{\text{km/hr}} \cr} $$

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

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

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

$$\eqalign{ & 2\,kmph = \left( {2 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{9}\,{\text{m/sec}} \cr & 4\,kmph = \left( {4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{9}\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}\, \cr & {\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{be}}\,y\,{\text{m/sec}} \cr & {\text{Then}},\, {\frac{x}{{y - \frac{5}{9}}}} = 9\,{\text{and}}\, {\frac{x}{{y - \frac{{10}}{9}}}} = 10 \cr & \therefore 9y - 5 = x\,{\text{and}}\,10\left( {9y - 10} \right) = 9x \cr & \Rightarrow 9y - x = 5\,{\text{and}}\,90y - 9x = 100 \cr & {\text{On}}\,{\text{solving,}}\,{\text{we}}\,{\text{get}}:\,x = 50 \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{is}}\,50\,m \cr} $$

$$\eqalign{ & 4.5\,{\text{km/hr}} = \left( {4.5 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4}\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.25\,{\text{m/sec,}}\,{\text{and}} \cr & 5.4\,km/hr = \left( {5.4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{2}\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.5\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{Then}},\,\left( {x - 1.25} \right) \times 8.4 = \left( {x - 1.5} \right) \times 8.5 \cr & \Rightarrow 8.4x - 10.5 = 8.5x - 12.75 \cr & \Rightarrow 0.1x = 2.25 \cr & \Rightarrow x = 22.5 \cr & \therefore {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \left( {22.5 \times \frac{{18}}{5}} \right)\,{\text{km/hr}} \cr & = 81\,{\text{km/hr}} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then,}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{is}}\, {\frac{x}{2}} \,{\text{metres}} \cr & {\text{Relative}}\,{\text{speed}} = \left( {48 + 42} \right)\,{\text{kmph}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {90 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25\,{\text{m/sec}} \cr & \therefore \frac{{ {x + \left( {x/2} \right)} }}{{25}} = 12 \cr & or\,\frac{{3x}}{2} = 300 \cr & or\,x = 200 \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{first}}\,{\text{train}} = 200\,{\text{m}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{platform}}\,{\text{be}}\,y\,{\text{metres}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr & = \left( {48 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \frac{{40}}{3}\,{\text{m/sec}} \cr & \therefore \left( {200 + y} \right) \times \frac{3}{{40}} = 45 \cr & \Rightarrow 600 + 3y = 1800 \cr & \Rightarrow y = 400\,{\text{m}} \cr} $$

$$\eqalign{ & {\text{Suppose}}\,{\text{they}}\,{\text{meet}}\,x\,{\text{hours}}\,{\text{after}}\,{\text{7}}\,{\text{a}}{\text{.m}}. \cr & {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{A}}\, \cr & {\text{in}}\,x\,{\text{hours = 20x}}\,{\text{km}}{\text{.}} \cr & {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{B}} \cr & \,{\text{in}}\,\left( {x - 1} \right)\,{\text{hours}} = 25\left( {x - 1} \right)\,km \cr & \therefore 20x + 25\left( {x - 1} \right) = 110 \cr & \Rightarrow 45x = 135 \cr & \Rightarrow x = 3 \cr & {\text{So,}}\,{\text{they}}\,{\text{meet}}\,{\text{at}}\,{\text{10}}\,{\text{a}}{\text{.m}}{\text{.}}\, \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{us}}\,{\text{name}}\,{\text{the}}\,{\text{trains}}\,{\text{as}}\,{\text{A}}\,{\text{and}}\,{\text{B}}{\text{.}}\,{\text{Then}}, \cr & \left( {{\text{A's}}\,{\text{speed}}} \right):\left( {{\text{B's}}\,{\text{speed}}} \right) \cr & = \sqrt b :\sqrt a \cr & = \sqrt {16} :\sqrt 9 \cr & = 4:3\, \cr} $$

$$\eqalign{ & {\text{Speed }} \cr & {\text{ = }}\left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{3}{\text{ m/sec}} \cr & {\text{Total distance covered}} \cr & {\text{ = (100 + 140) m = 240 m}} \cr & \therefore {\text{Required time}} \cr & {\text{ = }}\left( {240 \times \frac{3}{{50}}} \right){\text{sec}} \cr & {\text{ = }}\frac{{72}}{5}{\text{sec}} \cr & {\text{ = 14}}{\text{.4 sec}} \cr} $$

$$\eqalign{ & {\text{Speed}} \cr & {\text{ = }}\left( {\frac{{132}}{6}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {22 \times \frac{{18}}{5}} \right){\text{km/sec}} \cr & {\text{ = 79}}{\text{.2 km/hr}} \cr} $$

$$\eqalign{ & {\text{Speed}} \cr & {\text{ = }}\left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{3}{\text{m/sec}} \cr & {\text{Time = 27 sec}}{\text{.}} \cr & {\text{Let the length of the train be }}x{\text{ metres}}{\text{.}} \cr & {\text{Then,}}\frac{{x + 200}}{{27}}{\text{ = }}\frac{{50}}{3}{\text{ }} \cr & \Leftrightarrow x + 200 = \left( {\frac{{50}}{3} \times 27} \right) = 450 \cr & \Leftrightarrow x = 450 - 200 = 250{\text{ metres}} \cr} $$

$$\eqalign{ & {\text{Let the length of the train be x metres}}{\text{.}} \cr & {\text{Then, length of the platform = (2}}x{\text{) metres}}{\text{.}} \cr & {\text{Speed of the train}} \cr & {\text{ = }}\left( {90 \times \frac{5}{{18}}} \right)m/\sec \cr & = 25m/sec \cr & \therefore \frac{{x + 2x}}{{25}} = 36 \cr & \Rightarrow 3x = 900 \cr & \Rightarrow x = 300 \cr & {\text{Hence, length of platform}} \cr & {\text{ = }}2x = \left( {2 \times 300} \right){\text{m}} = 600{\text{m}} \cr} $$

$$\eqalign{ & {\text{Speed = }}\left( {\frac{{150 + 300}}{{40.5}}} \right)m/\sec \cr & = \left( {\frac{{450}}{{40.5}} \times \frac{{18}}{5}} \right)km/hr \cr & = 40km/hr \cr} $$

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

$$\eqalign{ & {\text{Speed of train relative to man}} \cr & {\text{ = }}\left( {60 + 6} \right){\text{km/hr}} \cr & = 66\,{\text{km/hr}} \cr & = \left( {66 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{55}}{3}} \right)m/\sec \cr & \therefore {\text{Time taken to pass the man}} \cr & = \left( {110 \times \frac{3}{{55}}} \right)\sec \cr & = 6\,\sec \cr} $$

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (72}} - {\text{60) km/hr}} \cr & {\text{ = 12 km/hr}} \cr & = \left( {12 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{10}}{3}} \right)m/\sec \cr & {\text{Total distance covered}} \cr & {\text{ = Sum of lengths of trains}} \cr & {\text{ = (240 + 240) m}} \cr & {\text{ = 480 m}} \cr & {\text{Time taken}} \cr & {\text{ = }}\left( {480 \times \frac{3}{{10}}} \right)\sec \cr & = 144\sec \cr & = 2\min \,24sec \cr} $$

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (60 + 90) km/hr}} \cr & {\text{ = }}\left( {150 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{{125}}{3}} \right){\text{m/sec}} \cr & {\text{Distance coverd}} \cr & {\text{ = (1}}{\text{.10 + 0}}{\text{.9)km}} \cr & {\text{ = 2 km}} \cr & {\text{ = 2000 m}}{\text{}} \cr & {\text{Required time}} \cr & {\text{ = }}\left( {2000 \times \frac{3}{{125}}} \right)\sec \cr & = 48{\text{ sec}}\cr} $$