Question
A train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is
Answer: Option A
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr
& {\text{Then,}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{is}}\, {\frac{x}{2}} \,{\text{metres}} \cr
& {\text{Relative}}\,{\text{speed}} = \left( {48 + 42} \right)\,{\text{kmph}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {90 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25\,{\text{m/sec}} \cr
& \therefore \frac{{ {x + \left( {x/2} \right)} }}{{25}} = 12 \cr
& or\,\frac{{3x}}{2} = 300 \cr
& or\,x = 200 \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{first}}\,{\text{train}} = 200\,{\text{m}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{platform}}\,{\text{be}}\,y\,{\text{metres}} \cr
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr
& = \left( {48 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& = \frac{{40}}{3}\,{\text{m/sec}} \cr
& \therefore \left( {200 + y} \right) \times \frac{3}{{40}} = 45 \cr
& \Rightarrow 600 + 3y = 1800 \cr
& \Rightarrow y = 400\,{\text{m}} \cr} $$
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$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr
& {\text{Then,}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{is}}\, {\frac{x}{2}} \,{\text{metres}} \cr
& {\text{Relative}}\,{\text{speed}} = \left( {48 + 42} \right)\,{\text{kmph}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {90 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25\,{\text{m/sec}} \cr
& \therefore \frac{{ {x + \left( {x/2} \right)} }}{{25}} = 12 \cr
& or\,\frac{{3x}}{2} = 300 \cr
& or\,x = 200 \cr
& \therefore {\text{Length}}\,{\text{of}}\,{\text{first}}\,{\text{train}} = 200\,{\text{m}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{platform}}\,{\text{be}}\,y\,{\text{metres}} \cr
& {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr
& = \left( {48 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& = \frac{{40}}{3}\,{\text{m/sec}} \cr
& \therefore \left( {200 + y} \right) \times \frac{3}{{40}} = 45 \cr
& \Rightarrow 600 + 3y = 1800 \cr
& \Rightarrow y = 400\,{\text{m}} \cr} $$
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