Question
A train, 240 m long, crosses a man walking alone the line in opposite direction at the rate of 3 kmph in 10 seconds. The speed of the train is?
Answer: Option C
$$\eqalign{
& {\text{Speed of the train relative to man}} \cr
& {\text{ = }}\left( {\frac{{240}}{{10}}} \right){\text{m/sec}} \cr
& {\text{ = 24 m/sec}} \cr
& {\text{ = }}\left( {24 \times \frac{{18}}{5}} \right){\text{ km/sec}} \cr
& {\text{ = }}\frac{{432}}{5}{\text{km/hr}} \cr
& {\text{Let the speed of the train be x kmph}}{\text{.}} \cr
& {\text{Then relative speed = }}\left( {x + 3} \right){\text{kmph}} \cr
& \therefore x{\text{ + 3 = }}\frac{{432}}{5} \cr
& \Rightarrow x = \frac{{432}}{5} - 3 \cr
& \Rightarrow x = \frac{{417}}{5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 83.4\,{\text{kmph}} \cr} $$
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$$\eqalign{
& {\text{Speed of the train relative to man}} \cr
& {\text{ = }}\left( {\frac{{240}}{{10}}} \right){\text{m/sec}} \cr
& {\text{ = 24 m/sec}} \cr
& {\text{ = }}\left( {24 \times \frac{{18}}{5}} \right){\text{ km/sec}} \cr
& {\text{ = }}\frac{{432}}{5}{\text{km/hr}} \cr
& {\text{Let the speed of the train be x kmph}}{\text{.}} \cr
& {\text{Then relative speed = }}\left( {x + 3} \right){\text{kmph}} \cr
& \therefore x{\text{ + 3 = }}\frac{{432}}{5} \cr
& \Rightarrow x = \frac{{432}}{5} - 3 \cr
& \Rightarrow x = \frac{{417}}{5} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 83.4\,{\text{kmph}} \cr} $$
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