Question
A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
Answer: Option B
$$\eqalign{
& {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr
& = \left( {60 + 6} \right)\,{\text{km/hr}} \cr
& = 66\,{\text{km/hr}} \cr
& = {66 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr
& = {\frac{{55}}{3}} \,{\text{m/sec}} \cr
& \therefore {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr
& = {110 \times \frac{3}{{55}}} {\text{sec}} = 6\,{\text{sec}} \cr} $$
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$$\eqalign{
& {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr
& = \left( {60 + 6} \right)\,{\text{km/hr}} \cr
& = 66\,{\text{km/hr}} \cr
& = {66 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr
& = {\frac{{55}}{3}} \,{\text{m/sec}} \cr
& \therefore {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr
& = {110 \times \frac{3}{{55}}} {\text{sec}} = 6\,{\text{sec}} \cr} $$
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