Question
A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:
Answer: Option D
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr
& {\text{Relative}}\,{\text{speed}}\, \cr
& = \,\left( {x + 50} \right)\,{\text{km/hr}} \cr
& = \left[ {\left( {x + 50} \right) \times \frac{5}{{18}}} \right]\,{\text{m/sec}} \cr
& = {\frac{{250 + 5x}}{{18}}} \,{\text{m/sec}} \cr
& {\text{Distance}}\,{\text{covered}} \cr
& = \left( {108 + 112} \right) = 220\,m \cr
& \therefore \frac{{220}}{{ {\frac{{250 + 5x}}{{18}}} }} = 6 \cr
& \Rightarrow 250 + 5x = 660 \cr
& \Rightarrow x = 82\,{\text{km/hr}} \cr} $$
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$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr
& {\text{Relative}}\,{\text{speed}}\, \cr
& = \,\left( {x + 50} \right)\,{\text{km/hr}} \cr
& = \left[ {\left( {x + 50} \right) \times \frac{5}{{18}}} \right]\,{\text{m/sec}} \cr
& = {\frac{{250 + 5x}}{{18}}} \,{\text{m/sec}} \cr
& {\text{Distance}}\,{\text{covered}} \cr
& = \left( {108 + 112} \right) = 220\,m \cr
& \therefore \frac{{220}}{{ {\frac{{250 + 5x}}{{18}}} }} = 6 \cr
& \Rightarrow 250 + 5x = 660 \cr
& \Rightarrow x = 82\,{\text{km/hr}} \cr} $$
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