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