a) 45 m
b) 50 m
c) 54 m
d) 72 m
Answer: b
Explanation:
$$\eqalign{
& 2\,kmph = \left( {2 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{9}\,{\text{m/sec}} \cr
& 4\,kmph = \left( {4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{9}\,{\text{m/sec}} \cr
& {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}\, \cr
& {\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{be}}\,y\,{\text{m/sec}} \cr
& {\text{Then}},\, {\frac{x}{{y – \frac{5}{9}}}} = 9\,{\text{and}}\, {\frac{x}{{y – \frac{{10}}{9}}}} = 10 \cr
& 9y – 5 = x\,{\text{and}}\,10\left( {9y – 10} \right) = 9x \cr
& \Rightarrow 9y – x = 5\,{\text{and}}\,90y – 9x = 100 \cr
& {\text{On}}\,{\text{solving,}}\,{\text{we}}\,{\text{get}}:\,x = 50 \cr
& {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{is}}\,50\,m \cr} $$.
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