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45 meter

50 meter

55 meter

60 meter

B. 50 meter

First person speed = 2*(5/18) = 5/9 m/sec

Second person speed = 4*(5/18) = 10/9 m/sec

Let the length of train is x metre and speed is y m/sec

then,

$\frac{x}{y-{\displaystyle \frac{5}{9}}}=9\phantom{\rule{0ex}{0ex}}\Rightarrow 9y-5=x\phantom{\rule{0ex}{0ex}}\Rightarrow 9y-x=5...\left(i\right)\phantom{\rule{0ex}{0ex}}Also,\phantom{\rule{0ex}{0ex}}\frac{x}{y-{\displaystyle \frac{10}{9}}}=10\phantom{\rule{0ex}{0ex}}90y-9x=100...\left(ii\right)\phantom{\rule{0ex}{0ex}}form\left(i\right)and\left(ii\right),weget,\phantom{\rule{0ex}{0ex}}x=50$

So length of train is 50 metre