Indian Institute of Foreign Trade 2012 Solved Paper

Given that,$\left(45-\mathit{x}\right)\times \sqrt{\mathit{n}}$
$\Rightarrow \left(45-\mathit{x}\right)=\mathit{k}\times \sqrt{\mathit{n}}$
Where x is the speed of the train, k is the proportionality constant and n is the number of wagon wheels attached.
Now given that when $\mathit{n}=9,\mathit{x}=30$
Thus,$15=\mathit{k}\times 3$
Thus, $\mathit{k}=5$
Thus,$45-\mathit{x}=5\sqrt{\mathit{n}}$
Thus, when $\mathit{n}=81,\mathit{x}=0$
Thus, for the train to move the maximum number of wagon wheels = 80
Hence, option C is the correct answer.