The fuel charges for running a train are proportional to the square of the speed generated in km per hour and cost Rs.100 per hour at 20km/hr. If the fixed charge amount to Rs.10,000 per hour, the most economical speed in km per hour is equal to K, then  $\frac{K}{50}=$

Let f  represents fuel charges 
$f=k{v}^2\to \left(1\right)f=100,v=20$
$k=\frac{100}{400}=\frac{1}{4}\Rightarrow f=\frac{v^2}{4}$
Let the train covers  $\lambda km\mathrm{in}thourst=\frac{\lambda }{v}fuelch\arg e$
In time t =  $\frac{v^2}{4}\times \frac{\lambda }{v}=\frac{\lambda v}{4}$
Total cost in running the train ,$c=\frac{\lambda v}{4}+\frac{10000\lambda }{v}$ 
$$\begin{gathered} \therefore \frac{dc}{dv}=\frac{\lambda }{4}-\frac{10000\lambda }{v^2};\frac{dc}{dv}=0 \\ \Rightarrow \frac{\lambda }{4}=\frac{10000\lambda }{v^2};v=\sqrt{40000}=200 \\ \therefore k=200,\frac{k}{50}=4 \end{gathered}$$