The ratio of the distance carried away by the water current, downstream, in crossing a river, by a person, making same angle with downstream and upstream is 2 : 1. The ratio of the speed of person to the water current cannot be less than 

 Motion of the person making an angle (say $\alpha$) with the downstream

The time taken to cross the river $=\frac{\mathrm{d}}{\mathrm{vsin}\mathrm{\alpha }}$. 

The distance carried away downstream in the same time speed $\times$ time. 

  ${\mathrm{x}}_1=(\mathrm{u}+\mathrm{vcos}\mathrm{\alpha })\frac{\mathrm{d}}{\mathrm{vsin}\mathrm{\alpha }}$ ………..(i)

Motion of the person making $\alpha$ angle with upstream.

The time taken to cross the river is equal to $\frac{\mathrm{d}}{\mathrm{vsin}\mathrm{\alpha }}$. 

Distance carried away downstream in the same time 

$\begin{array}{l} {\mathrm{x}}_2=\left[\mathrm{u}+\mathrm{vcos}\left({180}^{\circ }-\mathrm{\alpha }\right)\right]\frac{\mathrm{d}}{\mathrm{vsin}\mathrm{\alpha }}\\ \Rightarrow {\mathrm{x}}_2=(\mathrm{u}-\mathrm{vcos}\mathrm{\alpha })\frac{\mathrm{d}}{\mathrm{vsin}\mathrm{\alpha }} ........\left(\mathrm{ii}\right)\\ \mathrm{Given} \frac{(\mathrm{u}+\mathrm{vcos}\mathrm{\alpha })\frac{\mathrm{d}}{\mathrm{vsin}\mathrm{\alpha }}}{(\mathrm{u}-\mathrm{vcos}\mathrm{\alpha })\frac{\mathrm{d}}{\mathrm{vsin}\mathrm{\alpha }}}=\frac{2}{1}\\ \frac{(\mathrm{u}+\mathrm{vcos}\mathrm{\alpha })}{(\mathrm{u}-\mathrm{vcos}\mathrm{\alpha })}=\frac{2}{1}\Rightarrow 3\mathrm{vcos}\mathrm{\alpha }=\mathrm{u}\\ \Rightarrow \frac{\mathrm{v}}{\mathrm{u}}=\frac{\sec \mathrm{\alpha }}{3} .....\left(\mathrm{iii}\right)\\ \sec \mathrm{\alpha }\ge 1\Rightarrow \frac{\sec \mathrm{\alpha }}{3}\ge \frac{1}{3}\end{array}$

From Eq. (iii), $\frac{\mathrm{v}}{\mathrm{u}}\ge \frac{1}{3}$

So, v/u cannot be less than 1/3.