Let $v_r$ = velocity of river

$v_{\mathrm{br}}$ = velocity of river in still water and

$w$ = width of river

Given,  $t_{\min }= 10 \min$

or   $\frac{w}{v_{\text{br}}}=10$      ...(i)

Drift, in this case, will be,

$x=v_{\mathrm{r}}t$

$\therefore$   $120=10v_{\mathrm{r}}$     ...(ii)





The shortest path is taken when v_b is along $\mathrm{AB}$. In this case,

$v_{\text{b}}=\sqrt{v_{\text{br}}^2-v_r^2}$

Now,      $12\text{.}5=\frac{w}{v_{\text{b}}}=\frac{w}{\sqrt{v_{\text{br}}^2-v_{\text{r}}^2}}$         ...(iii)

Solving these three equations, we get

$v_{\mathrm{br}} = 20 \mathrm{m} {\min }^{-1}$,

$v_{\mathrm{r}} = 12 \mathrm{m} {\min }^{-1}$

and   $w= 200 \mathrm{m}$