For upstream, Speed $\Rightarrow v-u$
(where $v \rightarrow$ man and $u \rightarrow$ water)
For downstream, Speed $\Rightarrow v+u$
$t_{ up }=\frac{d}{v-u}$
$t_2=\frac{d}{v-u}$
$\Rightarrow d=(v-u) t_2 \ldots (i)$
$t_{ up }=\frac{d}{v-u}$
$t_2=\frac{d}{v-u}$
$\Rightarrow d=(v-u) t_2 \ldots (ii)$
$t_{ still }=\frac{d}{v}$
$t_{\text {still }}=\frac{2 t_1 t_2}{t_1+t_2}$
On equating $(i)$ and $(ii)$
$(v-u) t_2=(v+u) t_1$
$\Rightarrow v t_2-u t_2=v t_1+u t_1$
$\Rightarrow v\left(t_2-t_1\right)=u\left(t_1+t_2\right)$
$\Rightarrow u=\frac{v\left(t_2-t_1\right)}{t_2+t_1}$
So, $d=\left(v-\frac{v\left(t_2-t_1\right)}{t_1+t_2}\right) t_2=v t_2\left(\frac{t_1+t_2-t_2+t_1}{t_1+t_2}\right)$
$\frac{d}{v}=\frac{2 t_1 t_2}{t_1+t_2} \rightarrow$ Remember as shortcut
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
