NIFT UG 2013 Question Paper with solutions

Let the place is at distance = d km.
Effective velocity along the river flow $=\mathit{u}+\mathit{v}=9.5+2.5=12$ km/h
Effective velocity against the river flow $=\mathit{u}\mathit{–}\mathit{v}=9.5\mathit{–}2.5=7\mathit{k}\frac{\mathit{m}}{\mathit{h}}$
Here, u and v are velocity with which a man can row in still water and velocity of still water respectively.
Again, let time of row along the stream = t hr, so, time of row against the stream $=\left(1.58\mathit{–}\mathit{t}\right)$hr.
According to the question, $12.\mathit{t}=\mathit{d}$
$\Rightarrow \mathit{t}=\frac{\mathit{d}}{12}$
and $7\left(1.58-\mathit{t}\right)=\mathit{d}$
$\Rightarrow 1.58-\mathit{t}=\frac{\mathit{d}}{7}$
therefore $1.58=\frac{\mathit{d}}{12}+\frac{\mathit{d}}{7}$
$1.58=\frac{19\mathit{d}}{84}$
$\mathit{d}=\frac{1.58\times 84}{19}=7$km