A motorboat covers a given distance in $6\text{hours}$ moving downstream in a river. It covers the same distance in $10\text{hours}$ moving upstream. The time taken to cover the same distance in still water is

The correct option is B $7.5\text{hours}$
Let distance be ${}^{\prime }{d}^{\prime }$
let $u$ be velocity of the stream.
$v$ be velocity of boat in still water.

velocity of boat in downstream is $v+u$
velocity of boat in upstream is $v-u$
Given,
Time take by boat in downstream $t_{\text{down}}=6\text{hours}$
Time take by boat in upstream$t_{\text{up}}=10\text{hours}$


So, $\frac{d}{t_{\text{down}}}=v+u=\frac{d}{6}$
Similarly, $\frac{d}{t_{\text{up}}}=v-u=\frac{d}{10}$

On adding, we get $2v=\frac{d}{6}+\frac{d}{10}=\frac{16d}{60}$

$v=\frac{8d}{60}=\frac{d}{7.5}$
In still water $(u=0)$
$\Rightarrow t=\frac{d}{v}=\frac{d}{\frac{d}{7.5}}=7.5\text{hours}$