Question

A boat covers certain distance between two spots in a river taking $t_{1}$ hrs going downstream and $t_{2}$ hrs going upstream. What time will be taken by boat to cover same distance in still water?

• A. $\frac{t_{1}+t_{2}}{2}$
• B. $2\left(t_{2}-t_{1}\right)$
• C. $\frac{2 t_{1} t_{2}}{t_{1}+t_{2}}$
• D. $\sqrt{t_{1} t_{2}}$

Answer 1

Answer: (C)

For upstream, Speed $\Rightarrow v-u$
For downstream, Speed $\Rightarrow v+$
$t_{ up }=\frac{d}{v-u}$
$t_{2}=\frac{d}{v-u}$
$\Rightarrow d=(v-u) t_{2}$ ...(i)
(where $v \rightarrow$ man and $u \rightarrow$ water)
$t_{\text {down }}=\frac{d}{v +u}$
$t_{1}=\frac{d}{v +u}$
$\Rightarrow d=(v+ u) t_{1}$ ...(ii)
$t_{\text {still }}=\frac{d}{v}$
$-t_{\text{still}}=\frac{2 t_{1} t_{2}}{t_{1}+t_{2}}$
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, $d=\left(v-\frac{v\left(t_{2}-t_{1}\right)}{t_{1}+t_{2}}\right) t_{2}$
$=v t_{2}\left(\frac{t_{1}+t_{2}-t_{2}+t_{1}}{t_{1}+t_{2}}\right)$
$\frac{d}{v}=\frac{2 t_{1} t_{2}}{t_{1}+t_{2}} \mid \rightarrow$ Remember as shortcut