Question

A sailor goes $8$ km downstream in $40$ minutes and returns in $1$ hour. Determine the speed of the sailor in still water and the speed of the current.

Answer 1

Let the speed of the sailor in still water be $x$ km/hr and speed of the current be $y$ km/hr.

Then, speed of sailor downstream $=x+y$ km/hr and speed of sailor upstream $=x-y$ km/hr.

Distance covered $=8$ km.

Also, $Time=\frac{Distance}{Speed}$

From the given information, we have,

$\frac{8}{x+y}=\frac{40}{60}and\frac{8}{x-y}=1or,\frac{8}{x+y}=\frac{40}{60}=>\frac{8}{x+y}=\frac{2}{3}=>24=2(x+y)=>x+y=\mathrm{12....}\left(i\right)Also,\frac{8}{x-y}=1=>x=8+y....\left(ii\right)$

Substituting (ii) in (i), we get,

$x+y=12=>8+y+y=12=>2y=4=>y=2$

Substituting $y=2$ in equation (ii), we get,

$x=8+y=>x=8+2=>x=10$

Therefore, speed of sailor in still water $=x=10$ km/hr

and speed of the current $=y=2$ km/hr.