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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.
Given:
A sailor goes 8 km downstream in 40 minutes and returns in 1 hour.
To do:
We have to determine the speed of the sailor in still water and the speed of the current.
Solution:
Let the speed of the sailor in still water be $x$ km/hr and the speed of the current be $y$ km/hr.
This implies,
Speed of the boat downstream$=x+y$ km/hr
Speed of the boat upstream$=x-y$ km/hr
Time taken by the boat to go 8 km downstream$=\frac{8}{x+y}$ hours.
$\frac{8}{x+y}=\frac{40}{60}$ ($1\ hour=60\ minutes$)
$\frac{8}{x+y}=\frac{2}{3}$
$3(8)=2(x+y)$ (On cross multiplication)
$x+y=12$.....(i)
Time taken by the boat to go 8 km upstream$=\frac{8}{y-x}$ hours
$\frac{8}{y-x}=1$
$1(y-x)=8$ (On cross multiplication)
$y-x=8$.....(ii)
Adding equations (i) and (ii), we get,
$x+y+y-x=12+8$
$2y=20$
$y=\frac{20}{2}$
$y=10$
Substituting $y=10$ in equation (i), we get,
$x+y=12$
$x+10=12$
$x=12-10$
$x=2$
The speed of the sailor in still water is $2$ km/hr and the speed of the current is $10$ km/hr.