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. 

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Updated on: 10-Oct-2022

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