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Hint- Downstream means with the flow of the current and upstream means opposite to the flow of current waves.

Let the speed of the sailor in still water $ = x{\text{ }}\dfrac{{km}}{h}$

And the speed of current $ = y{\text{ }}\dfrac{{km}}{h}$

Then, speed of sailor downstream $ = \left( {x + y} \right){\text{ }}\dfrac{{km}}{h}$ (As he is going with the flow of current thus eventually the overall speed adds up)

Speed of sailor upstream $ = \left( {x - y} \right){\text{ }}\dfrac{{km}}{h}$ (As he is going against the flow of current thus eventually the overall speed reduces)

Now, let’s calculate the speed of the sailor by using the formula $speed = \dfrac{{dis\tan ce}}{{Time}}$………………………. (1)

Now the speed when sailor goes downstream it is given that he goes 8 km in 40 min, thus using this data in equation (1)

$\left( {x + y} \right) = \dfrac{8}{{\dfrac{{40}}{{60}}}} = \dfrac{{8 \times 60}}{{40}}$ (1hr=60min)

$ \Rightarrow \left( {x + y} \right) = 12$………………………… (2)

Now the speed when sailor goes upstream it is given that he returns in 1 hour , this means he covers a distance of 8 km in 1 hour in this case , thus using this data in equation (1)

$\left( {x - y} \right) = \dfrac{8}{1}$

$ \Rightarrow x = 8 + y$…………………………… (3)

Now substitute (3) in equation (2) we get

$

\left( {x + y} \right) = 12 \Rightarrow \left( {8 + y + y} \right) = 12 \\

\Rightarrow 2y = 12 - 8 = 4 \\

\Rightarrow y = 2 \\

$

Substituting the value of y in equation (3) we get

$

\Rightarrow x = 8 + y \\

\Rightarrow x = 8 + 2 \\

\Rightarrow x = 10 \\

$

Therefore the speed of the sailor in still water = $x = 10{\text{ }}\dfrac{{km}}{h}$

And speed of the current =$y = 2{\text{ }}\dfrac{{km}}{h}$

Note-When we calculate the speed of any object in water, we have to keep in mind whether the object is moving with flow of water or opposite. If the object is moving with the flow of water then the speed of moving water will be added to the speed of the object else it (speed of water) will restrict that object that means the speed will be subtracted from the speed of the object.

Let the speed of the sailor in still water $ = x{\text{ }}\dfrac{{km}}{h}$

And the speed of current $ = y{\text{ }}\dfrac{{km}}{h}$

Then, speed of sailor downstream $ = \left( {x + y} \right){\text{ }}\dfrac{{km}}{h}$ (As he is going with the flow of current thus eventually the overall speed adds up)

Speed of sailor upstream $ = \left( {x - y} \right){\text{ }}\dfrac{{km}}{h}$ (As he is going against the flow of current thus eventually the overall speed reduces)

Now, let’s calculate the speed of the sailor by using the formula $speed = \dfrac{{dis\tan ce}}{{Time}}$………………………. (1)

Now the speed when sailor goes downstream it is given that he goes 8 km in 40 min, thus using this data in equation (1)

$\left( {x + y} \right) = \dfrac{8}{{\dfrac{{40}}{{60}}}} = \dfrac{{8 \times 60}}{{40}}$ (1hr=60min)

$ \Rightarrow \left( {x + y} \right) = 12$………………………… (2)

Now the speed when sailor goes upstream it is given that he returns in 1 hour , this means he covers a distance of 8 km in 1 hour in this case , thus using this data in equation (1)

$\left( {x - y} \right) = \dfrac{8}{1}$

$ \Rightarrow x = 8 + y$…………………………… (3)

Now substitute (3) in equation (2) we get

$

\left( {x + y} \right) = 12 \Rightarrow \left( {8 + y + y} \right) = 12 \\

\Rightarrow 2y = 12 - 8 = 4 \\

\Rightarrow y = 2 \\

$

Substituting the value of y in equation (3) we get

$

\Rightarrow x = 8 + y \\

\Rightarrow x = 8 + 2 \\

\Rightarrow x = 10 \\

$

Therefore the speed of the sailor in still water = $x = 10{\text{ }}\dfrac{{km}}{h}$

And speed of the current =$y = 2{\text{ }}\dfrac{{km}}{h}$

Note-When we calculate the speed of any object in water, we have to keep in mind whether the object is moving with flow of water or opposite. If the object is moving with the flow of water then the speed of moving water will be added to the speed of the object else it (speed of water) will restrict that object that means the speed will be subtracted from the speed of the object.

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