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

Let the speed of the boat be $x$ $km/h$ and the speed of the current be $y$ $km/h$.
$\therefore$ Speed downstream $=(x+y)$ $km/h$
Speed upstream $=(x-y)$ $km/h$

I am unable to carry on further.

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since, $$speed = \frac{distance}{time}$$ $$x+y = \frac{8km}{4/6hr}$$ and $$x-y = \frac{8km}{1hr}$$ then solve both the equations.

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  • $\begingroup$ Wow! Thank you very much! $\endgroup$ – Abhishekstudent Sep 22 '15 at 6:04
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Hint:

I suppose uniform motion, so you have $v=s/t$. And , using $v_1=x+y$ and $v_2=x-y$ you have a system of two linear equations. Can you write and solve this system?

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If the boat goes downstream $8$km in $40$minutes, then that means in $1$ hour, the boat would travel $12$km, by solving $8/40 = z/60$. So the speed of the boat downstream is $12$ km/h

Given in the question, we know that upstream, the speed of the boat is $8$ km/h.

Now using your equations, and plugging in these values, we have

$12 = x+y$ and

$8 = x-y$

Adding these equations together gives

$20 = 2x$, solving for $x$ gives $x = 10$. Plugging $x$ back in gives $12 = 10+y$ therefore $y =2$.

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