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With same amount of time, if a boat moves upstream, it can move $80 \text{ m}$ and if that moves downstream it can also move $120 \text{ m}$. If the average speed of the boat is always $20 \text{ miles/h}$ without any influence with the water speed, what is the average speed of water in miles per hour? $$\text{(A) } 2\quad\text{ (B) } 4\quad\text{ (C) } 6\quad\text{ (D) } 8\quad\text{ (E) } 10$$

I tried setting up the problem with the distance, rate, time setup, but ended up confusing myself. Can somebody please explain?

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Using the Distance, Rate, Time formula, $T=\frac{D}{R}$ and knowing the time is the same for both legs, you can say, $\frac{D}{R}=\frac{D}{R}$ or $\frac{80}{20-x}=\frac{120}{20+x}$ and solve for x.

(Because on the 120 meter leg, the stream speed is adding to your total speed, while on the 80 meter leg, the stream speed is subtracting from your total speed.)

$$\frac{80}{20-x}=\frac{120}{20+x}\\120(20-x)=80(20+x)$$

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The ratio of the speed of the boat upstream to the speed downstream is $\dfrac{2}{3}$. We have the equation

$$\frac{20-S}{20+S}=\frac{2}{3}$$

where $S$ is the speed of the stream.

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Let $\Delta t$ denote the length of time passed in the problem. Let $R$ denote the river speed. Then moving upstream, the speed of the boat (measured from the shore) would be $20-R$, while moving downstream it would be $20+R$.

Therefore we have the two equations

$\Delta t(20-R)=80$

and

$\Delta t(20+R)=120$.

You can solve these for $\Delta t$ and $R$, and $R$ will be your answer.

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    $\begingroup$ It looks like you intended to denote the river's speed by $R$. $\endgroup$ – N. F. Taussig Dec 17 '14 at 2:56

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