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#### Algebra Solutions

##### Topics || Problems

A boat, which can travel 8 miles per hour in still water, travels upstream for a certain time in a river whose current flows 2 miles per hour. Then, the boat returns to its starting point. If the trip, up and back, consumed 8 hours, how long did the boat travel up-stream?

Let $$v_b$$ be the speed of the boat in still water and $$v_w$$ be the speed of water.

Travelling downstream: $$V = v_b + v_w$$

$$V_{down} = 8 + 2$$

$$V_{down} = 10$$

Travelling upstream: $$V_{up}= v_b - v_w$$

$$V_{up} = 8 - 2$$

$$V_{up} = 6$$

$$V = \frac{distance}{time}$$

Let $$t$$ be the total travel time. Thus,
$$t = t_{up} + t_{down}$$
$$8 = t_{up} + t_{down}$$.

The distance travelled in upstream is same distance travelled downstream, $$d$$.

$$V_{down} = 10 = \frac{d}{t_{down}}$$

$$d = 10 t_{down}$$

$$V_{up} = 6 = \frac{d}{t_{up}}$$

$$d = 6 t_{up}$$

$$6 t_{up} = 10 t_{down}$$

$$t_{down} = \frac{3}{5} t_{up}$$ and $$8 = t_{up} + t_{down}$$.

$$8 = t_{up} +\frac{3}{5} t_{up}$$.

$$8 = \frac{8}{5} t_{up}$$.

$$\frac{40}{8} = t_{up}$$.

$$t_{up} = 5 ~hrs$$. Answer