# It took a crew 80 minutes to row 3km upstream and back again. If the rate of flow of the stream was 3km/h, what was the rowing rate of the crew?

Jun 10, 2016

$- \frac{9}{4} + \frac{5 \sqrt{7}}{4} \textcolor{w h i t e}{. .} \frac{K m}{h}$ as an exact value

$1.057 \textcolor{w h i t e}{. .} \frac{K m}{h} \text{ }$ ( to 3 decimal places) as an approximate value

#### Explanation:

It is important to keep the units all the same.

As unit time for velocities is in hours:
Total time = 80 minutes $\to \frac{80}{60} h o u r s$

Given that distance 1 way is 3Km

Let rowing velocity be $r$

Let time to row against current be ${t}_{a}$
Let time to row with current be ${t}_{w}$

Thus ${t}_{w} + {t}_{a} = \frac{80}{60}$

Known: distance is velocity x time
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Thus
For 'with current' $\text{ "3Km = (r+3)t_w" "->" } {t}_{w} = \frac{3}{r + 3}$
For against current$\text{ "3Km=(r-3)t_a" "->" } {t}_{a} = \frac{3}{r - 3}$

But ${t}_{w} + {t}_{a} = \frac{80}{60}$

$\implies \frac{3}{r - 3} + \frac{3}{r + 3} = \frac{80}{60}$
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Consider that ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
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$\implies \frac{3 \left(r + 3\right) + 3 \left(r - 3\right)}{\left(r - 3\right) \left(r + 3\right)} \text{ "->" } \frac{6 r}{{r}^{2} - 9} = \frac{80}{60}$

$\implies \frac{360 r}{80} = {r}^{2} - 9$

$\implies {r}^{2} - \frac{360 r}{80} - 9 = 0 \text{ note that } \left(\frac{360}{80} \equiv \frac{9}{2}\right)$
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Compare to $y = a {x}^{2} + b x + c \text{ where } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$r = \frac{- \frac{9}{2} \pm \sqrt{\frac{81}{4} - 4 \left(1\right) \left(- 9\right)}}{2 \left(1\right)}$

$r = \frac{- \frac{9}{2} \pm \sqrt{\frac{81}{4} + 36}}{2}$

$r = \frac{- \frac{9}{2} \pm \sqrt{\frac{225}{4}}}{2}$

$r = \frac{- \frac{9}{2} \pm \frac{\sqrt{{5}^{2} \times 7}}{2}}{2}$

$r = - \frac{9}{4} \pm \frac{5 \sqrt{7}}{4}$

$\implies r \approx 1.057 \text{ and "-3.077 " } \frac{K m}{h}$

The negative solution is not logical so

Rowing speed is:

$- \frac{9}{4} + \frac{5 \sqrt{7}}{4} \textcolor{w h i t e}{. .} \frac{K m}{h}$ as an exact value

$1.057 \textcolor{w h i t e}{. .} \frac{K m}{h} \text{ }$ ( to 3 decimal places) as an approximate value