Question 182578
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Relative to the bank of the river, the boat travels the speed in still water <i><b>plus</b></i> the speed of the current when going downstream, and speed in still water <i><b>minus</b></i> the speed of the current when going upstream.


The speed of the current is 5 mph, so if the speed of the boat in still water is <i>r</i>, the downstream speed is <i>r</i> + 5, and the upstream speed is <i>r</i> - 5.


Since distance equals rate times time, *[tex \Large d = rt], we also know that time equals distance divided by rate, *[tex \Large t = \frac{d}{r}].


So the time for the downstream trip must be:


*[tex \LARGE \text{          }\math t = \frac{66}{r + 5}]


While the time for the upstream trip must be:


*[tex \LARGE \text{          }\math t = \frac{33}{r - 5}]


But these times are equal so:


*[tex \LARGE \text{          }\math \frac{66}{r + 5} = \frac{33}{r - 5}]


Giving us a simple proportion.  Cross-multiply:


*[tex \LARGE \text{          }\math 66(r - 5) = 33(r + 5)]


Divide both sides by 33:


*[tex \LARGE \text{          }\math 2(r - 5) = r + 5]


Distribute and collect:


*[tex \LARGE \text{          }\math 2r - r = 5 + 10]


*[tex \LARGE \text{          }\math r = 15]


Checking the answer is left as an exercise for the student.




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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