Question 250597
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Let *[tex \Large r_c] represent the rate of the current.


When the boat is going upstream, against the current, it is going 33 mph through the water, but *[tex \Large 33\ -\ r_c] mph with respect to the riverbed or the bank of the river or whatever fixed thing you are using to measure the distance traveled against.  Likewise, when the boat is going downstream, it is travelling *[tex \Large 33\ +\ r_c] mph with respect to the riverbed.


Remember that distance equals rate times time, or *[tex \Large d\ =\ rt].  Solving this equation time in terms of distance and rate yields:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{r}]


Now let's substitute some values from the problem.  To describe the upstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{300}{33\ -\ r_c}]


And for the downstream trip:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{360}{33\ +\ r_c}]


But since we know that the time for each of the trips was the same ("It takes as long..."), we can now write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{300}{33\ -\ r_c}\ =\ \frac{360}{33\ +\ r_c}]


Now all you have to do is cross-multiply the proportion and solve for *[tex \Large r_c].



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