Question 199336
<font face="Garamond" size="+2">

Let *[tex \Large r] represent the rate in still water.


Let *[tex \Large t] represent the elapsed time for the upstream part of the trip.


When the boat is going upstream, that is, against the 3 km/hr current, the speed relative to the bank of the river must be *[tex \Large r - 3].  Conversely, the speed relative to the bank of the river must be *[tex \Large r + 3] on the downstream part of the trip.


The total elapsed time for both legs of the trip was 2 hours and 40 minutes.  Since 40 minutes is *[tex \Large \frac{2}{3}] hour, the total trip was *[tex \Large 2\frac{2}{3} = \frac{8}{3}] hour.  That means that if the upstream trip took *[tex \Large t] hours, then the downstream trip must have taken *[tex \Large \frac{8}{3} - t] hours.


Since we know that distance equals rate times time, or *[tex \Large d = rt], which can be expressed as *[tex \Large t = \frac{d}{r}], and we know that the distance for both the upstream and downstream trips is 6 km, we can describe the upstream trip by the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{6}{r-3}]


Similarly, the downstream trip can be described:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8}{3}-t = \frac{6}{r+3}]


which can be arranged so that it is expressed in terms of *[tex \Large t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{6}{r + 3} - \frac{8}{3} = \frac{8r+6}{3(r + 3)}]


Now that we have two different expressions in *[tex \Large r] that are equal to *[tex \Large t], we can set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6}{r-3} =  \frac{8r+6}{3(r + 3)}]


Solve the proportion by first cross-multiplying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8r^2 - 18r - 18 = 18r + 54]


Collect terms, put the quadratic into standard form, and eliminate the factor common to the coefficients, i.e. 4:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r^2 - 9t - 18 = 0]


Now all you have to do is solve the quadratic by ordinary means (this one factors rather tidily).  You will find two roots.  Exclude the negative root because negative rate makes little sense in this regard.  The positive root will be the answer to the question posed.


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