Question 222066
<font face="Garamond" size="+2">

If the boat went 12 miles in 2 hours, its rate against the current is 12 divided by 2 = 6 mph.  Likewise, the rate with the current is 9 mph.


The rate against the current is the still water speed minus the current speed.  The rate with the current is the still water speed plus the current speed, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{r_s-r_c\ =\ 6\cr r_s+r_c\ =\ 9\right]


Add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r_s + 0r_c = 15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_s = 7.5]


So the speed of the boat in still water is 7.5 mph.  The speed of the current is then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7.5 - r_c = 6\ \Rightarrow\ r_c = 1.5]


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