Question 222430
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<font size="+4">This is the solution to problem #222066 which is almost exactly the same as your problem.  Just change the numbers.</font>


<i>A row boat traveled 12 miles in 2 hours against the current. It traveled 18 miles in the same amount of time with the current. Find the rate of the boat in still water and ther rate of the current?</i>


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]
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