Question 1183926
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With the current a motorboat can travel 84 miles in two hours. 
Against the current the same trip takes three hours. 
Find the average rate of the boat in {{{cross(Stillwater)}}} <U>still water</U> and the average rate of the current.
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;As usual, &nbsp;the setup in the post by @josgarithmetic is &nbsp;TOTALLY &nbsp;WRONG.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I came to bring a correct solution.



<pre>
Let r be the motorboat rate in still water and let c be the rate of the current.


Then you have these two equations


    r + c = 42    (1)    (for the effective rate moving with the current;  42 = 84/2)

    r - c = 28    (2)    (for the effective rate moving against the current;  28 = 84/3)


Adding equations, you get

    2r = 42 + 28 = 70;  --->  r = 70/2 = 35.


Then from equation (1),  c = 42 - r = 42 - 35 = 7.


<U>ANSWER</U>.  The rate of the motorboat in still water is 35 miles per hour;

         The current rate is 7 miles per hour.
</pre>

Solved.



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For your info:  &nbsp;&nbsp;<U>still water</U> &nbsp;is not the same as &nbsp;<U>Stillwater</U>.



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After seeing my post, &nbsp;@josgarithmetic rewrote his exactly as my, &nbsp;so now his post is safe &nbsp;(although useless).


At least, &nbsp;I forced him to make it safe . . .