Question 288541
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A motorboat takes 4 hours to travel 256 mi going upstream. The return trip takes 2 hours going downstream. 
What is the rate of the boat in still water and what is the rate of the current?
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<pre>
Let u be the rate of the motorboat in still water (in miles per hour)
and v be the rate of the current (in the same units).


Then the effective rate of the motorboat downstream is  u + v  miles per hour,
and  the effective rate of the motorboat   upstream is  u - v  miles per hour.


From the problem, the effective rate of the motorboat downstream is the distance of 256 miles 
divided by the time of 2 hours  {{{256/2}}} = 128 mph.

                  The effective rate of the motorboat upstream is the distance of 256 miles 
divided by the time of 4 hours  {{{256/4}}} = 64 mph.


So, we have two equations to find 'u' and 'v'

    u + v = 128,    (1)

    u - v =  64.    (2)


To solve, add equations (1) and (2).  The terms 'v' and '-v' will cancel each other, and you will get

    2u = 128 + 64 = 192  --->   u = 192/2 = 96.

Now from equation (1)

     v = 128 - u = 128 - 96 = 32.


<U>ANSWER</U>.  Under given conditions, the rate of the motorboat in still water is 96 mph.  
         The rate of the current is 32 mph km/h.
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Solved.