Question 1167838
<pre>
He's right but I think I'd do it this way:

Let t be that same time that it takes him to go 11 miles upstream
that it takes him to go downstream 7 miles.  So make this chart 

                       distance   rate    time
downstream 1st time       11               t  
upstream 1st time          7               t
downstream 2nd time                        3  
upstream 2nd time                          4

Use rate = distance/time to fill in the rates

                       distance   rate    time
downstream 1st time       11      11/t     t  
upstream 1st time          7       7/t     t
downstream 2nd time               11/t     3  
upstream 2nd time                  7/t     4

Use distance = rate x times to fill in the other two distances:

                       distance   rate    time
downstream 1st time       11      11/t     t  
upstream 1st time          7       7/t     t
downstream 2nd time      33/t     11/t     3  
upstream 2nd time        28/t      7/t     4 

So the two distances are 5 miles apart, so

33/t - 28/t = 5
        5/t = 5
          5 = 5t
          1 = t

So his rate upstream is 11/1 = 11 mph and his rate downstream is 7/1 = 7 mph

The average of those is 9 mph, his speed in still water.  So the speed of
the current is how much the current speeds him up going with it, and how much
it slows him down going against it, which is 2 mph.

Edwin</pre>