Question 322140
This one is slightly trickier than most of these up and down stream problems.

When Gary rows with the stream (downstream) he is moving at a rate equal to his rate in still water plus the rate of the current.

When he rows upstream, against the water, he is moving at a rate equal to his rate in still water minus the rate of the current.

Let r=his rate in still water and c=rate of current.

Therefore, since d=rt, and the distances up and back are the same, we have

17(r+c)=26.5(r-c)

But, we are told that c=r-6.8

17(r+(r-6.8))=26.5(r-(r-6.8))

r=8.7 is his rate in still water

Thus, the rate of the current is 1.9