Question 330939
Wow, this is a tricky little problem. But, doable I am sure.

No doubt, there are other ways to tackle this, but I am going to let

x=the speed of boat A and y=the speed of boat B.

Let w=the river width (I assume you mean a river and not a 'driver' :):)).

As with many problems, we can set up a ratio.

Since d=rt, we have t=d/r.

The distance they meet from bank A is 800 yds, then the distance from bank B is

w-800 yds. So, we have: 

{{{(w-800)/y=800/x}}}

The next time they meet after they turn and head back, they are 400 yds from bank B. This time, boat B has turned from bank A and is heading back and meets boat A 400 yards from its shore, and boat A has turned at bank B and goes back 400 yards and meets boat B. Tricky to think about, huh?. We now have:

{{{(2*w-400)/y=(w+400)/x}}}

Now, we can hammer away at these until we have only w left.

Divide:

{{{(800/x)/((w+400)/x)=((w-800)/y)/((2w-400)/y)}}}

This simplifies down quite nicely. Equate:

{{{800/(w+400)=(w-800)/(2*w-400)}}}

Cross multiply:

{{{800(2*w-400)=(w-800)*(w+400)}}}

Expand:

{{{1600w-320000=w^2-400w-320000}}}

{{{2000w=w^2}}}

Assuming the river is not 0 yards wide, we have:

{{{w=2000}}} yards.

That's over a mile wide. Big River. 

I hope I didn't make a mistake after all that. WHEW! :):)