<PRE><B>Question 757381</B>
Debbie traveled by boat 5 miles upstream to fish. Because of the 4 mph current, it took her 40 minutes longer to get there than to return. How fast will her boat go in still water? 
-----
Upstream DATA: distance = 5 miles ; rate = b-4 mph ; time = d/r = 5/(b-4)
----
Downstream DATA: distance = 5 miles ; b+4 mph ; time = d/r = 5/(b+4)
------
Equation:
up time - down time = 2/3 hrs.
------
5/(b-4) - 5/(b+4) = 2/3
-------
Multiply thru by 3(b-4)(b+4) to get:
15(b+4) - 15(b-4) = 2(b^2-16)
60 + 60 = 2b^2-32
-----
60 = b^2 - 16
b^2 = 76
Positive solution:
b = 8.72 mph (speed of the boat in still water)
=========================
Cheers,
Stan H.
========================= </PRE>