Question 202323
Debbie traveled by boat 5 miles upstream to
fish in her favorite spot. Because of the 4-mph current, it
took her 20 minutes longer to get there than to return. How
fast will her boat go in still water?
-----------------------
Upstream DATA:
distance = 5 miles; time = x + 20  ; rate = d/t = 5/(x+2) hrs.
---------------------------
Downstream DATA:
distance = 5 miles ; time = x ; rate = d/t = 5/x
----------------------------
Equation:
rate upstream = b-4=5/(x+2)
rate downstrea= b+4=5/x
--------------------------
Subtract 1st from 2nd to get:
8 = 5/x - 5/(x+2)
Solve for "x":
 8x(x+2) = 5(x+2) - 5x
8x^2+16x = 10
4x^2+8x-5 = 0
4x^2+10x-2x-5 = 0
2x(2x+5)-(2x+5) = 0
(2x+5)(2x-1) = 0
Positive solution:
x = 1/2
---------------------
Solve for "b" when b+4 = 5/x
b+4 = 5/(1/2)
b+4 = 10
b = 6 mph (speed of the boat in still water)
=================================================
Cheers,
Stan H.