SOLUTION: Gone fishing. 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
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Question 202323This question is from textbook Elementary and Intermediate Algebra
: Gone fishing. 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?
This question is from textbook Elementary and Intermediate Algebra
Found 2 solutions by stanbon, scott8148:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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.
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
let s="speed in still water" ___ so upstream speed is s-4, and downstream is s+4
keeping the units consistent ___ 20min = 1/3 hr
distance is the same up and down ___ and, time = distance / rate
5 / (s-4) = [5 / (s+4)] + 1/3
multiplying by 3(s-4)(s+4) ___ 15s + 60 = 15s - 60 + s^2 - 16
subtracting 15s + 60 ___ 0 = s^2 - 136
s = sqrt(136)
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