<PRE><B>Question 171646</B>
Tim paddled his kayak 12 km upstream against a 3 km/h current and back again
 in 5 h 20 min. In that time how far could he have paddled in still water?
:
Let s = his speed in still water
then
(s-3) = his speed upstream
and
(s+3) = his speed downstream
;
Write a time equation: time = {{{distance/speed}}}
:
Change 5 hr 20 min to hrs: 5{{{1/3}}} or {{{16/3}}} hrs
:
Time upstream + time downstream = 5{{{1/3}}} hrs
{{{12/((s-3))}}} + {{{12/((s+3))}}} = {{{16/3}}}
Multiply equation by 3(s-3)(s+3)
3(s-3)(s+3)*{{{12/((s-3))}}} + 3(s-3)(s+3)*{{{12/((s+3))}}} = 3(s-3)(s+3)*{{{16/3}}}
Results:
36(s+3) + 36(s-3) = 16(s-3)(s+3)
:
36s + 108 + 36s - 108 = 16(s^2 - 9)
:
72s = 16s^2 - 144
Arrange as a quadratic equation:
16s^2 - 72s - 144 = 0
Simplify divide equation by 8
2s^2 - 9s - 18 = 0
Factor
(2s + 3)(s - 6) = 0
Positive solution: 
s = 6 mph; his still water speed
:
How far would he travel in 5{{{1/3}}} hrs at 6 mph?
6 * {{{16/3}}} = 32 miles
:
:
Check solution by find the times
12/9 = 1.33
12/3 = 4.0
Total= 5.33 hrs</PRE>