<PRE><B>Question 159972</B>
A boat heads upstream a distance of 30 miles on the Mississippi river, whose
 current is running at 5 miles per hour. If the trip back takes an hour less,
 what was the speed of the boat in still water?
:
Let s = speed of the boat in still water
then
(s-5) = speed upstream
and
(s+5) = speed downstream
:
:
Write a time equation: Time = {{{dist/speed}}}
:
Time up = time down - 1 hr
{{{30/((s-5))}}} = {{{30/((s+5))}}} + 1
:
Multiply equation by (s+5)(s-5); results:
30(s+5) = 30(s-5) + (s+5)(s-5)
:
30s + 150 = 30s - 150  + (s^2-25)
:
Combine like terms on the right
0 = s^2 + 30s - 30s - 25 - 150 - 150
:
0 = s^2 - 325 
:
s^2 = +325
s = {{{sqrt(325)}}}
s = 18 mph boat speed in still water
:
:
Check solution by finding the time for each
30/13 = 2.3 hr
30/23 = 1.3 hr
----------------
differ = 1 hr </PRE>