<PRE><B>Question 465553</B>
A river has a current of 5mph. It takes Al half an hour longer to paddle upstream
 1.2 miles than to paddle downstream the same distance.
 What is Al&#39;s rate in still water.
:
Let s = his paddling rate in still water
then
(s+5) = his effective speed downstream
and
(s-5) = his effective speed upstream
:
Write a time equation, time = dist/speed
:
upstream time = downstream time + .5 hrs
{{{1.2/((s-5))}}} = {{{1.2/((s+5))}}}
:
multiply by (s-5)(s+5) to clear the denominators, results:
1.2(s+5) = 1.2(s-5) + .5(s+5)(s-5)
1.2s + 6 = 1.2s - 6 + .5(s^2 - 25)
:
Combine on the right
0 = 1.2s - 1.2s - 6 - 6 + .5s^2 - 12.5
:
Combine like terms, arrange as a quadratic equation
.5s^2 - 24.5 = 0
:
multiply by 2 to clear the decimals
s^2 - 49 = 0
s^2 = +49
s = {{{sqrt(49)}}}
s = 7 mph in still water
:
:
See if this is true, find the times
1.2/(7-5) = .6 hrs upstream
1.2/(7+5) = .1 hrs down
-------------------
time dif: = .5 hrs</PRE>