<PRE><B>Question 1007344</B>
A river barge travels at an average of 8 miles per hour in still water.
 The barge travels 60 miles up the Mississippi River and returns 60 miles downriver in 16.5 hours.
 What is the average speed of the current, (the river), in this section of the Mississippi River? 
:
let c = the speed of the current
then
(8-c) = effective speed up-river
and
(8+c) = effective speed down-river
:
Time up + time down = 16.5 hrs
{{{60/(8-c)}}} + {{{60/(8+c)}}} = 16.5
multiply equation by (8-c)(8+c), cancel the denominators
60(8+c) + 60(8-c) = 16.5(8-c)(8+c)
480 + 60c + 480 - 60c = 16.5(64-c^2)
960 = 1056 - 16.5c^2
16.5c^2 = 1056 - 960
16.5c^2 = 96
c^2 = 96/16.5
c^2 = 5.82
c = {{{sqrt(5.82)}}}
c = 2.41 mph is the speed of he current
:
:
See if that checks out, find the actual time each way
{{{60/(8-2.41)}}} = 10.73 hrs
{{{60/(8+2.41)}}} =  5.76 hrs
-----------------------------
tot time: 16.49 hrs, close enough</PRE>