document.write( "Question 338451: in still water, a boat averages 7 mph. it takes the same amount of time to travel 20 miles downstream, with the current, as 8 miles upstream, against the current. what is the rate of the water's current? \n" ); document.write( "

Algebra.Com's Answer #242823 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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in still water, a boat averages 7 mph.
\n" ); document.write( " it takes the same amount of time to travel 20 miles downstream, with the
\n" ); document.write( "current, as 8 miles upstream, against the current.
\n" ); document.write( "what is the rate of the water's current?
\n" ); document.write( ":
\n" ); document.write( "Let c = rate of the current
\n" ); document.write( "then
\n" ); document.write( "(7+c) = effective rate downstream
\n" ); document.write( "and
\n" ); document.write( "(7-c) = effective rate upstream
\n" ); document.write( ":
\n" ); document.write( "Write a time equation: time = dist/rate
\n" ); document.write( ":
\n" ); document.write( "Time downstr = time upstr
\n" ); document.write( " $\"20%2F%287%2Bc%29\"$ = $\"8%2F%287-c%29\"$
\n" ); document.write( "Cross multiply
\n" ); document.write( "20(7-c) = 8(7+c)
\n" ); document.write( "140 - 20c = 56 + 8c
\n" ); document.write( "140 - 56 = 8c + 20c
\n" ); document.write( "84 = 28c
\n" ); document.write( "c = $\"84%2F28\"$
\n" ); document.write( "c = 3 mph is the current rate
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check this by finding the times
\n" ); document.write( "20/10 = 2 hrs
\n" ); document.write( "8/4 = 2 hrs \n" ); document.write( "

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