document.write( "Question 1193084: Leo swims at 2 miles per hour in still water. After he swims down a river a quarter of a mile, returning takes three times as long as swimming downstream. Find the rate of the current. \n" ); document.write( "

Algebra.Com's Answer #825059 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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Leo swims at 2 miles per hour in still water.
\n" ); document.write( " After he swims down a river a quarter of a mile, returning takes three times as long as swimming downstream.
\n" ); document.write( " Find the rate of the current.
\n" ); document.write( ":
\n" ); document.write( "let c = the rate of the current
\n" ); document.write( "then
\n" ); document.write( "(2+c) = his speed down stream
\n" ); document.write( "and
\n" ); document.write( "(2-c) = his speed upstream
\n" ); document.write( ":
\n" ); document.write( "write the dist as .25 mi
\n" ); document.write( "Write a time equation, time = dist/speed
\n" ); document.write( " $\".25%2F%282-c%29\"$ = $\"3%28.25%2F%282%2Bc%29%29\"$
\n" ); document.write( " $\".25%2F%282-c%29\"$ = $\"%28.75%2F%282%2Bc%29%29\"$
\n" ); document.write( "cross multiply
\n" ); document.write( ".25(2+c) = .75(2-c)
\n" ); document.write( ".5 + .25c = 1.5 - .75c
\n" ); document.write( ".25c + .75c = 1.5 - .5
\n" ); document.write( "1c = 1
\n" ); document.write( "c = 1 mph is the rate of the current
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check, find the actual time each way in minutes
\n" ); document.write( ".25/1 = .25*60 = 15 min upstream
\n" ); document.write( ".25/3 = .0833*60 = 5 min downstream
\n" ); document.write( " \n" ); document.write( "

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