document.write( "Question 387781: A speedboat takes 4 hours longer to go 80 miles up a river than to return. If the boat cruises at 15 miles per hour in still water, what is the rate of the current?
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Algebra.Com's Answer #276903 by robertb(5830) $\"\"$ $\"About$

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Let c = rate of the current.\r
\n" ); document.write( "\n" ); document.write( " Then 15+c = rate of speedboat down the river
\n" ); document.write( "15 - c = rate of speedboat up the river\r
\n" ); document.write( "\n" ); document.write( "From the Formula D = R*T, we get T = D/R, so
\n" ); document.write( " $\"80%2F%2815-c%29\"$ = time of speedboat up the river
\n" ); document.write( " $\"80%2F%2815%2Bc%29\"$ = time of speedboat down the river. From these we get
\n" ); document.write( " $\"80%2F%2815-c%29+-+80%2F%2815%2Bc%29+=+4\"$ , or $\"20%2F%2815-c%29+-+20%2F%2815%2Bc%29+=+1\"$ .\r
\n" ); document.write( "\n" ); document.write( "Combining and simplifying, we get
\n" ); document.write( " $\"40c+=+225+-+c%5E2\"$ , or $\"c%5E2+%2B+40c+-+225+=+0\"$ , or (c-5)(c+45) = 0.
\n" ); document.write( "Hence c = 5 miles per hour, the rate of the current. \n" ); document.write( "

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