document.write( "Question 132810: a boat travels 50km downstream in the same time that it takes to go 30km upstream. The speed of the boat in still water is 16 km/h. Find the speed upstream. \n" ); document.write( "
Algebra.Com's Answer #97080 by nycsharkman(136)\"\" \"About 
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Let s be the speed of the river.\r
\n" ); document.write( "\n" ); document.write( "The speed of the boat downstream is 16 + s (faster)
\n" ); document.write( "The speed of the boat upstream is 16 - s (slower)\r
\n" ); document.write( "\n" ); document.write( "D = RT\r
\n" ); document.write( "\n" ); document.write( "Downstream:
\n" ); document.write( "50 = (16 + s)t
\n" ); document.write( "t = 50/(16 + s)\r
\n" ); document.write( "\n" ); document.write( "Upstream:
\n" ); document.write( "30 = (16 - s)t
\n" ); document.write( "t = 30/(16 - s)\r
\n" ); document.write( "\n" ); document.write( "Since these times are the same, equate them and solve for s.
\n" ); document.write( "50/(16+s) = 30/(16-s)\r
\n" ); document.write( "\n" ); document.write( "Cross multiply:
\n" ); document.write( "50(16 - s) = 30(16 + s)
\n" ); document.write( "800 - 50s = 480 + 30s\r
\n" ); document.write( "\n" ); document.write( "Add 50s to both sides:
\n" ); document.write( "800 = 480 + 80s\r
\n" ); document.write( "\n" ); document.write( "Subtract 480 from both sides:
\n" ); document.write( "320 = 80s\r
\n" ); document.write( "\n" ); document.write( "Divide both sides by 80:
\n" ); document.write( "s = 320/80
\n" ); document.write( "s = 4\r
\n" ); document.write( "\n" ); document.write( "So the speed of the river is 4 km/h\r
\n" ); document.write( "\n" ); document.write( "The boat goes downstream at 16 + 4 = 20 km/h
\n" ); document.write( "The boat goes upstream at 16 - 4 = 12 km/h\r
\n" ); document.write( "\n" ); document.write( "Answer:
\n" ); document.write( "12 km/h upstream (against the 4 km/h current)
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