document.write( "Question 593007: 1. Mike can row in still water at a speed twice that of the current in a certain river. It takes Mike 2 hours more to row 10 km up the river than it takes him to row 15 km down the river. What is the speed of current in the river?\r
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Algebra.Com's Answer #376219 by ankor@dixie-net.com(22740)\"\" \"About 
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1. Mike can row in still water at a speed twice that of the current in a certain river.
\n" ); document.write( " It takes Mike 2 hours more to row 10 km up the river than it takes him to row 15 km down the river.
\n" ); document.write( " What is the speed of current in the river?
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\n" ); document.write( "Let r = speed of current of the river
\n" ); document.write( "It states, \"row in still water at a speed twice that of the current\", therefore
\n" ); document.write( "2r = his rowing speed in still water
\n" ); document.write( "Then
\n" ); document.write( "2r - r = r, effective speed upstream
\n" ); document.write( "2r + r = 3r, effective speed downstream
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\n" ); document.write( "Write time equation
\n" ); document.write( "Upriver time - downriver time = 2 hrs
\n" ); document.write( "\"10%2Fr\" - \"15%2F%283r%29\" = 2
\n" ); document.write( "multiply by 3r
\n" ); document.write( "3r*\"10%2Fr\" - 3r*\"15%2F%283r%29\" = 3r(2)
\n" ); document.write( "Cancel the denominators
\n" ); document.write( "3(10) - 15 = 6r
\n" ); document.write( "30 - 15 = 6r
\n" ); document.write( "15/6 = r
\n" ); document.write( "r = 2.5 km/hr is the speed of the current
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\n" ); document.write( "Check that out, rowing speed in still water will be 5 km/hr, find the times
\n" ); document.write( "10/2.5 = 4hrs
\n" ); document.write( "15/7.5 = 2hrs
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\n" ); document.write( "difference: 2 hrs
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