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)![]() ![]() You can put this solution on YOUR website! 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? \n" ); document.write( ": \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 \n" ); document.write( ": \n" ); document.write( "Write time equation \n" ); document.write( "Upriver time - downriver time = 2 hrs \n" ); document.write( " \n" ); document.write( "multiply by 3r \n" ); document.write( "3r* \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 \n" ); document.write( ": \n" ); document.write( ": \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 \n" ); document.write( "--------------- \n" ); document.write( "difference: 2 hrs \n" ); document.write( " |