document.write( "Question 74280: I need some help with this problem.\r
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document.write( "The speed of a mototboat in still water is 15 mi/hr. If it takes the motorboat the same time to go 15 mi upstream the river as it takes to go 30 mi
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document.write( "downstream, what is the speed of the current?\r
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document.write( "Thank You! \n" );
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Algebra.Com's Answer #53276 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The speed of a mototboat in still water is 15 mi/hr. If it takes the motorboat the same time to go 15 mi upstream the river as it takes to go 30 mi \n" ); document.write( "downstream, what is the speed of the current? \n" ); document.write( "-------------- \n" ); document.write( "Let the rate of the current be \"x\": \n" ); document.write( "Upstream DATA: \n" ); document.write( "Distance = 15 mi ; rate = 15-x ; time = d/r = 15(15-x) mph \n" ); document.write( "----------- \n" ); document.write( "Downstream DATA: \n" ); document.write( "Distance = 30 mi ; rate = 15+x ; time = d/r = 30/(15+x) mph \n" ); document.write( "------------- \n" ); document.write( "EQUATION: \n" ); document.write( "time up = time down \n" ); document.write( "15/(15-x) = 30/(15+x) \n" ); document.write( "Divide both sides by 15 to get: \n" ); document.write( "1/(15-x) = 2/(15+x) \n" ); document.write( "15+x = 2(15-x) \n" ); document.write( "15+x = 30-2x \n" ); document.write( "3x=15 \n" ); document.write( "x=5 mph \n" ); document.write( "The speed of the current is 5 mph \n" ); document.write( "--------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |