document.write( "Question 205939: Confusing. Help Pls.\r
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document.write( "How much water must be added to 30 liters of a 75% salt solution to reduce it to 15%? \n" );
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Algebra.Com's Answer #155559 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! How much water must be added to 30 liters of a 75% salt solution to reduce it to 15%? \n" ); document.write( ": \n" ); document.write( "Let x = amt of water required to this \n" ); document.write( ": \n" ); document.write( "the amt salt remains the same, only the % salt changes \n" ); document.write( "A simple equation \n" ); document.write( ": \n" ); document.write( ".75(30) = .15(x + 30) \n" ); document.write( ": \n" ); document.write( "22.5 = .15x + 4.5 \n" ); document.write( ": \n" ); document.write( "22.5 - 4.5 = .15x \n" ); document.write( ": \n" ); document.write( "18 = .15x \n" ); document.write( "x = \n" ); document.write( "x = 120 liters of water required \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution original equation (equal amt of salt) \n" ); document.write( ".75(30) = .15(120 + 30) \n" ); document.write( "22.5 = .15(150) \n" ); document.write( "22.5 = 22.5 \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Did this unconfuse you somewhat \n" ); document.write( " \n" ); document.write( " |