document.write( "Question 923292: How many liters of a 30% antifreeze solution and 60% antifreeze solution must be mixed to make 10 liters of a 39% antifreeze solution \n" ); document.write( "

Algebra.Com's Answer #559954 by ptaylor(2198) $\"\"$ $\"About$

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Let x=number of liters of the 30 % solution needed
\n" ); document.write( "Then 10-x=number of liters of the 60% solution needed
\n" ); document.write( "Now we know that the amount of pure antifreeze that exists before the mixture takes place(0.30X+0.60(10-x)) has to equal the amount of pure antifreeze that exists after the mixture takes place (0.39*10). Soooo our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.30x+0.60(10-x)=0.39*10 simplify
\n" ); document.write( "0.30x+6-0.60x=3.9 collect like terms
\n" ); document.write( "-0.30x=-6+3.9=-2.1
\n" ); document.write( "x=7 liters------amount of 30% solution needed
\n" ); document.write( "10-x=10-7=3 liters amount of 60% solution needed
\n" ); document.write( "CK
\n" ); document.write( "0.30*7+0.60*3=0.39*10
\n" ); document.write( "2.1+1.8=3.9
\n" ); document.write( "3.9=3.9\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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