document.write( "Question 133669: John runs a garage. He needs to mix antifreeze in such a way as to produce 20 quarts of 50% solution. He has on hand ample supplies of 25% and 75% solutions. How much of 25% and 75% solutions must he mix together to produce the needed 20 quarts of 50% solution? \n" ); document.write( "

Algebra.Com's Answer #97843 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let x=amount of 75% solution needed
\n" ); document.write( "Then 20-x=amount of 25% solution needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure antifreeze in the 25% solution (0.25(20-x)) plus the amount of pure antifreeze in the 75% solution (0.75x) has to equal the amount of pure antifreeze in the final mixture (0.50*20). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "0.25(20-x)+0.75x=0.50*20 simplify\r
\n" ); document.write( "\n" ); document.write( "5-0.25x+0.75x=10 subtract 5 from both sides
\n" ); document.write( "5-5-0.25x+0.75x=10-5 collect like terms\r
\n" ); document.write( "\n" ); document.write( "0.50x=5 divide both sides by 0.50
\n" ); document.write( "x=10 qts------------------------------amount of 75% solution needed
\n" ); document.write( "20-x=20-10=10 qts------------------------------amount of 25% solution needed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "0.25*10+0.75*10=0.50*20
\n" ); document.write( "2.5+7.5=10
\n" ); document.write( "10=10\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor \n" ); document.write( "

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