document.write( "Question 260651: A garage owner wants to fill a 55-gallon drum with a 20% winter mixture of antifreeze for his customers. How many gallons of 100% antifreeze should he mix with some 10% antifreeze mixtures in order to fill the drum? \n" ); document.write( "

Algebra.Com's Answer #192005 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "Let x=amount of 100% antifreeze needed
\n" ); document.write( "Then 55-x=amount of 10% antifreeze mixtures needed\r
\n" ); document.write( "\n" ); document.write( "Now we know that the amount of pure antifreeze used (x) plus the amount of pure antifreeze in the 10% antifreeze mixtures (0.10(55-x)) has to equal the amount of pure antifreeze in the final mixture (0.20*55). So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "x+0.10(55-x)=0.20*55 simplify
\n" ); document.write( "x+5.5-0.10x=11 subtract 5.5 from each side
\n" ); document.write( "x+5.5-5.5-0.10x=11-5.5 collect like terms
\n" ); document.write( "0.9x=5.5 divide each side by 0.9
\n" ); document.write( "x=6.1 gal-------------------------amount of 100% antifreeze needed
\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "6.1+0.10(48.9)=11
\n" ); document.write( "6.1+4.9=11
\n" ); document.write( "11=11\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( "

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