document.write( "Question 57310This question is from textbook Elementary and Intermediate algebra
\n" ); document.write( ": Your assistance is greatly appreciated. \r
\n" ); document.write( "\n" ); document.write( "Angela needs 20 quarts of 50% antifreeze solution in her radiator. She plans to obtain this by mixing some pure antifreeze with an appropriate amount of 40% antifreeze solution. How many quarts of each should she use? \n" ); document.write( "
Algebra.Com's Answer #42100 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Let x=amt of pure antifreeze that is mixed in
\n" ); document.write( "Then 20-x=amt of 40% antifreeze solution that is mixed in
\n" ); document.write( "We know that the amt of pure antifreeze that is mixed in (x)(1) plus the amt of pure antifreeze in the 40% solution that is mixed in (20-x)(.40) equals the amt of pure antifreeze in the final solution (20)(.50).
\n" ); document.write( "Thus, our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "(x)(1)+(20-x)(.40)=(20)(.50) Simplifying, we get:
\n" ); document.write( "x+8-.4x=10 or
\n" ); document.write( ".6x=2; 6x=20
\n" ); document.write( "x=3.3333 qts of pure antifreeze
\n" ); document.write( "20-x=20-3.3333=16.6667 qts of the 40% antifreeze solution\r
\n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor
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