document.write( "Question 47437This question is from textbook
\n" ); document.write( ": A certain of beer is 5% alcohol and another type of beer is 2% alcohol. How much of each type should be mixed to get 100L of beer which is 3.2% alcohol? \n" ); document.write( "
Algebra.Com's Answer #31344 by checkley71(8403)\"\" \"About 
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.05X+.02(100-X)=.032*100 OR .05X+2-.02X=3.2 OR .03X=1.2 OR X=1.2/.03 OR X=40L OF 5%\r
\n" ); document.write( "\n" ); document.write( "& 100-40=60L OF 2% \r
\n" ); document.write( "\n" ); document.write( "PROOF 40*.05+60*.02=100*.032 OR 2+1.2=3.2 OR 3.2=3.2\r
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