document.write( "Question 119381: \r
\n" ); document.write( "\n" ); document.write( "How much pure alcohol must be added to 40 oz of a 25%
\n" ); document.write( "solution to produce a mixture that is 40% alcohol?
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Algebra.Com's Answer #87455 by stanbon(75887)\"\" \"About 
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How much pure alcohol must be added to 40 oz of a 25%
\n" ); document.write( "solution to produce a mixture that is 40% alcohol?
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\n" ); document.write( "25% solution DATA:
\n" ); document.write( "Amt = 40 oz ; amt of active ingredient = 0.25*40 = 10 oz
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\n" ); document.write( "Pure solution DATA:
\n" ); document.write( "Amt = x oz ; amt of active ingredient = 100%*x = x oz
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\n" ); document.write( "40% Mixture DATA:
\n" ); document.write( "amt. = 40+x oz ; amt of active ingredient = 0.40(40+x)=16+0.4x oz
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\n" ); document.write( "EQUATION:
\n" ); document.write( "active + active = active
\n" ); document.write( "10 + x = 16 + 0.4x
\n" ); document.write( "0.6x = 6
\n" ); document.write( "x = 10 oz (amt. of 100% alcohol needed for the mixture)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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