document.write( "Question 137312: How many ounces of pure water must be added to 80 oz of a 9% salt solution to make a 6% salt solution?
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\n" ); document.write( " A.
\n" ); document.write( " 120 oz
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\n" ); document.write( "B.
\n" ); document.write( " 60 oz
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\n" ); document.write( "C.
\n" ); document.write( " 50 oz
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\n" ); document.write( "D.
\n" ); document.write( " 40 oz
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Algebra.Com's Answer #100452 by stanbon(75887)\"\" \"About 
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How many ounces of pure water must be added to 80 oz of a 9% salt solution to make a 6% salt solution?
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\n" ); document.write( "9% solution DATA:
\n" ); document.write( "Amt. = 80 oz ; salt = 0.09*80 = 7.2 oz
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\n" ); document.write( "Water DATA:
\n" ); document.write( "Amt. = x oz ; salt = 0
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\n" ); document.write( "Mixture DATA:
\n" ); document.write( "Amt. = (80+x)oz ; salt = 0.06(80+x) = 4.8 + 0.06x
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\n" ); document.write( "EQUATION:
\n" ); document.write( "salt + salt = salt
\n" ); document.write( "7.2 + 0 = 4.8+0.06x
\n" ); document.write( "0.06x = 2.4
\n" ); document.write( "x = 40 oz (amt. of pure water that must be added)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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