document.write( "Question 1049826: Dan mixed 11 liters of cranberry juice into 12 liters of apple juice that had 34 percent sugar. if the cranapple mixture was 43 percent sugar, what was the percent of sugar in the cranberry juice?\r
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Algebra.Com's Answer #665403 by josgarithmetic(39617) $\"\"$ $\"About$

You can put this solution on YOUR website!
The emphasis is on concentration of sugar.\r
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document.write( "                   percent sugar        VOLUME       pure sugar\r\n" );
document.write( "cranberryjuice          p                11          0.01*11*p\r\n" );
document.write( "applejuice             34                12          0.34*12\r\n" );
document.write( "MIXTURE                43                23          0.43*23\r\n" );
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\n" ); document.write( "\n" ); document.write( "One way to use this is to account for the amount of pure sugar although the actual way
\n" ); document.write( "the concentrations are used is not too clear. Was that as mass per volume? This would
\n" ); document.write( "not be appropriate with such high concentrations of sugar. The rest of this problem-solving
\n" ); document.write( "may be faulty because of this need for the right kind of percentage. Sugar itself would
\n" ); document.write( "be a solid, so counting it in mass would be used. Your description did not say. The concentration
\n" ); document.write( "must either be volume per volume (which makes no sense here), or mass per mass (but no density were
\n" ); document.write( "given), or mass per volume (but this seems poor to use for such high sugar concentrations).\r
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\n" ); document.write( "\n" ); document.write( "Going ahead with just the arithmetic, accounting for the sugar:
\n" ); document.write( " $\"highlight_green%280.01%2A11p%2B0.34%2A12=0.43%2A23%29\"$
\n" ); document.write( "Solve for p.
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