document.write( "Question 103295: How many pounds of 70% Columbian coffee must be added to ten pounds of 90% Columbian coffee to have p% Columbian coffee? \n" ); document.write( "

Algebra.Com's Answer #75153 by HyperBrain(694) $\"\"$ $\"About$

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Let x= the amt. of 70% columbian coffee that MUST be added\r
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\n" ); document.write( "\n" ); document.write( " $\"10%2A90%2B70x=p%2A%28x%2B10%29\"$
\n" ); document.write( " $\"900%2B70x=px%2B10p\"$
\n" ); document.write( " $\"%2870-p%29x=10p-900\"$
\n" ); document.write( " $\"x=10p-900%2F%2870-p%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Therefore, $\"10p-900%2F%2870-p%29\"$ pounds must be added \n" ); document.write( "

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