document.write( "Question 798204: Lee blends coffee for a living. He needs to prepare 140 pounds of blended coffee beans selling for $4.11 per pound. He plans to do this by blending together a high-quality bean costing $5.00 per pound and a cheaper bean at $2.50 per pound. To the nearest pound, how much high-quality coffee bean and how much cheaper coffee bean he should blend? \n" ); document.write( "
Algebra.Com's Answer #482094 by stanbon(75887)\"\" \"About 
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Lee blends coffee for a living. He needs to prepare 140 pounds of blended coffee beans selling for $4.11 per pound. He plans to do this by blending together a high-quality bean costing $5.00 per pound and a cheaper bean at $2.50 per pound. To the nearest pound, how much high-quality coffee bean and how much cheaper coffee bean he should blend?
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\n" ); document.write( "Equations:
\n" ); document.write( "Quantity Eq.::: lower + higher = 140 lbs
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\n" ); document.write( "Value Eq:::::: 2.5L +5H = 4.11*140 dollars
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\n" ); document.write( "Modify:
\n" ); document.write( "250L + 250H = 250*140
\n" ); document.write( "250L + 500H = 411*140
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\n" ); document.write( "Solve for H:
\n" ); document.write( "250H = 161*140
\n" ); document.write( "H= (14/25)161
\n" ); document.write( "H = 90.16 lbs (amt. of high-quality beans needed)
\n" ); document.write( "L = 140-H = 49.84 lbs (amt. of lower-quality beans needed)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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