document.write( "Question 1150335: a supermarket mixes coffee that sells for $3.60 per pond with coffee that sells for $7.20 per pound to obtain 40m pounds of coffee selling for $6.00 per pound. how much of each type of coffee should be used?
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Algebra.Com's Answer #771743 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "By the traditional algebraic method for solving mixture problems like this, the solution could start like this:

\n" ); document.write( "x pounds of coffee at $3.60 per pound, plus (40-x) pounds of coffee at $7.20 per pound, makes 40 pounds of coffee at $6.00 per pound.

\n" ); document.write( "\"x%283.60%29%2B%2840-x%29%287.20%29+=+40%286.00%29\"

\n" ); document.write( "That equation is solved by basic algebra, but the actual calculations are a bit tedious.

\n" ); document.write( "Here is a much faster and easier way to solve a mixture problem like this, where only two things are being mixed.

\n" ); document.write( "(1) The per-pound price of the mixture, $6.00, is two-thirds of the way from $3.60 per pound to $7.20 per pound. (If it helps, look at the three numbers 3.60, 6.00, and 7.20 on a number line. From 3.60 to 7.20 is 3.60; from 3.60 to 6.00 is 2.40. 2.40 is two-thirds of 3.60; 6.00 is two-thirds of the way from 3.60 to 7.20.)

\n" ); document.write( "(2) That means two-thirds of the mixture must be the higher priced coffee.

\n" ); document.write( "Unfortunately, the numbers don't work out \"nicely\" in this problem, because 2/3 of 40 pounds is not a whole number....

\n" ); document.write( "ANSWER: 80/3 pounds, or 26 2/3 pounds, of the $7.20 per pound coffee and 40/3 pounds or 13 1/3 pounds, of the $3.60 per pound coffee.

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