Question 1142695
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I don't know what kind of "model" one would use for this problem....<br>
Here is a solution to your problem using common sense and logical reasoning that is far less work than the formal algebraic solution method shown by the other tutor.<br>
(1) The per-pound cost of the mixture, $1.02, is 1/3 of the way from 93 cents to $1.20.  (93 cents to $1.20 is 27 cents; 93 cents to $1.02 is 9 cents; 9/27 = 1/3)<br>
(2) Therefore 1/3 of the mixture must be the more expensive coffee.<br>
ANSWER: 1/3 of 30 pounds, or 10 pounds, of the $1.20 per pound coffee; and 20 pounds of the $0.93 per pound coffee.