Question

Coffee cafe makes 90 pounds of coffee that costs $6 per pound. The types of coffee used to make this mixture cost $7 per pound and $4 per pound. How many pounds of $7-per-pound coffee should be used in this mixture?

Answer 1

60 pounds

Explanation:

For mixture problems, it often works well to let a variable represent the quantity of the highest-cost contributor to the mix. Here, that quantity is what the problem is asking for.

The problem tells us the quantity of lower-cost contributor will be (90-x), where the quantities are measured in pounds. Then the relation that tells us the total value of the mix will be one we can solve for the quantity of interest.

7x +4(90 -x) = 6(90) . . . . . value is total of quantities times their values

3x = 180 . . . . . . . . . . . . . . subtract 360, collect terms

x = 60 . . . . . . . . . . . . . divide by 3

60 pounds of the $7/lb coffee should be used in the mixture.

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Additional coment

The higher-value contributor is found to be the fraction of the mix that is ...

(M -L)/(H -L) . . . . . fraction that is highest-value contributor

where L, M, H are the values of the lowest-value contributor, the mix, and the highest-value contributor. Here, that is ...

(6 -4)/(7 -4) = 2/3 . . . the fraction of the mix that is $7/lb coffee