Question

Julia owns a coffee shop. She experimented with mixing City Roast Colombian coffee that costs $7.80 per pound with French Roast Colombian coffee that costs $8.10 per pound to make a 20 -pound blend. Her blend should cost her $7.92 per pound. How much of each type of coffee should she buy?

Answer 1

Answer:

8 lb of French Roast and 12 lb of the City Roast type of coffee

Step-by-step explanation:

The two unknowns in the problem are : the number of pounds of City Roast coffee (let's assign to such the letter "C"); and the number of pounds of French Roast coffee (let's identify it with the letter "F")

We write two equations,

1) one for the total number of pounds of coffee (20 lb total), which should come from C pounds of City Roast coffee and F pounds of French Roast coffee:

first equation    C + F = 20

2) another equation that takes into account the cost associated with the mixture. Since each pound of City Roast costs $7.80, then C pounds of it will cost $7.80 * C

Similarly, with the French Roast coffee, since each pound of it costs $8.10, F pounds of it will cost $8.10 * F

We want the addition of these two partial costs give the total cost we would like for the 20 lb mixture: $7.92 * 20 = $158.4

Now we write the equation for the total cost as:

$7.80 * C + $8.10 * F = $158.4

We use our first equation to solve for one of the unknowns in termsof the other, let's say solve for "C":

C + F = 20

C = 20 - F

and use it to replace "C) in the second equation:

7.80 (20 - F) + 8.10 F = 158.4

7.80 * 20 - 7.80 F + 8.10 F = 158.4

where we can combine the two terms in F to get + 0.3 F

156 + 0.3 F = 158.4

Now isolating F we get:

0.3 F = 158.4 - 156 = 2.4

dividing by 0.3 both sides:

F = 2.4 / 0.3 = 8

Which tells us that we should include 8 pounds of the French Roast.

Therefore the rest of the mixture (12 lb to complete a total of 20 lb) should be of the City Roast.