Question

Julia and her husband own a coffee shop. They experimented with mixing a City Roast Columbian coffee that cost $7.00 per pound with French Roast Columbian coffee that cost $7.80 per pound to make a 8-pound blend. Their blend should cost them $7.10 per pound. How much of each type of coffee should they buy?

Answer 1

We have two types of coffee:

• City roast (lets call it C), with a cost of $7.00 per pound.

,

• French roast (lets call it F), with a cost of $7.80 per pound.

They make a 8-pound blend, so the sum of the pounds of each coffee is equal to 8 pounds. We can express it as:

$C+F=8$

where C: pounds of City roast, and F: pounds of French roast.

The blend should cost $7.10 per pound.

The total cost of the blend is then 8*7.10 = $ 56.8.

This has to be equal to the sum of the price of City roast coffee times the quantity, 7.00*C, and the price of French roast coffe times the quantity, 7.80*F.

We can express this as:

$7.00\cdot C+7.80\cdot F=56.80$

We have a system of equations with two unknowns: C and F.

We can use the first equation to express C in function of F:

$\begin{gathered} C+F=8 \\ C=8-F \end{gathered}$

Now, we replace C in the second equation and solve for F:

$\begin{gathered} 7.00C+7.80F=56.80 \\ 7.00(8-F)+7.80F=56.80 \\ 56-7F+7.80F=56.80 \\ 56+0.8F=56.80 \\ 0.8F=56.8-56 \\ 0.8F=0.8 \\ F=\frac{0.8}{0.8} \\ F=1 \end{gathered}$

Then, C can be found as:

$\begin{gathered} C=8-F \\ C=8-1 \\ C=7 \end{gathered}$

Answer:

They should buy 7 pounds of City roast and 1 pound of French roast per 8-pound blend.