Question

The owner of The Daily Grind coffee shop mixes French roast coffee worth $9.00 per pound with Kenyan coffee worth $7.50 per pound in order to get a 10 pound mixture worth $8.40 per pound. How much of each type of coffee was used?

Answer 1

Answer on Question #43681, Math, Abstract Algebra

Problem. The owner of The Daily Grind coffee shop mixes French roast coffee worth $9.00 per pound with Kenyan coffee worth $7.50 per pound in order to get a 10 pound mixture worth $8.40 per pound. How much of each type of coffee was used?

Solution. Assume that the owner used $x$ pounds of French coffee and $y$ pounds of Kenyan coffee. Then $x + y = 10$ (equation concerning weight). Also $9x + 7.5y = 8.4 \cdot 10$ (equation concerning price). Now we solve the system of equations:

$$\left\{ \begin{array}{c} x + y = 10, \\ 9x + 7.5y = 84; \end{array} \right. \left\{ \begin{array}{c} x + y = 10, \\ 6x + 5y = 52; \end{array} \right. \left\{ \begin{array}{c} x + y = 10, \\ x + 5(x + y) = 52; \end{array} \right. \left\{ \begin{array}{c} x + y = 10, \\ x + 5 \cdot 10 = 52; \end{array} \right. \left\{ \begin{array}{c} x + y = 8, \\ x = 2; \end{array} \right.$$

Thus, $x = 2$, $y = 8$ and we are done.

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