Mixture Problems

A coffee company makes a product which is a mixture of two coffees, using a coffee that costs $10.20 per pound and another coffee that costs $6.80 per pound. In order to make 20 pounds of a mixture that costs $8.50 per pound, how much of each type of coffee should it use?

Let $m$ be the amount of the $10.20 coffee, and let $n$ be the amount needed of the $6.80 coffee. Since we want 20 pounds of coffee that costs $8.50 per pound, the total cost for all 20 pounds is $20⋅$8.50=$170$. The cost for the 20 pounds of mixture is equal to the cost of each type of coffee added together: $10.20⋅m+6.8⋅n=170.$

Also, the amount of each type of coffee added together equals 20 pounds: $m+n=20$.

The system is: $\left\{\begin{array}{c} m+n=20 \\ 10.20 \cdot m+6.8 \cdot n=170 \end{array}\right.$

We can isolate one variable and use substitution to solve the system:

$m=20−n$

Now solve for $n$.

$\begin{aligned} 10.20(20-n)+6.8 n &=170 & \\ 204-10.20 n+6.8 n &=0170 && \text {   Distributive Property } \\ 204-3.4 n &=170 && \text {   Add like terms. } \\ -3.4 n &=-34 && \text {   Subtract 204. } \\ n &=10 && \text {   Divide by }-3.4 \end{aligned}$

Since $n = 10$, we can plug that into $m + n = 20$.

$m+10=20 \Rightarrow m=10$

The coffee company needs to use 10 pounds of each type of coffee in order to have a 20 pound mixture that costs $8.50 per pound.