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.