A company sells a container of mixed nuts that is 12% peanuts and another container that is 60% peanuts. How many cups of each mixture would be needed to make 12 cups that is 40% peanuts?
Let X be the number of cups of 12% peanuts mixture needed.
Then the number of cups of 60% peanuts mixture needed would be 12 - X.
The amount of peanuts in X cups of 12% peanuts mixture is 0.12X, and the amount of peanuts in (12 - X) cups of 60% peanuts mixture is 0.6(12 - X).
To make 12 cups of 40% peanuts mixture, the total amount of peanuts should be 0.4(12) = 4.8 cups.
Therefore, we can write the equation:
0.12X + 0.6(12 - X) = 4.8
Simplifying and solving for X, we get:
0.12X + 7.2 - 0.6X = 4.8
-0.48X = -2.4
X = 5 cups
So, 5 cups of 12% peanuts mixture and 7 cups of 60% peanuts mixture would be needed to make 12 cups that is 40% peanuts.