9. 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?

1 answer

Let x be the number of cups of the 12% peanut mixture needed and y be the number of cups of the 60% peanut mixture needed to make 12 cups that is 40% peanuts.

We can start by setting up two equations based on the information given:

Equation 1: The total number of cups needed is 12.
x + y = 12

Equation 2: The combined mixture should be 40% peanuts.
0.12x + 0.60y = 0.40(12)

Simplifying equation 2:

0.12x + 0.60y = 4.8
12x + 60y = 480
x + 5y = 40

We now have a system of two equations:

x + y = 12
x + 5y = 40

We can solve this by subtracting equation 1 from equation 2:

4y = 28

y = 7

Substitute y = 7 into equation 1 to find x:

x + 7 = 12

x = 5

Therefore, we would need 5 cups of the 12% peanut mixture and 7 cups of the 60% peanut mixture to make 12 cups that is 40% peanuts.