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?
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.
To determine how many cups of each mixture would be needed, we can set up a system of equations based on the given information.
Let's assume x cups of the 12% peanuts mixture are needed and y cups of the 60% peanuts mixture are needed.
Based on the percentage of peanuts, we can establish the following equations:
Equation 1: (0.12)x + (0.6)y = (0.4)(12)
Equation 2: x + y = 12
First, let's solve Equation 2 for x:
x = 12 - y
Now, substitute the value of x in Equation 1:
(0.12)(12 - y) + (0.6)y = (0.4)(12)
1.44 - 0.12y + 0.6y = 4.8
Combine like terms:
0.48y = 3.36
Divide both sides by 0.48:
y = 7
Now, substitute the value of y back into Equation 2 to solve for x:
x + 7 = 12
x = 12 - 7
x = 5
Therefore, you would need 5 cups of the 12% peanuts mixture and 7 cups of the 60% peanuts mixture to make 12 cups that are 40% peanuts.