Milk A contains 10% butterfat and Milk B contains 60% butterfat. How many liters of each must be taken to obtain a mixture of 100 liters that will be 45% butterfat?
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Milk A contains 10% butterfat and Milk B contains 60% butterfat. How many liters of each must be taken to obtain a mixture of 100 liters that will be 45% butterfat?
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Answer:
Let's assume that x liters of Milk A is needed to make a mixture of 100 liters and y liters of Milk B is needed for the same.
From the problem, we know that the total mixture required is 100 liters, so we can write:
x + y = 100 (Equation 1)
Also, we need to find out the amount of butterfat in the final mixture. We are given that the final mixture should contain 45% butterfat.
So, the amount of butterfat in Milk A and Milk B combined should be:
0.10x + 0.60y = 0.45(100) (Equation 2)
Simplifying Equation 2, we get:
0.10x + 0.60y = 45
Multiplying both sides by 10, we get:
x + 6y = 450 (Equation 3)
Now we can solve the equations (1) and (3) simultaneously to find the values of x and y.
Multiplying Equation 1 by 6, we get:
6x + 6y = 600 (Equation 4)
Subtracting Equation 3 from Equation 4, we get:
5x = 150
x = 30
Substituting the value of x in Equation 1, we get:
30 + y = 100
y = 70
Therefore, we need 30 liters of Milk A and 70 liters of Milk B to obtain 100 liters of mixture containing 45% butterfat.