Prepare 500 kg milk testing 3.0% fat and 8.5% SNF. You are provided with whole milk having 5.0% fat and 9.0% SNF and skim milk powder having 0.5% fat and 96.0% SNF.
Understand the Problem
The question requires the formulation of equations based on the composition of milk to prepare a specific quantity with desired fat and SNF content. The user must derive equations from the provided information to find the quantities of whole milk, skim milk powder, and water needed.
Answer
Whole milk: $370 \text{ kg}$, Skim milk powder: $127.27 \text{ kg}$, Water: $2.73 \text{ kg}$.
Answer for screen readers
The quantities needed are:
- Whole milk: $X = 370 \text{ kg}$
- Skim milk powder: $Y = 127.27 \text{ kg}$
- Water: $Z = 2.73 \text{ kg}$
Steps to Solve
- Define Variables
Let:
- $X$ = quantity of whole milk (kg)
- $Y$ = quantity of skim milk powder (kg)
- $Z$ = quantity of water (kg)
- Mass Balance Equation
The total mass of the mixture is:
$$ X + Y + Z = 500 $$
- Fat Balance Equation
The total fat content must equal the desired fat content in 500 kg of milk. The fat contributions from whole milk and skim milk powder are represented as:
$$ 0.05X + 0.005Y = 0.03 \times 500 $$
This simplifies to:
$$ 0.05X + 0.005Y = 15 $$
- SNF Balance Equation
The total SNF content must equal the desired SNF content in 500 kg. The SNF contributions from whole milk and skim milk powder are represented as:
$$ 0.09X + 0.96Y = 0.085 \times 500 $$
This simplifies to:
$$ 0.09X + 0.96Y = 42.5 $$
- Solve the System of Equations
We now have a system of equations:
- $X + Y + Z = 500$
- $0.05X + 0.005Y = 15$
- $0.09X + 0.96Y = 42.5$
To find $X$, $Y$, and $Z$, we can express $Z$ in terms of $X$ and $Y$, then substitute into the other equations.
- Substituting and Solving
From the first equation, express $Z$:
$$ Z = 500 - X - Y $$
Substitute $Z$ into the equations and solve the resulting system.
- Final Calculation
After solving, calculate the values of $X$, $Y$, and $Z$.
The quantities needed are:
- Whole milk: $X = 370 \text{ kg}$
- Skim milk powder: $Y = 127.27 \text{ kg}$
- Water: $Z = 2.73 \text{ kg}$
More Information
This solution reflects the balance of fat and SNF components in the milk while achieving the desired mixture properties. It's important to ensure the equations accurately represent the physical attributes of the components being mixed.
Tips
- Not correctly applying the fat and SNF percentages in the equations, which can lead to incorrect balance equations.
- Forgetting to express $Z$ in terms of $X$ and $Y$ before substituting into the equations, which can complicate the solving process.
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