Question

Given 10,000 kg of 7% fat containing milk, cream of 40% fat is separated from the milk. Skim milk tests 0.1% fat while butter contains 83.33% fat and buttermilk contains 0.5% fat. Determine the quantity of buttermilk recovered.

• A. 683.7 kg
• B. 849.3 kg
• C. 729.5 kg
• D. 899.3 kg

Hint:

Material balance: Fat balance is key.
Use simultaneous equations for butter and buttermilk.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Material balance is used to determine product quantities.
Fat balance is applied to solve the problem.

Step 2: Key Formula or Approach:
Fat balance: Fat in milk = Fat in cream + Fat in skim milk.
Cream quantity: \(C = \frac{F_m - F_s}{F_c - F_s} \times M\).
Butter and buttermilk balance from cream.

Step 3: Detailed Explanation:
Milk = 10,000 kg, Fat = 7%, Skim fat = 0.1%, Cream fat = 40%.
Cream quantity = \(\frac{7 - 0.1}{40 - 0.1} \times 10000 = \frac{6.9}{39.9} \times 10000 = 1729.3\) kg.
Skim milk = 10000 - 1729.3 = 8270.7 kg.
Butter fat = 83.33%, Buttermilk fat = 0.5%.
Fat in cream = 1729.3 \(\times\) 0.40 = 691.7 kg.
Let butter = B kg, buttermilk = BM kg.
B + BM = 1729.3.
Fat: 0.8333B + 0.005BM = 691.7.
Solving: B = (691.7 - 0.005 \(\times\) 1729.3) (0.8333 - 0.005)
B = (691.7 - 8.65) 0.8283 = 683.05 0.8283 = 824.6 kg.
Buttermilk = 1729.3 - 824.6 = 904.7 kg.
Closest option is 899.3 kg.

Step 4: Final Answer:
The quantity of buttermilk recovered is approximately 899.3 kg.
Hence, the correct option is (D).