The problem involves repeatedly removing 2 liters of a 10-liter solution of milk and water and replacing it with pure milk. Let’s break this down step by step.

Initial setup:

• Initial amount of milk: 8 liters in a 10-liter solution.
• After removing 2 liters, we replace it with pure milk.

Formula:

At each step, $\frac{2}{10} = \frac{1}{5}$ of the milk is removed. Hence, $\frac{4}{5}$ of the milk remains. This is repeated three times.

The formula for the amount of milk after $n$ repetitions is: $M_n = M_0 \times \left(\frac{4}{5}\right)^n + (10 \times \left(1 - \left(\frac{4}{5}\right)^n\right))$ Where $M_0 = 8$ liters (initial milk content).

We will calculate the amount of milk after 3 replacements. Let me perform the calculation.After three repetitions of the process, the amount of milk in the mixture is approximately 8.976 liters.

Therefore, the correct option is:

Option 1: 8.976

Would you like further details or have any questions? Here are some related questions you might explore:

1. How would the final amount of milk change if the solution initially had 7 liters of milk instead of 8?
2. What if the process was repeated four times instead of three?
3. How does increasing the volume of solution removed (e.g., 3 liters instead of 2 liters) affect the outcome?
4. What is the mathematical explanation for why milk approaches a certain limit as the process repeats?
5. How would the answer change if the initial solution contained a different total volume, such as 15 liters?

Tip: Repeated dilution problems often involve exponential decay, where each step reduces the remaining amount by a fixed fraction.