Step 1: Conceptual Understanding:
This problem involves a mixture scenario where a portion of the original substance is removed and substituted with another substance through repeated operations. The core task is to determine the residual fraction of the original substance after each operation.
Step 2: Governing Formula/Methodology:
The quantity of the original substance remaining after 'n' sequential replacements is calculated using the following formula:
\[ \text{Final Quantity} = \text{Initial Quantity} \times \left(1 - \frac{\text{Quantity Replaced}}{\text{Total Volume}}\right)^n \]
Here, 'n' represents the total count of replacement operations performed.
Step 3: Procedural Breakdown:
Identified parameters:
Initial milk volume = 60 litres.
Total mixture volume = 60 litres (constant throughout).
Volume replaced per operation = 6 litres.
The operation was conducted once, followed by two additional repetitions. Consequently, the total number of operations is \(n = 1 + 2 = 3\).
Applying the formula:
\[ \text{Amount of milk left} = 60 \times \left(1 - \frac{6}{60}\right)^3 \]
Simplify the fractional term within the parentheses:
\[ \frac{6}{60} = \frac{1}{10} = 0.1 \]
The expression transforms to:
\[ \text{Amount of milk left} = 60 \times (1 - 0.1)^3 \]
\[ = 60 \times (0.9)^3 \]
Calculate the value of \( (0.9)^3 \):
\[ (0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.9 = 0.729 \]
Finally, multiply this result by the initial volume:
\[ \text{Amount of milk left} = 60 \times 0.729 \]
\[ = 43.74 \]
Step 4: Conclusive Result:
Following three operations, the residual milk volume in the container is 43.74 litres.