This problem involves the concept of successive dilution. Let's walk through it step-by-step to determine how much milk remains after three cycles of replacement.

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Step 1: Basic Concept of Replacement

When $x$ litres of liquid is removed from a solution of volume $V$, and replaced by another liquid, the concentration of the original liquid decreases exponentially. The remaining amount of the original liquid after $n$ replacements is given by the formula:

$\text{Remaining quantity} = V \left( 1 - \frac{x}{V} \right)^n$

In our case:

• Total volume $V = 100$ litres
• Replaced volume $x = 20$ litres
• Number of replacements $n = 3$

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Step 2: Calculate the Remaining Milk After 3 Replacements

Using the formula:

$\text{Remaining milk} = 100 \left( 1 - \frac{20}{100} \right)^3$

$\text{Remaining milk} = 100 \left( \frac{80}{100} \right)^3 = 100 \times 0.8^3$

$0.8^3 = 0.512$

$\text{Remaining milk} = 100 \times 0.512 = 51.2 \, \text{litres}$

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Step 3: Final Answer

After three successive replacements, 51.2 litres of milk remains in the vessel.

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Do you want further details or have any questions?

Related Questions:

1. What if the replacement volume was different each time?
2. How does the formula change if we don't refill to full capacity after each replacement?
3. What is the amount of water in the vessel after the three replacements?
4. How does increasing the number of replacements affect the remaining milk quantity?
5. Can we generalize this to infinite replacement steps?

Tip: Successive dilution problems become easier if you recognize the pattern in geometric progression for the concentration changes.