A container initially has 960 litres of pure milk. If 48 litres of milk is removed and replaced with 48 litres of water, and this replacement process is repeated two more times (total 3 replacements), what will be the amount of pure milk left in the container after the third replacement?

Introduction:
This is a classic repeated replacement (dilution) problem. When a fixed volume is removed from a mixture and replaced with another liquid, the remaining amount of the original liquid decreases by a constant factor each time. The same logic applies whether we talk about milk, alcohol, or any solution. Here, the container starts with pure milk, and each operation removes some milk and replaces it with water. After multiple operations, we compute how much pure milk remains.

Given Data / Assumptions:

• Initial volume = 960 litres (pure milk)• Each time removed = 48 litres• Each time replaced with = 48 litres water• Number of replacements = 3

Concept / Approach:
After each replacement, the fraction of the original liquid that remains is (1 - removed/total). After n identical operations, remaining original liquid = initial * (1 - removed/total)^n. This works because each removal takes away the same fraction of whatever milk remains at that moment.

Step-by-Step Solution:
Step 1: Compute the fraction remaining after one replacement.removed/total = 48/960 = 0.05remaining fraction = 1 - 0.05 = 0.95Step 2: Apply the repeated replacement formula for n = 3.Milk left = 960 * (0.95)^3Step 3: Calculate (0.95)^3.0.95^2 = 0.90250.95^3 = 0.9025 * 0.95 = 0.857375Step 4: Multiply by 960.Milk left = 960 * 0.857375 = 823.08 litres (approx)

Verification / Alternative check:
Because each operation removes 5% of the mixture, it removes 5% of the milk present at that time. So milk decreases by a factor of 0.95 each step. Applying the factor three times (0.95 * 0.95 * 0.95) confirms the final factor 0.857375. Multiplying by 960 gives the same result, so the calculation is consistent.

Why Other Options Are Wrong:
901.54 ltr is too high; it would correspond to fewer replacements or a smaller removed fraction.821.54 ltr and 719.64 ltr are too low, implying more dilution than actually performed.857.38 ltr matches 960*0.8931 approximately, not the correct 0.95^3 factor.

Common Pitfalls:
• Subtracting 48 litres three times directly from 960, which is incorrect because removed liquid is partly water after the first step.• Using n = 2 instead of n = 3.• Forgetting to use the fraction removed (48/960) rather than absolute litres in the formula.

Final Answer:
The amount of pure milk left after three replacements is 823.08 litres.