From a container full of pure wine, a thief steals 15 litres of wine and replaces it with 15 litres of water. He repeats this process two more times, each time removing 15 litres of the current mixture and replacing it with 15 litres of water. After three such operations, the ratio of wine to water in the container is 343 : 169. What was the initial quantity of wine in the container (in litres)?

Introduction:
This is a classic repeated replacement mixture problem, where a fixed quantity of liquid is removed and replaced with another liquid several times. The question provides the final ratio of wine to water and asks for the initial volume of wine. Understanding the replacement formula for mixtures is crucial in solving this efficiently.

Given Data / Assumptions:

• The container initially contains only pure wine.
• In each operation, 15 litres of the mixture are removed.
• These 15 litres are then replaced with 15 litres of water.
• This process is repeated three times in total.
• After three operations, the ratio of wine to water is 343 : 169.
• The total volume of the liquid in the container remains constant.

Concept / Approach:
Let V be the initial volume of wine in the container. In each operation, a fraction 15 / V of the current mixture is removed and replaced by water. After one operation, the fraction of wine left is (1 - 15 / V). After n repeated operations, the fraction of wine remaining is (1 - 15 / V)^n. Here n = 3. This fraction must equal the final wine fraction obtained from the ratio 343 : 169.

Step-by-Step Solution:
Step 1: Let the initial volume of wine be V litres. Step 2: After each replacement, the fraction of wine remaining is (1 - 15 / V). Step 3: After three such operations, the fraction of wine remaining is (1 - 15 / V)^3. Step 4: The final ratio of wine to water is 343 : 169, so total parts = 343 + 169 = 512. Step 5: The fraction of wine finally is 343 / 512. Step 6: Set up the equation: (1 - 15 / V)^3 = 343 / 512. Step 7: Take cube roots: 1 - 15 / V = 7 / 8 (we discard the negative root by logic). Step 8: Solve for V: 15 / V = 1 - 7 / 8 = 1 / 8. Step 9: Therefore V = 15 * 8 = 120 litres.

Verification / Alternative check:
Start with 120 litres of wine. After first operation, wine left = 120 * (1 - 15 / 120) = 120 * (105 / 120) = 105 litres. After second operation, wine left = 105 * (105 / 120) = 105^2 / 120. After third operation, wine left = (105^3) / (120^2). The fraction of wine is (105 / 120)^3. Simplify 105 / 120 to 7 / 8. So final fraction = (7 / 8)^3 = 343 / 512, matching the given ratio, confirming that V = 120 litres is correct.

Why Other Options Are Wrong:
75 litres, 100 litres, and 150 litres produce different fractions of wine after three operations and do not yield the ratio 343 : 169.
90 litres similarly does not satisfy the equation (1 - 15 / V)^3 = 343 / 512.

Common Pitfalls:
A common mistake is to subtract 15 litres of wine each time directly, forgetting that after the first replacement the container holds a mixture, not pure wine. Another frequent error is to ignore the exponential nature of repeated replacement and simply multiply or subtract linearly. It is also easy to miscalculate the final ratio by treating 343 : 169 as a difference rather than a ratio.

Final Answer:
The initial amount of wine in the container was 120 litres.