A petrol tank contains 200 litres of pure petrol. Each time the seller sells 40 litres of petrol (or the current petrol-kerosene mixture), and then immediately replaces the sold 40 litres by adding 40 litres of kerosene back into the tank (so the tank remains 200 litres). After repeating this sell-and-replace process 4 times, how many litres of kerosene will be present in the tank?

Introduction / Context:
This is a repeated replacement (successive dilution) problem. Each time 40 litres are removed from a 200 litre tank, the fraction removed is 40/200 = 1/5. Whatever petrol remains is multiplied by (1 - 1/5) = 4/5 after each operation. Kerosene amount then equals total volume minus remaining petrol.

Given Data / Assumptions:

• Initial petrol = 200 L (pure)
• Each operation: remove 40 L of current mixture
• Replace removed 40 L with 40 L kerosene
• Number of operations = 4
• Total tank volume stays 200 L throughout

Concept / Approach:
Petrol left after n operations = initial petrol * (1 - removed/total)^n. Then kerosene = total - petrol left.

Step-by-Step Solution:

Verification / Alternative check:
As operations increase, petrol decreases and kerosene increases. After 4 rounds, petrol is 81.92 L (less than half), so kerosene 118.08 L is reasonable and consistent with successive replacement behavior.

Why Other Options Are Wrong:

Common Pitfalls:
Many learners subtract 40 litres of petrol each time and assume petrol reduces linearly (200 - 4*40 = 40), which is wrong because later removals contain kerosene too. Another mistake is using (0.2)^4 instead of (0.8)^4 for remaining fraction.

Final Answer:
118.08 L