A container initially contains 120 litres of pure diesel. From the container, 12 litres of the current mixture is removed and replaced with 12 litres of kerosene. This remove-and-replace operation is repeated 3 times in total (assume perfect mixing before each removal). How many litres of diesel will remain in the container after the 3rd replacement?

Introduction / Context:
This is a successive replacement problem. Each time you remove a fixed fraction of the mixture and replace it with another liquid, the amount of the original liquid (diesel) reduces by the same fraction repeatedly. The remaining diesel follows an exponential pattern: diesel left = initial diesel * (remaining fraction)^number of operations.

Given Data / Assumptions:

• Initial diesel = 120 L (pure)
• Each time removed = 12 L of current mixture
• Each time replaced = 12 L kerosene
• Number of repetitions = 3
• Perfect mixing before each removal

Concept / Approach:
Diesel left after n operations = initial diesel * (1 - removed/total)^n. Here removed/total = 12/120 = 0.1, so remaining fraction is 0.9 each time.

Step-by-Step Solution:

Verification / Alternative check:
After 1st: 120*0.9 = 108 L. After 2nd: 108*0.9 = 97.2 L. After 3rd: 97.2*0.9 = 87.48 L. The step-by-step multiplication confirms the same result.

Why Other Options Are Wrong:

Common Pitfalls:
Do not subtract 12 litres diesel each time (120-36=84), because after the first step the removed 12 litres contains kerosene too. Another mistake is using (0.1)^3 instead of (0.9)^3. Remember: 0.9 is what remains, not what is removed.

Final Answer:
87.48 litres