Successive replacement in a 960-litre milk container:\nPure milk initially. Replace 48 litres thrice (each time milk removed, same volume of water added). Find remaining pure milk.

Introduction / Context:
Successive replacement problems use the dilution formula: remaining fraction after one draw-replace cycle is (1 − drawn/total). Repeating multiplies the factor.

Given Data / Assumptions:

• Total V = 960 litres; draw d = 48 litres each time.
• Number of operations n = 3; initially pure milk.

Concept / Approach:
Remaining pure milk = V * (1 − d/V)^n.

Step-by-Step Solution:

Verification / Alternative check:
Compute sequentially: 960→912 (milk left) then 912*(19/20)=866.4; then 866.4*(19/20)=823.08.

Why Other Options Are Wrong:
Other values correspond to single/double replacement or arithmetic slips.

Common Pitfalls:
Removing fixed 48 litres of milk each time instead of using the fraction method; the composition changes each step.

Final Answer:
823.08 ltr