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

Difficulty: Medium

Correct Answer: 823.08 ltr

Explanation:

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:

Factor = (1 − 48/960) = (1 − 1/20) = 19/20 = 0.95.After 3 operations: remaining = 960 * (0.95)^3.(0.95)^3 = 0.857375; amount = 960*0.857375 = 823.08 litres (approx).

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

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