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.
Aptitude
Alligation or Mixture
Difficulty: Medium
Choose an option
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A901.54 ltr
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B821.54 ltr
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C719.64 ltr
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D823.08 ltr
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ENone of these
Answer
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