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Successive Replacement (Compound Type)

Introduction

Successive Replacement (Compound Type) problems involve repeated cycles where portions of a mixture are removed and replaced by different substances in a sequence - for example, remove A and add B, then remove B and add C, and so on.

This pattern is important because such compound operations appear in advanced dilution, alloy, and chemical-mixture problems. Recognizing the compound structure lets you combine stepwise effects into compact formulas or simple repeated calculations.

Pattern: Successive Replacement (Compound Type)

Pattern

Key concept: Treat each replacement as a multiplier on the remaining fraction of the substance of interest. For compound sequences, apply each operation in order (multiply the retained fractions) and track additions separately.

Steps to follow:
1. For each removal of x litres from total T, the fraction of the component that remains after that removal = (1 - x/T).
2. If different components are removed/added in sequence, compute the retained fraction step-by-step and multiply retained fractions for repeated removals of the same component.
3. When a different substance is added (not pure), include its pure-part contribution (added amount × concentration) to the numerator at the appropriate step.
4. After completing all steps, compute final amounts by combining retained original parts and all added contributions; then divide by total to get final concentration or composition.

Step-by-Step Example

Question

A 100-litre tank contains pure milk. 20 litres are removed and replaced with water. Then 30 litres of this mixture are removed and replaced with a 50% milk solution. Find the final quantity of milk in the tank.

Solution

  1. Step 1: Record initial data

    Total volume T = 100 L. Initially milk = 100 L (pure milk). First removal = 20 L (replaced by water). Second removal = 30 L (replaced by 50% milk solution).

  2. Step 2: After first replacement (remove 20 L, add water)

    Fraction of milk remaining after removing 20 L = (1 - 20/100) = 0.80. Milk left = 100 × 0.80 = 80 L. After adding 20 L water, total stays 100 L.

  3. Step 3: Prepare for second removal (mixture now: 80 L milk + 20 L water)

    When 30 L of this mixture is removed, it carries milk in the same proportion as current mixture: milk fraction = 80/100 = 0.80. Milk removed = 0.80 × 30 = 24 L. Milk left after removal = 80 - 24 = 56 L.

  4. Step 4: Add 30 L of 50% milk solution

    Added milk = 0.50 × 30 = 15 L. New milk total = 56 + 15 = 71 L. Total volume = 100 L.

  5. Final Answer:

    The tank contains 71 litres of milk after the compound operations.

  6. Quick Check:

    Track numerically: start 100 → after first step milk 80 → remove 30% of current (24) → 56 → add 15 → 71. All steps conserve total volume and give consistent milk amount ✅

Quick Variations

1. Remove x, add pure Y, then remove y and add pure Z - handle each step sequentially and sum added pure parts.

2. Repeatedly remove same fraction and add different concentrations - multiply retained fraction for each removal and add contributions separately.

3. Replacement with mixtures of known concentrations - treat added mixture like an injection of its pure-part amount.

4. Alloy problems: removal of alloy portions and addition of components follow identical bookkeeping but with component ratios.

Trick to Always Use

  • Step 1: Work sequentially - compute retained fraction and removed pure amount at each step.
  • Step 2: Multiply retained fractions when the same original component is removed multiple times.
  • Step 3: Add contributions from non-pure additions directly to the running total of the component.
  • Step 4: Use a one-line numeric check: follow values step-by-step to ensure totals and parts add up correctly.

Summary

Summary

In the Successive Replacement (Compound Type) pattern:

  • Treat each removal as multiplying the current amount by (1 - removed/total).
  • For compound sequences, apply operations in order and multiply retained fractions when appropriate.
  • When adding a mixture, compute its pure contribution (volume × concentration) and add to the running total of that component.
  • Always perform a step-by-step numeric check to confirm conservation of total volume and correctness of component amounts.

Practice

(1/5)
1. A 100-litre vessel is full of milk. 20 litres are removed and replaced with water. Then 20 litres of the resulting mixture are removed and replaced again with pure milk. Find the amount of milk in the final mixture.
easy
A. 80 L
B. 84 L
C. 88 L
D. 90 L

Solution

  1. Step 1: After first removal and replacement

    Remove 20 L from 100 L pure milk → remaining milk = 100 - 20 = 80 L. After adding 20 L water, total = 100 L (milk = 80 L).

  2. Step 2: Second removal (20 L of the current mixture)

    Milk fraction now = 80/100 = 0.8. Milk removed = 0.8 × 20 = 16 L. Milk left = 80 - 16 = 64 L.

  3. Step 3: Add 20 L pure milk

    Milk added = 20 L → Final milk = 64 + 20 = 84 L.

  4. Final Answer:

    84 L → Option B.

  5. Quick Check:

    Sequence: 100 → 80 → remove 16 → 64 → add 20 → 84 ✅

Hint: Handle each step in order: compute retained part, subtract removed milk, then add the pure portion introduced.
Common Mistakes: Multiplying fractions without adding the newly introduced pure portion at the last step.
2. A 60-litre vessel contains 100% milk. 15 litres are removed and replaced with water. Then again 15 litres are removed and replaced with water. Find the quantity of milk left after 2 operations.
easy
A. 33.75 L
B. 40.5 L
C. 42.25 L
D. 45 L

Solution

  1. Step 1: Fraction retained each time

    Fraction retained after removing 15 L from 60 L = 1 - 15/60 = 3/4.

  2. Step 2: After two identical operations

    Retained fraction = (3/4)² = 9/16. Amount of milk left = 60 × 9/16 = 33.75 L.

  3. Final Answer:

    33.75 L → Option A.

  4. Quick Check:

    (3/4)² = 0.5625; 0.5625 × 60 = 33.75 ✅

Hint: Use (1 - removed/total)^n × total when the same fraction is removed repeatedly and replaced by water.
Common Mistakes: Subtracting the removed volume twice instead of using the power rule.
3. An 80-litre vessel contains pure milk. 20 litres are removed and replaced by a solution containing 50% milk. Find the quantity of milk in the final mixture.
easy
A. 60 L
B. 65 L
C. 70 L
D. 75 L

Solution

  1. Step 1: After removal of 20 L

    Milk removed = 20 L (pure) → milk left = 80 - 20 = 60 L.

  2. Step 2: Add 20 L of 50% milk solution

    Milk added = 0.5 × 20 = 10 L. Final milk = 60 + 10 = 70 L.

  3. Final Answer:

    70 L → Option C.

  4. Quick Check:

    Net milk change = -20 + 10 = -10 → 80 - 10 = 70 ✅

Hint: Calculate milk removed, then add the milk portion of the replacement solution (volume × concentration).
Common Mistakes: Treating the replacement volume as fully milk instead of its actual concentration.
4. A 120-litre solution contains 75% milk. 30 litres are removed and replaced with pure milk. Find the final percentage of milk in the mixture.
medium
A. 85%
B. 87.5%
C. 80%
D. 81.25%

Solution

  1. Step 1: Compute initial milk

    Milk initially = 0.75 × 120 = 90 L.

  2. Step 2: Remove 30 L of the mixture

    Milk removed = 0.75 × 30 = 22.5 L → milk left = 90 - 22.5 = 67.5 L.

  3. Step 3: Add 30 L pure milk

    Milk added = 30 L → new milk total = 67.5 + 30 = 97.5 L. Total volume = 120 L.

  4. Step 4: Compute final %

    Final percentage = (97.5 ÷ 120) × 100 = 81.25%.

  5. Final Answer:

    81.25% → Option D.

  6. Quick Check:

    Work: 90 → remove 22.5 → 67.5 → add 30 → 97.5; 97.5/120 = 0.8125 ✅

Hint: When replacing with pure component, subtract removed portion (current fraction × removed volume) then add the pure amount.
Common Mistakes: Using percentage differences instead of computing actual litres removed and added.
5. A 100-litre vessel has 40% acid. 25 litres are removed and replaced by pure acid. Then again 25 litres are removed and replaced with water. Find the final percentage of acid.
medium
A. 41.25%
B. 45.25%
C. 50.25%
D. 55.25%

Solution

  1. Step 1: Initial acid amount

    Initial acid = 0.40 × 100 = 40 L.

  2. Step 2: First removal (25 L)

    Acid removed = 0.40 × 25 = 10 L → acid left = 40 - 10 = 30 L.

  3. Step 3: Add 25 L pure acid

    Acid added = 25 L → total acid = 30 + 25 = 55 L.

  4. Step 4: Second removal (25 L)

    Concentration before removal = 55/100 = 0.55. Acid removed = 0.55 × 25 = 13.75 L → acid left = 55 - 13.75 = 41.25 L.

  5. Step 5: Add 25 L water

    Total volume = 100 L; acid = 41.25 L → final % = (41.25/100) × 100 = 41.25%.

  6. Final Answer:

    41.25% → Option A.

  7. Quick Check:

    Track litres: 40 → 30 → +25 = 55 → remove 13.75 → 41.25 → final % = 41.25% ✅

Hint: Always compute actual litres removed/added at each step; convert to percentage only at the end.
Common Mistakes: Using percentages directly without converting to absolute amounts for removal/addition steps.

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