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Replacement or Repeated Dilution

Introduction

In some mixture problems, a part of the solution is removed and replaced repeatedly - either with pure liquid or with another mixture. This causes gradual change in concentration after every replacement.

The Replacement or Repeated Dilution pattern helps calculate the final concentration or quantity after multiple cycles of removal and replacement without solving step by step each time.

Pattern: Replacement or Repeated Dilution

Pattern

Key idea: After each replacement, the fraction of the original liquid that remains is (1 - x / total), where x is the amount replaced.

For n replacements, the quantity of the original liquid remaining is:
Remaining = Initial × (1 - x / Total)n

From this, you can find the final concentration of the liquid or the amount of the other liquid added.

Step-by-Step Example

Question

A vessel contains 80 L of milk. 20 L of milk is taken out and replaced with water. The process is repeated twice more. Find the amount of milk left in the vessel after 3 such operations.

Solution

  1. Step 1: Identify given data

    Total = 80 L; Replaced each time = 20 L; Replacements = 3.

  2. Step 2: Apply dilution formula

    Fraction remaining after each replacement = (1 - 20/80) = (3/4).
    After 3 replacements, remaining fraction = (3/4)3 = 27/64.

  3. Step 3: Calculate final amount

    Milk left = 80 × (27/64) = 33.75 L.

  4. Final Answer:

    33.75 L of milk remains in the vessel.

  5. Quick Check:

    Each cycle removes ¼ of milk → remaining after 3 cycles ≈ (¾)³ = 0.422 → 42.2% of 80 = 33.76 ✅

Quick Variations

1. Replacement with water → concentration of original substance decreases.

2. Replacement with same concentration → no change in ratio.

3. Replacement with another solution → use weighted ratio or stepwise composition.

4. The same logic applies to solid mixtures or alloy replacement problems.

Trick to Always Use

  • Step 1: Compute (1 - replaced / total) → fraction left after one replacement.
  • Step 2: Raise that fraction to the power of n for n operations.
  • Step 3: Multiply by the initial quantity to get the remaining amount.
  • Step 4: Subtract from total if you need the amount replaced by another liquid.

Summary

Summary

In the Replacement or Repeated Dilution pattern:

  • The key formula is Remaining = Initial × (1 - x / Total)n.
  • Each operation removes a fixed fraction of the remaining liquid.
  • The concentration decreases exponentially with each replacement.
  • Quick check: The smaller the replaced fraction, the slower the dilution.

Practice

(1/5)
1. A 60-litre vessel is full of milk. 10 litres of milk are removed and replaced with water. The process is repeated once more. Find the amount of milk left after 2 operations.
easy
A. 41.67 L
B. 45 L
C. 42.25 L
D. 48 L

Solution

  1. Step 1: Note the data

    Total = 60 L; replaced each time = 10 L; repetitions = 2.

  2. Step 2: Use the dilution formula

    Fraction remaining after one replacement = 1 - (10/60) = 5/6. After 2 ops: (5/6)^2.

  3. Step 3: Calculate amount left

    Milk left = 60 × (5/6)^2 = 60 × 25/36 = 1500/36 = 41.666... L ≈ 41.67 L.

  4. Final Answer:

    41.67 L → Option A.

  5. Quick Check:

    After 1st op milk = 50 L; after 2nd op remove 10 L of mixture (milk fraction 50/60 = 5/6) → milk removed ≈ 8.333 → left ≈ 41.667 L ✅

Hint: Use Remaining = Total × (1 - replaced/total)^n.
Common Mistakes: Subtracting replaced volume directly each time instead of using fractional removal.
2. A 100-litre tank of acid solution has 10 litres replaced by water each time. The operation is performed 3 times. Find the fraction of the original acid left.
easy
A. 0.70
B. 0.729
C. 0.80
D. 0.81

Solution

  1. Step 1: Fraction retained per operation

    Per op retained fraction = 1 - (10/100) = 0.9.

  2. Step 2: Apply power for 3 operations

    Fraction left = (0.9)^3 = 0.729.

  3. Final Answer:

    0.729 → Option B.

  4. Quick Check:

    0.9×0.9×0.9 = 0.729 ✅

Hint: Raise the retained fraction to the power n: (1 - x/Total)^n.
Common Mistakes: Multiplying linearly instead of using exponential power.
3. A vessel contains 80 L of milk. 20 L of milk is removed and replaced with water. The process is repeated twice. How much milk remains after the two operations?
easy
A. 33.75 L
B. 36 L
C. 45 L
D. 30 L

Solution

  1. Step 1: Find retained fraction per op

    Retained fraction = 1 - (20/80) = 3/4.

  2. Step 2: After 2 operations

    Fraction left = (3/4)^2 = 9/16.

  3. Step 3: Amount of milk left

    Milk left = 80 × 9/16 = 80 × 0.5625 = 45 L.

  4. Final Answer:

    45 L → Option C.

  5. Quick Check:

    After 1st op milk = 60 L; after 2nd op remove 20 L of mixture (milk fraction 60/80 = 3/4) → milk removed = 15 → left = 45 L ✅

Hint: Use (1 - replaced/total)^n then multiply by initial volume.
Common Mistakes: Treating each operation as removing the same absolute amount of the pure substance.
4. A 40-litre vessel contains pure alcohol. 8 litres are taken out and replaced with water. The operation is repeated twice. Find the amount of alcohol left.
medium
A. 25.60 L
B. 23.04 L
C. 24.00 L
D. 20.50 L

Solution

  1. Step 1: Compute retained fraction per op

    Retained fraction = 1 - (8/40) = 0.80.

  2. Step 2: Apply for 2 operations

    Fraction left = (0.80)^2 = 0.64.

  3. Step 3: Amount left

    Alcohol left = 40 × 0.64 = 25.6 L.

  4. Final Answer:

    25.60 L → Option A.

  5. Quick Check:

    After 1st op alcohol = 32 L; after 2nd op remove 8 L of mixture (alcohol fraction 32/40 = 0.8) → removed alcohol = 6.4 → left = 25.6 L ✅

Hint: Multiply the initial amount by (1 - x/Total)^n for n replacements.
Common Mistakes: Rounding too early - keep decimals until final step.
5. A 90-litre mixture contains 80% alcohol. 30 litres of the mixture are removed and replaced with water. Find the new concentration of alcohol (to two decimal places).
medium
A. 50.00%
B. 60.00%
C. 70.00%
D. 53.33%

Solution

  1. Step 1: Compute initial alcohol (litres)

    Initial alcohol = 0.80 × 90 = 72 L.

  2. Step 2: Fraction retained after removal

    Removed fraction = 30/90 = 1/3 → retained fraction = 2/3.

    3. Step 3: Alcohol left in litres

    Alcohol left = 72 × (2/3) = 48 L.

  3. Step 4: New concentration

    New % = (48 ÷ 90) × 100 = 53.333...% → 53.33%.

  4. Final Answer:

    53.33% → Option D.

  5. Quick Check:

    After removal, alcohol = 48 L; total remains 90 L → 48/90 = 0.5333 → 53.33% ✅

Hint: Alcohol left = initial_alcohol × (1 - removed/total); divide by total for %.
Common Mistakes: Forgetting that replacement reduces the pure component proportionally to the removed fraction.

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