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Chain Mixture Problems

Introduction

Chain Mixture Problems involve sequential mixing or replacement steps where each step changes the composition for the next step. They appear when several operations are performed one after another (e.g., mix A with B, then mix result with C, or successive replacements).

This pattern is important because it trains you to track absolute quantities through multiple stages, use retained-fraction formulas when appropriate, and combine algebraic reasoning with stepwise bookkeeping.

Pattern: Chain Mixture Problems

Pattern

Key concept: Treat each step as a separate operation - compute absolute quantities (litres/weights) of the component before and after each step, or use retained-fraction formulas for repeated identical removals.

Two common approaches:
1. Stepwise bookkeeping: For each operation, compute component removed/added in absolute terms, update totals, then move to the next step.
2. Retained-fraction formula: When the same fraction is removed repeatedly and replaced with the same component (e.g., remove x L from V L, replace with water), use: Remaining component = Initial amount × (1 - removed/total)ⁿ where n = number of identical operations.

Step-by-Step Example

Question

A vessel initially contains 100 L of a 30% salt solution. First, 20 L of this mixture is removed and replaced with pure water. Then 30 L of the new mixture is removed and replaced with a 50% salt solution. Find the final concentration of salt.

Solution

  1. Step 1: Compute initial salt amount

    Initial salt = 30% of 100 L = 30 L.

  2. Step 2: First operation - remove 20 L and replace with water

    Salt removed = 30% of 20 = 0.30 × 20 = 6 L. Salt left = 30 - 6 = 24 L. Total volume after replacement = 100 L (salt = 24 L).

  3. Step 3: Compute salt fraction after first operation

    Salt fraction = 24/100 = 0.24 (24%).

  4. Step 4: Second operation - remove 30 L of the current mixture

    Salt removed = 0.24 × 30 = 7.2 L. Salt left after removal = 24 - 7.2 = 16.8 L. Volume after removal = 70 L.

  5. Step 5: Replace 30 L with 50% salt solution

    Salt added = 0.50 × 30 = 15 L. Final salt amount = 16.8 + 15 = 31.8 L. Final total volume = 100 L.

  6. Step 6: Compute final concentration

    Final concentration = (31.8 ÷ 100) × 100 = 31.8%.

  7. Final Answer:

    31.8% salt concentration.

  8. Quick Check:

    Track litres: salt 30 → -6 = 24 → -7.2 = 16.8 → +15 = 31.8; volume stays 100 L → 31.8% ✅

Quick Variations

1. Repeated identical removal & replacement (use retained-fraction formula).

2. Chain of different concentrated additions (use stepwise bookkeeping).

3. Mixing several solutions sequentially - treat each mix as a stage and record component totals.

4. Problems asking for original amount given final concentration - work backwards by reversing steps algebraically.

Trick to Always Use

  • Step 1: Work in absolute units (litres or kg) not percentages for intermediate steps.
  • Step 2: For identical repeated removals, use Remaining = Initial × (1 - r)ⁿ where r = removed/total.
  • Step 3: When replacing with a new concentrated solution, compute the actual amount of component added (replacement volume × concentration) and add to the remaining amount.
  • Step 4: For reverse problems, substitute variables and unwind steps algebraically (invert each forward step carefully).

Summary

Summary

Chain Mixture Problems require careful sequential tracking:

  • Prefer absolute amounts over percentages while computing intermediate steps.
  • Use the retained-fraction formula for repeated identical removals: Remaining = Initial × (1 - removed/total)ⁿ.
  • When replacement is with a concentrated solution, add the exact amount of component introduced (replacement volume × concentration).
  • For backward/unknown-original problems, set variables and reverse each stage algebraically, checking that final concentration lies within feasible bounds.

Practice

(1/5)
1. A vessel contains 100 L of 40% salt solution. 20 L of the mixture is removed and replaced with water. What is the final concentration of salt?
easy
A. 32%
B. 36%
C. 30%
D. 34%

Solution

  1. Step 1: Initial salt amount

    Salt = 40% of 100 = 40 L.

  2. Step 2: Salt removed with 20 L

    Salt removed = 0.40 × 20 = 8 L. Salt left = 40 - 8 = 32 L.

  3. Step 3: Replace with water (total volume returns to 100 L)

    Final salt = 32 L in 100 L → 32%.

  4. Final Answer:

    32% → Option A.

  5. Quick Check:

    40 - 8 = 32; 32/100 = 0.32 → 32% ✅

Hint: Remaining = Initial × (1 - replaced/total) → 40 × (80/100) = 32.
Common Mistakes: Forgetting that total volume stays constant after replacement.
2. A vessel contains 80 L of 25% alcohol solution. 20 L are removed and replaced with pure alcohol. What is the final concentration of alcohol?
easy
A. 35%
B. 37.5%
C. 40%
D. 43.75%

Solution

  1. Step 1: Initial alcohol amount

    Alcohol = 0.25 × 80 = 20 L.

  2. Step 2: Remove 20 L of mixture

    Alcohol removed = 0.25 × 20 = 5 L. Remaining alcohol = 20 - 5 = 15 L.

  3. Step 3: Add 20 L pure alcohol

    Alcohol added = 20 L → final alcohol = 15 + 20 = 35 L.

  4. Step 4: Final concentration

    35/80 = 0.4375 → 43.75%.

  5. Final Answer:

    43.75% → Option D.

  6. Quick Check:

    20 - 5 + 20 = 35; 35/80 = 0.4375 → 43.75% ✅

Hint: Subtract removed component (current%×removed) then add the pure amount.
Common Mistakes: Using the original percentage on the changed total without updating litres.
3. A vessel has 100 L of 20% sugar solution. 25 L are removed and replaced with 40% sugar solution. What is the new concentration?
easy
A. 24%
B. 25%
C. 26%
D. 28%

Solution

  1. Step 1: Initial sugar amount

    Sugar = 0.20 × 100 = 20 L.

  2. Step 2: Remove 25 L of mixture

    Sugar removed = 0.20 × 25 = 5 L. Sugar left = 20 - 5 = 15 L.

  3. Step 3: Add 25 L of 40% solution

    Sugar added = 0.40 × 25 = 10 L. Final sugar = 15 + 10 = 25 L.

  4. Step 4: Final concentration

    25/100 = 25%.

  5. Final Answer:

    25% → Option B.

  6. Quick Check:

    20 - 5 + 10 = 25; 25/100 = 0.25 → 25% ✅

Hint: Track absolute sugar litres through each step, then divide by total volume.
Common Mistakes: Averaging percentages instead of doing mass balance.
4. A container has 100 L of 30% acid. 20 L are replaced by pure acid, then 20 L more are replaced with water. What is the final concentration of acid?
medium
A. 35.2%
B. 36.2%
C. 32.2%
D. 28.2%

Solution

  1. Step 1: Initial acid

    Acid = 0.30 × 100 = 30 L.

  2. Step 2: First replacement (20 L pure acid)

    Acid removed = 0.30 × 20 = 6 L. Acid left = 30 - 6 = 24 L. Add 20 L pure acid → acid = 24 + 20 = 44 L.

  3. Step 3: Second replacement (remove 20 L, add water)

    Concentration before removal = 44/100 = 0.44. Acid removed = 0.44 × 20 = 8.8 L. Acid left = 44 - 8.8 = 35.2 L. After adding water total = 100 L.

  4. Step 4: Final concentration

    35.2/100 = 35.2%.

  5. Final Answer:

    35.2% → Option A.

  6. Quick Check:

    30 - 6 + 20 - 8.8 = 35.2; 35.2/100 = 0.352 → 35.2% ✅

Hint: After each step recompute absolute litres of component before moving on.
Common Mistakes: Using original percentage for the second removal instead of updated concentration.
5. A vessel has 120 L of 25% salt solution. 40 L are replaced with 60% salt solution and then 20 L are replaced with water. Find the final salt concentration (rounded to two decimals).
medium
A. 35.36%
B. 38.26%
C. 30.56%
D. 32.76%

Solution

  1. Step 1: Initial salt

    Salt = 0.25 × 120 = 30 L.

  2. Step 2: Replace 40 L with 60% solution

    Salt removed = 0.25 × 40 = 10 L. Remaining = 30 - 10 = 20 L. Salt added = 0.60 × 40 = 24 L. New salt = 20 + 24 = 44 L.

  3. Step 3: Replace 20 L with water

    Salt removed = (44/120) × 20 = 7.333333... L. Salt left = 44 - 7.333333... = 36.6666667 L.

  4. Step 4: Final concentration

    36.6666667 / 120 = 0.30555556 → 30.56% (rounded to two decimals).

  5. Final Answer:

    30.56% → Option C.

  6. Quick Check:

    30 - 10 + 24 - 7.3333 = 36.6667; 36.6667/120 ≈ 0.3056 → 30.56% ✅

Hint: Handle each replacement step-by-step using absolute litres; round final percent at the end.
Common Mistakes: Rounding intermediate values too early instead of at final step.

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