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Chain Mixture Problems

Introduction

Chain Mixture Problems में sequential mixing या replacement steps होते हैं, जहाँ हर step अगली step के लिए composition बदल देता है। ये तब आते हैं जब कई operations एक के बाद एक किए जाते हैं (जैसे पहले mix A with B, फिर result को C के साथ mix करना, या successive replacements करना)।

यह pattern इसलिए ज़रूरी है क्योंकि यह आपको multiple stages में absolute quantities को track करना सिखाता है, repeated operations में retained-fraction formulas का सही उपयोग करवाता है, और algebraic reasoning को stepwise bookkeeping के साथ combine करता है।

Pattern: Chain Mixture Problems

Pattern

मुख्य concept: हर step को एक अलग operation की तरह treat करें - हर बार component की absolute मात्रा (litres/weights) को step से पहले और step के बाद compute करें, या repeated identical removals के cases में retained-fraction formulas का इस्तेमाल करें।

दो common approaches:
1. Stepwise bookkeeping: हर operation में remove/add की गई component की absolute मात्रा निकालें, totals update करें, और फिर अगले step पर जाएँ।
2. Retained-fraction formula: जब एक ही fraction बार-बार remove की जाती है और उसी component से replace किया जाता है (जैसे remove x L from V L, replace with water), तब formula उपयोग करें: Remaining component = Initial amount × (1 - removed/total)ⁿ जहाँ n = identical operations की संख्या।

Step-by-Step Example

Question

एक vessel में शुरू में 100 L का 30% salt solution है। पहले 20 L mixture हटाकर उतनी ही मात्रा में pure water डाल दिया जाता है। फिर नई mixture में से 30 L हटाकर 50% salt solution डाला जाता है। Final salt concentration निकालें।

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. Replacement के बाद total volume = 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. Removal के बाद volume = 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 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:

    Litres tracking: salt 30 → -6 = 24 → -7.2 = 16.8 → +15 = 31.8; volume 100 L रहता है → 31.8% ✅

Quick Variations

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

2. अलग-अलग concentrations को chain में add करना (stepwise bookkeeping)।

3. कई solutions को sequentially mix करना - हर stage को अलग step मानकर component totals record करें।

4. Final concentration दी हो और original amount चाहिए - steps को reverse करके algebraically solve करें।

Trick to Always Use

  • Step 1: Intermediate steps में percentages की बजाय absolute units (litres या kg) में work करें।
  • Step 2: Identical repeated removals में formula उपयोग करें: Remaining = Initial × (1 - r)ⁿ जहाँ r = removed/total।
  • Step 3: जब replacement किसी concentrated solution से हो, तो component की actual added मात्रा निकालें (replacement volume × concentration) और उसे remaining amount में जोड़ें।
  • Step 4: Reverse problems में variables लेकर हर forward step को उलट कर algebraically solve करें।

Summary

Summary

Chain Mixture Problems में sequential steps को ध्यान से track करना ज़रूरी है:

  • Intermediate steps में हमेशा absolute amounts का उपयोग करें।
  • Repeated removals के लिए retained-fraction formula: Remaining = Initial × (1 - removed/total)ⁿ.
  • Concentrated replacement में added component (replacement volume × concentration) को सही से जोड़ें।
  • Backward problems में steps को reverse कर algebraically solve करें और देखें कि final concentration feasible range में हो।

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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