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Concentration / Percentage Mixture

Introduction

कई real-world mixture problems में liquids या substances किसी component (जैसे salt, alcohol, acid) का एक fixed concentration या percentage रखते हैं। यह pattern आपको बताता है कि desired concentration पाने के लिए कितना pure substance या solution जोड़ना या निकालना पड़ेगा।

Dilution, strengthening और concentration adjustment जैसे सवाल तेजी से हल करने के लिए यह pattern बहुत ज़रूरी है।

Pattern: Concentration / Percentage Mixture

Pattern

मुख्य विचार: Pure substance की मात्रा = (Total quantity × Concentration%) / 100

जब भी mixtures को मिलाया या बदला जाता है:
1. हमेशा pure substance की मात्रा को mixing से पहले और बाद में track करें।
2. Concentration (%) = (Pure substance ÷ Total mixture quantity) × 100.
3. Conservation rule: Pure before = Pure after.

Step-by-Step Example

Question

40-litre solution में 25% alcohol है। Solution को 40% alcohol बनाने के लिए कितना pure alcohol मिलाना पड़ेगा?

Solution

  1. Step 1: दिए गए डेटा को पहचानें

    Total solution = 40 L, Alcohol concentration = 25%, New concentration = 40%.

  2. Step 2: Pure alcohol की मौजूदा मात्रा

    Pure alcohol = 25% of 40 = (25/100) × 40 = 10 L.

  3. Step 3: मान लें x litres pure alcohol जोड़ा जा रहा है

    तो total alcohol = (10 + x) L और total solution = (40 + x) L।

  4. Step 4: Concentration formula लागू करें

    (10 + x)/(40 + x) × 100 = 40 → (10 + x) = 0.4(40 + x).

  5. Step 5: Equation simplify करें

    10 + x = 16 + 0.4x → 0.6x = 6 → x = 10 L.

  6. Final Answer:

    10 litres pure alcohol जोड़ना होगा।

  7. Quick Check:

    Total = 40 + 10 = 50 L → Alcohol = 10 + 10 = 20 L → (20/50)×100 = 40% ✅

Quick Variations

1. Pure substance के बजाय पानी जोड़ने से concentration घटती है।

2. Pure substance जोड़ने से concentration बढ़ती है।

3. अगर mixture का कुछ हिस्सा replace किया जा रहा हो, तो replacement formula या successive dilution rule उपयोग करें।

Trick to Always Use

  • Step 1 → लिखें “pure = total × percentage / 100”.
  • Step 2 → Conservation rule लागू करें (pure before = pure after)।
  • Step 3 → Equation बनाकर unknown quantity निकालें।

Summary

Summary

Concentration / Percentage Mixture pattern में:

  • Concentration (%) = (Pure quantity ÷ Total quantity) × 100.
  • Mixing या adding में pure substance की मात्रा बराबर रखनी होती है (as per situation)।
  • Pure substance जोड़ने से concentration बढ़ती है; solvent जोड़ने से घटती है।
  • Quick check: नई total quantity और pure quantity से concentration verify करें।

Practice

(1/5)
1. A 50-litre solution contains 20% sugar. How much pure sugar must be added to make the solution 40% sugar?
easy
A. 10.33 L
B. 16.67 L
C. 13.33 L
D. 15.67 L

Solution

  1. Step 1: Identify data

    Total = 50 L; current concentration = 20%; target = 40%.

  2. Step 2: Find current pure sugar

    Pure sugar = 20% of 50 = 10 L.

  3. Step 3: Let x L pure sugar be added

    New pure sugar = 10 + x; new total = 50 + x.

  4. Step 4: Set up equation

    (10 + x)/(50 + x) = 0.40 → 10 + x = 20 + 0.4x → 0.6x = 10 → x = 50/3 L = 16.67 L.

  5. Final Answer:

    Add 16.67 L (≈ 50/3 L) → Option B.

  6. Quick Check:

    Using exact value: (10 + 50/3) / (50 + 50/3) = (80/3) / (200/3) = 80/200 = 0.40 → 40% ✅
Hint: Set (pure before + added) ÷ (total before + added) = target fraction and solve for x.
Common Mistakes: Choosing the nearest option without checking exact fraction; forgetting to increase total volume.
2. A 60-litre acid solution contains 30% acid. How much pure water should be added to reduce acid concentration to 15%?
easy
A. 40 L
B. 60 L
C. 80 L
D. 90 L

Solution

  1. Step 1: Identify data

    Total = 60 L, Acid = 30%, Target = 15%.

  2. Step 2: Find pure acid

    (30/100)×60 = 18 L.

  3. Step 3: Let x L water be added

    Total = 60 + x; acid = 18 L.

  4. Step 4: Apply formula

    18/(60 + x) = 0.15 → 18 = 9 + 0.15x → 0.15x = 9 → x = 60 L.

  5. Final Answer:

    Add 60 L water → Option B.

  6. Quick Check:

    18/(60 + 60)=18/120=0.15=15% ✅
Hint: When diluting, pure amount is fixed - solve for total after dilution.
Common Mistakes: Reducing acid quantity when adding only water.
3. A 40-litre mixture contains 10% alcohol. How much pure alcohol should be added to make it 25% alcohol?
easy
A. 6 L
B. 8 L
C. 10 L
D. 12 L

Solution

  1. Step 1: Identify data

    Total = 40 L, Alcohol = 10%, Target = 25%.

  2. Step 2: Find current alcohol

    (10/100)×40 = 4 L.

  3. Step 3: Let x L pure alcohol be added

    Total = 40 + x; alcohol = 4 + x.

  4. Step 4: Apply formula

    (4 + x)/(40 + x) = 0.25 → 4 + x = 10 + 0.25x → 0.75x = 6 → x = 8 L.

  5. Final Answer:

    Add 8 L → Option B.

  6. Quick Check:

    (4 + 8)/(40 + 8)=12/48=25% ✅
Hint: Solve (pure before + x) ÷ (total + x) = target decimal.
Common Mistakes: Taking average instead of solving proportionally.
4. A 30-litre salt solution has 50% salt. How much of this must be replaced by pure water to make it 30% salt?
medium
A. 10 L
B. 12 L
C. 15 L
D. 18 L

Solution

  1. Step 1: Identify data

    Total = 30 L, Salt = 50%, Target = 30%.

  2. Step 2: Salt initially

    (50/100)×30 = 15 L.

  3. Step 3: Let x L be replaced with pure water

    Salt left = 15 - 0.5x; salt added = 0 (water has none).

  4. Step 4: Apply formula

    (15 - 0.5x)/30 = 0.30 → 15 - 0.5x = 9 → 0.5x = 6 → x = 12 L.

  5. Final Answer:

    Replace 12 L → Option B.

  6. Quick Check:

    (15 - 6)/30 = 9/30 = 30% ✅
Hint: Subtract removed pure part and divide remaining by total volume.
Common Mistakes: Forgetting to reduce pure substance after removal.
5. A 25-litre solution contains 40% acid. How much water should be added to reduce acid concentration to 25%?
medium
A. 10 L
B. 12 L
C. 15 L
D. 20 L

Solution

  1. Step 1: Identify data

    Total = 25 L, Acid = 40%, Target = 25%.

  2. Step 2: Find pure acid

    (40/100)×25 = 10 L.

  3. Step 3: Let x L water be added

    Total = 25 + x; acid = 10 L.

  4. Step 4: Apply formula

    10/(25 + x) = 0.25 → 10 = 6.25 + 0.25x → 0.25x = 3.75 → x = 15 L.

  5. Final Answer:

    Add 15 L water → Option C.

  6. Quick Check:

    10/(25 + 15)=10/40=25% ✅
Hint: Keep pure part same, increase total → use proportion to solve.
Common Mistakes: Using percent difference instead of fraction equation.

Mock Test

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