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Different Quantity Mixtures

Introduction

कई mixture problems में दोनों components की quantities बराबर नहीं होतीं। कई बार एक quantity fixed होती है और दूसरी quantity को desired concentration या value पाने के लिए निकालना पड़ता है।

यह pattern unequal quantities को संभालने में मदद करता है-हर component के “pure” हिस्से पर ध्यान देकर simple equations बनती हैं जो पूरे mixture को represent करती हैं।

Pattern: Different Quantity Mixtures

Pattern

मुख्य विचार: प्रत्येक component की quantity को उसकी concentration से multiply करें ताकि उसका pure part मिले। Quantities अलग हों तो weighted-average या equation बनाकर unknown quantity निकालें।

Steps:
1. किसी भी known mixture का pure part = quantity × (percentage ÷ 100).
2. अगर एक quantity unknown हो, तो उसे x मानें और उसका pure part = x × (percentage ÷ 100).
3. Equation बनाएँ: (Sum of pure parts) ÷ (Total quantity) = desired concentration (fraction में).
4. x निकालें और weighted average से verify करें।

Step-by-Step Example

Question

आपके पास 20% salt solution के 30 L हैं। Final mixture को 30% बनाने के लिए आपको 50% salt solution के कितने litres जोड़ने होंगे?

Solution

  1. Step 1: Given data लिखें

    Known solution = 30 L of 20% salt → pure salt = 30 × 0.20 = 6 L.

    x = litres of 50% solution → pure salt = x × 0.50 = 0.5x L.

    Final concentration चाहिए = 30%.

  2. Step 2: Total equation बनाएँ

    Total pure salt = 6 + 0.5x
    Total volume = 30 + x
    Equation: (6 + 0.5x) ÷ (30 + x) = 0.30

  3. Step 3: Equation solve करें

    6 + 0.5x = 0.30(30 + x)
    → 6 + 0.5x = 9 + 0.3x
    → 0.5x - 0.3x = 9 - 6
    → 0.2x = 3
    → x = 15 L.

  4. Step 4: Final result

    आपको 15 litres of 50% salt solution जोड़ना होगा ताकि mixture 30% बन जाए।

  5. Quick Check:

    Pure salt = 6 + (0.5×15) = 13.5 L
    Total volume = 30 + 15 = 45 L
    Percentage = (13.5 ÷ 45) × 100 = 30% ✅

Quick Variations

1. एक quantity fixed और दूसरी unknown हो - equation बनाकर हल करें।

2. दोनों quantities known हों तो weighted-average formula सीधे लागू करें।

3. Replacement या dilution वाले questions में भी total और pure amounts को update करके यही तरीका चलता है।

4. Percentages के बदले अगर cost हो, तो “concentration” की जगह “price” रखकर वही logic follow करें।

Trick to Always Use

  • Step 1: हर percentage को decimal में बदलें (जैसे 20% → 0.20)।
  • Step 2: Pure part = quantity × fraction निकालें।
  • Step 3: Equation बनाएँ: (sum of pure parts) ÷ (total quantity) = desired fraction।
  • Step 4: Unknown quantity निकालें और final percentage दोबारा check करें।

Summary

Summary

Different Quantity Mixtures pattern में:

  • मात्रा को उसके pure हिस्से में बदलें: percentage ÷ 100।
  • Total pure part और total volume balance करने के लिए equation बनाएं।
  • Unknown quantity algebra से निकालें।
  • Weighted-average या percentage check से answer verify करें।

Practice

(1/5)
1. A 50-litre solution contains 10% sugar. How many litres of a 30% sugar solution must be added to make the resulting solution 16% sugar?
easy
A. 21.43 L
B. 20.57 L
C. 22.33 L
D. 25.68 L

Solution

  1. Step 1: Write down knowns

    Initial volume = 50 L; initial sugar = 10% → pure sugar = 0.10 × 50 = 5 L. Let x = litres of 30% solution added → pure sugar added = 0.30x.

  2. Step 2: Form the concentration equation

    Total sugar after mixing = 5 + 0.30x. Total volume = 50 + x. Set (5 + 0.30x)/(50 + x) = 0.16.

  3. Step 3: Solve for x

    5 + 0.30x = 0.16(50 + x) → 5 + 0.30x = 8 + 0.16x → 0.14x = 3 → x = 3 / 0.14 = 21.42857… L ≈ 21.43 L.

  4. Final Answer:

    Approximately 21.43 L → Option A.

  5. Quick Check:

    Pure sugar ≈ 5 + 0.30×21.43 = 11.429 L. Total ≈ 71.43 L. 11.429/71.43 ≈ 0.16 → 16% ✅

Hint: Set (pure_before + pure_added) ÷ (total_before + added) = target fraction and solve for x.
Common Mistakes: Forgetting that total volume increases after adding the second solution.
2. A 20-litre acid solution contains 25% acid. How much pure water must be added to reduce the acid concentration to 10%?
easy
A. 30 L
B. 25 L
C. 20 L
D. 15 L

Solution

  1. Step 1: Compute current pure acid

    Pure acid = 0.25 × 20 = 5 L. Let x = litres of water added. After adding water pure acid remains 5 L; total volume = 20 + x.

  2. Step 2: Form equation for desired concentration

    We want 5/(20 + x) = 0.10 → 5 = 0.10(20 + x).

  3. Step 3: Solve for x

    5 = 2 + 0.10x → 0.10x = 3 → x = 30 L.

  4. Final Answer:

    Add 30 L water → Option A.

  5. Quick Check:

    5/(20 + 30) = 5/50 = 0.10 → 10% ✅

Hint: When diluting, the pure amount stays fixed; solve for the new total volume needed for the target %.
Common Mistakes: Altering the pure substance amount when only water is added.
3. You mix 30 L of a 40% solution with 50 L of a 20% solution. What is the concentration of the final mixture?
easy
A. 26.5%
B. 27.5%
C. 28.5%
D. 30.5%

Solution

  1. Step 1: Compute pure parts

    Pure in first = 0.40 × 30 = 12 L. Pure in second = 0.20 × 50 = 10 L.

  2. Step 2: Add pure parts and volumes

    Total pure = 12 + 10 = 22 L. Total volume = 30 + 50 = 80 L.

  3. Step 3: Compute concentration

    Concentration = (22 ÷ 80) × 100 = 0.275 × 100 = 27.5%.

  4. Final Answer:

    27.5% → Option B.

  5. Quick Check:

    22/80 = 0.275 → 27.5% ✅

Hint: Always convert to actual pure amounts first (quantity × fraction), then combine.
Common Mistakes: Averaging percentages without weighting by quantities.
4. A 40-litre solution of 15% alcohol is to be strengthened to 30% by adding some 60% alcohol. How many litres of 60% alcohol are needed?
medium
A. 16 L
B. 18 L
C. 20 L
D. 24 L

Solution

  1. Step 1: Write known pure amounts

    Pure alcohol initially = 0.15 × 40 = 6 L. Let x = litres of 60% alcohol added → pure added = 0.60x.

  2. Step 2: Form the target equation

    Total pure after = 6 + 0.60x; total volume = 40 + x. Desired concentration: (6 + 0.60x)/(40 + x) = 0.30.

  3. Step 3: Solve

    6 + 0.60x = 0.30(40 + x) → 6 + 0.60x = 12 + 0.30x → 0.30x = 6 → x = 20 L.

  4. Final Answer:

    Add 20 L of 60% alcohol → Option C.

  5. Quick Check:

    Pure after = 6 + 0.6×20 = 6 + 12 = 18 L. Total = 60 L. 18/60 = 0.30 → 30% ✅

Hint: Keep the unknown symbolic, form one clean equation: (sum of pure parts)/(sum of volumes) = target fraction.
Common Mistakes: Forgetting to include the added volume in the denominator.
5. A 90-litre solution contains 25% salt. Some quantity of the solution is removed and the same amount of pure salt is added. If the final concentration becomes 30%, how many litres were removed and replaced?
medium
A. 5 L
B. 7 L
C. 4 L
D. 6 L

Solution

  1. Step 1: Compute initial pure salt

    Initial pure salt = 0.25 × 90 = 22.5 L. Let x = litres removed and replaced by pure salt.

  2. Step 2: Account for removal and replacement

    Salt removed with x litres = 0.25x. Salt left after removal = 22.5 - 0.25x. After adding x litres of pure salt, new pure salt = 22.5 - 0.25x + x = 22.5 + 0.75x. Total volume remains 90 L.

  3. Step 3: Form equation for final concentration

    (22.5 + 0.75x)/90 = 0.30 → 22.5 + 0.75x = 27 → 0.75x = 4.5 → x = 6 L.

  4. Final Answer:

    Replace 6 litres → Option D.

  5. Quick Check:

    New pure salt = 22.5 + 0.75×6 = 27 L → 27/90 = 0.30 → 30% ✅

Hint: When replacing with pure substance, add x to pure and subtract the removed fraction (original% × x) first.
Common Mistakes: Forgetting that removed portion carries away some pure substance.

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