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Different Quantity Mixtures

Introduction

In many mixture problems, the two components are not in equal quantities. Sometimes one quantity is fixed, and the other needs to be found to achieve a certain concentration or value.

This pattern helps you deal with unequal quantities by focusing on the amount of the “pure” substance in each component and forming simple equations that represent the total mixture.

Pattern: Different Quantity Mixtures

Pattern

Key concept: Multiply each component’s quantity by its concentration to get the pure part. When quantities differ, use a weighted-average or equation to find the required unknown quantity.

Steps to follow:
1. Calculate the pure part for any known mixture = quantity × (percentage ÷ 100).
2. If one quantity is unknown, let it be x and express its pure part as x × (percentage ÷ 100).
3. Form the equation: (Sum of pure parts) ÷ (Total quantity) = desired concentration (as a fraction).
4. Solve for x and verify by checking the weighted average.

Step-by-Step Example

Question

You have 30 L of a 20% salt solution. How many litres of a 50% salt solution must be added so that the final mixture is 30% salt?

Solution

  1. Step 1: Write down what is given

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

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

    We need a final mixture of 30% salt.

  2. Step 2: Form the total equation

    Total pure salt after mixing = 6 + 0.5x. Total mixture volume = 30 + x. So, (6 + 0.5x) ÷ (30 + x) = 0.30.

  3. Step 3: Solve the equation

    (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: Write the result clearly

    You must add 15 litres of the 50% salt solution to make the mixture 30% salt.

  5. Quick Check:

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

Quick Variations

1. One quantity is fixed, and the other is unknown - form an equation as shown above.

2. If both quantities are known, use the weighted-average formula directly.

3. Replacement or dilution problems can also use this approach by updating the total and pure amounts.

4. When costs are involved instead of percentages, replace “concentration” with “price” and follow the same logic.

Trick to Always Use

  • Step 1: Convert each percentage to a decimal (e.g., 20% → 0.20).
  • Step 2: Find the pure part = quantity × fraction.
  • Step 3: Form one equation: (sum of pure parts) ÷ (total quantity) = desired fraction.
  • Step 4: Solve for the unknown and double-check by recalculating the total concentration.

Summary

Summary

In the Different Quantity Mixtures pattern:

  • Convert each quantity to its pure part using percentage ÷ 100.
  • Form an equation balancing total pure part and total mixture volume.
  • Use algebra to find the unknown quantity.
  • Verify your answer with a weighted-sum or percentage check.

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.

Mock Test

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