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Equalizing Strength / Ratio-Based Equations

Introduction

Equalizing Strength or Ratio-Based Equation problems involve mixing two or more solutions (or alloys) of different strengths or concentrations so that the final mixture achieves a desired uniform strength. These questions are important because they help you convert percentage or ratio-based word statements into clear algebraic equations that can be solved systematically.

You will frequently encounter such problems in mixture, alloy, and concentration questions - they test your ability to balance quantities and apply proportional reasoning correctly.

Pattern: Equalizing Strength / Ratio-Based Equations

Pattern

Key concept: Express the total amount of active substance from each component, divide by the total quantity to find overall concentration, and set it equal to the target strength or another condition.

Steps to follow:
1. Convert all percentages or ratios into decimal fractions.
2. Represent unknown quantities using variables.
3. Write a total composition (mass-balance) equation: (Sum of component parts) ÷ (Total quantity) = Target concentration
4. Simplify and solve for the unknown.

Step-by-Step Example

Question

Two solutions A and B contain 30% and 70% acid respectively. They are to be mixed so that the final mixture has a concentration of 50%. If 12 litres of solution A are used, find the quantity of solution B required.

Solution

  1. Step 1: Define variables

    Let the quantity of solution B be x litres.

  2. Step 2: Express total acid from both solutions

    Acid from A = 30% of 12 = 0.30 × 12 = 3.6 L.
    Acid from B = 70% of x = 0.70 × x = 0.7x L.

  3. Step 3: Write the equation for the target concentration

    Total acid / Total volume = Target concentration ⇒ (3.6 + 0.7x) / (12 + x) = 0.50

  4. Step 4: Simplify and solve

    3.6 + 0.7x = 0.50(12 + x) ⇒ 3.6 + 0.7x = 6 + 0.5x ⇒ 0.2x = 2.4 ⇒ x = 12 L.

  5. Step 5: Verify the result

    Total acid = 3.6 + 0.7×12 = 3.6 + 8.4 = 12 L. Total volume = 12 + 12 = 24 L. 12 ÷ 24 = 0.50 → 50% ✅

  6. Final Answer:

    Solution B required = 12 litres

  7. Quick Check:

    Both solutions equal in quantity give an exact 50% concentration, since it lies midway between 30% and 70% ✅

Quick Variations

1. Finding the quantity of one mixture when target concentration is given.

2. Adjusting a stronger or weaker solution to achieve a specific strength.

3. Expressing unknowns as ratios (e.g., A : B = x : y) and solving for x or y.

4. Handling multi-step equalization when more than two mixtures are involved.

Trick to Always Use

  • Step 1: Always convert % into decimal form before applying formulas.
  • Step 2: Write one clear equation: (Sum of acid from each) ÷ (Total volume) = Target %.
  • Step 3: Cross multiply to remove denominators, then solve for the unknown quantity.
  • Step 4: Always verify the result - final % should lie between given % values.

Summary

Summary

The Equalizing Strength / Ratio-Based Equations pattern is used to make concentrations or strengths equal through algebraic balance.

  • Convert all strengths to fractional or decimal form.
  • Use a single formula: (Sum of parts) ÷ (Total mixture) = Target concentration.
  • Cross-multiply and simplify to find unknown quantities.
  • Check your answer - the final strength must lie between the original strengths.

Practice

(1/5)
1. Two acid solutions contain 30% and 60% acid respectively. In what ratio should they be mixed to obtain a 45% acid solution?
easy
A. 1 : 1
B. 2 : 1
C. 1 : 2
D. 3 : 2

Solution

  1. Step 1: Note concentrations

    Lower = 30%, Higher = 60%, Target = 45%.

  2. Step 2: Apply alligation

    (Higher - Target) : (Target - Lower) = (60 - 45) : (45 - 30) = 15 : 15 = 1 : 1.

  3. Final Answer:

    Mix in ratio 1 : 1 → Option A.

  4. Quick Check:

    Equal parts of 30% and 60% give (30 + 60)/2 = 45% ✅

Hint: Use (higher - target) : (target - lower) to get the ratio directly.
Common Mistakes: Swapping the two differences or forgetting to simplify the ratio.
2. A 20% salt solution is mixed with a 40% salt solution to get a 25% solution. Find the ratio of the two solutions.
easy
A. 1 : 3
B. 3 : 1
C. 2 : 3
D. 1 : 2

Solution

  1. Step 1: Note concentrations

    Lower = 20%, Higher = 40%, Target = 25%.

  2. Step 2: Apply alligation

    (Higher - Target) : (Target - Lower) = (40 - 25) : (25 - 20) = 15 : 5 = 3 : 1.

  3. Final Answer:

    Ratio (20% : 40%) = 3 : 1 → Option B.

  4. Quick Check:

    (3×20 + 1×40)/4 = (60 + 40)/4 = 100/4 = 25% ✅ - more of the 20% solution is needed because the target (25%) is closer to 20%.

Hint: Alligation gives parts of lower : higher as (higher - target) : (target - lower).
Common Mistakes: Reversing which difference corresponds to which component.
3. How many litres of 60% sugar solution must be mixed with 40 litres of 20% sugar solution to get a 40% sugar solution?
easy
A. 20 L
B. 30 L
C. 40 L
D. 50 L

Solution

  1. Step 1: Define variable

    Let x = litres of 60% solution.

  2. Step 2: Write concentration equation

    (0.6x + 0.2×40) / (x + 40) = 0.4

  3. Step 3: Simplify and solve

    0.6x + 8 = 0.4x + 16 → 0.2x = 8 → x = 40 L.

  4. Final Answer:

    40 L → Option C.

  5. Quick Check:

    (0.6×40 + 0.2×40)/(80) = (24 + 8)/80 = 32/80 = 0.4 = 40% ✅

Hint: Use (C1V1 + C2V2) / (V1 + V2) = Target to form a linear equation.
Common Mistakes: Forgetting to multiply concentrations by their volumes before dividing by total volume.
4. A 15% acid solution is mixed with a 60% acid solution so that the resulting mixture is 30% acid. If 12 litres of the 15% solution are used, find how much of the 60% solution is required.
medium
A. 3 L
B. 4 L
C. 5 L
D. 6 L

Solution

  1. Step 1: Let x = litres of 60% solution

  2. Step 2: Form equation

    (0.15×12 + 0.60×x) / (12 + x) = 0.30

  3. Step 3: Solve

    1.8 + 0.6x = 0.30(12 + x) → 1.8 + 0.6x = 3.6 + 0.3x → 0.3x = 1.8 → x = 6 L.

  4. Final Answer:

    6 L → Option D.

  5. Quick Check:

    Total acid = 1.8 + 3.6 = 5.4 L; total volume = 18 L; 5.4/18 = 0.30 → 30% ✅

Hint: Multiply each concentration by its volume, add, then divide by total volume and set equal to target.
Common Mistakes: Not multiplying the target by the total volume when rearranging the equation.
5. In what ratio must a 25% sugar solution be mixed with a 75% sugar solution to get a 60% solution?
medium
A. 3 : 7
B. 7 : 3
C. 1 : 3
D. 2 : 3

Solution

  1. Step 1: Apply alligation

    (Higher - Target) : (Target - Lower) = (75 - 60) : (60 - 25) = 15 : 35 = 3 : 7.

  2. Step 2: Interpret

    Parts correspond to (25% : 75%) = 3 : 7.

    3. Final Answer:

    3 : 7 → Option A.

  3. Quick Check:

    (25×3 + 75×7)/(3+7) = (75 + 525)/10 = 600/10 = 60% ✅

Hint: Alligation gives lower : higher as (higher - target) : (target - lower).
Common Mistakes: Reversing the parts or failing to simplify the ratio to smallest integers.

Mock Test

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