Mixture and Alligation — SSC CGL Study Notes

Overview

Mixture and Alligation is a high-yield topic in SSC CGL Quantitative Aptitude, typically contributing 1–2 questions per paper. The topic deals with combining two or more components in specific ratios and understanding how their individual properties (price, concentration, strength) blend to create a mixture with averaged characteristics.

Mastery of this topic is essential because questions appear in both direct ratio-calculation form and as word problems involving replacement, repeated mixing, or finding optimal combinations. The alligation rule provides a shortcut method that eliminates lengthy algebraic manipulation, making it invaluable under time pressure. Students must develop comfort with the visual cross-method diagram and recognize when to apply the weighted average concept versus the alligation shortcut.

The core competency tested is your ability to set up ratio relationships correctly, handle percentage-to-ratio conversions, and work backward from a mixture's properties to find constituent quantities or their proportions.

Key Concepts

• **Mixture**: A combination of two or more components (liquids, metals, ingredients) mixed together. The resulting mixture has properties that are the weighted average of its components.

• **Alligation**: A method to determine the ratio in which two or more ingredients at different prices or concentrations must be mixed to produce a mixture of a desired mean value. The alligation rule uses the cross-difference of values.

• **Mean Price/Concentration**: When components A and B with values a and b are mixed in ratio m:n, the mixture's value = (m×a + n×b)/(m+n). This is the weighted average principle.

• **Alligation Rule (Cross-Method)**: To mix two components with values C₁ and C₂ to get a mixture with mean value M, the ratio of quantities is (M − C₂):(C₁ − M). Visually represented as a cross diagram where you subtract diagonally.

• **Replacement**: When a portion of a mixture is removed and replaced with one of its pure components, the concentration of that component changes. After n replacements of fraction f each time, final quantity = initial × (1 − f)ⁿ for the non-replaced component.

• **Percentage to Ratio**: A mixture with 40% A and 60% B has A:B = 40:60 = 2:3. Always simplify ratios to lowest terms for easier calculation.

• **Successive Mixing**: When mixtures are combined in stages, apply the weighted average principle at each stage or use alligation iteratively.

Formulas / Key Facts

**1. Alligation Formula**: To obtain mean M from C₁ and C₂, mix in ratio = (M − C₂):(C₁ − M).

**2. Weighted Average**: If quantities Q₁ and Q₂ with values V₁ and V₂ are mixed, mean value = (Q₁V₁ + Q₂V₂)/(Q₁ + Q₂).

**3. Single Replacement Formula**: Final quantity of a component after one replacement = Initial quantity × (1 − fraction removed).

**4. Multiple Replacements**: After n identical replacements of fraction f, remaining quantity of non-replaced component = Initial × (1 − f)ⁿ.

**5. Milk-Water Ratio Conversion**: If milk:water = a:b, then milk% = [a/(a+b)]×100 and water% = [b/(a+b)]×100.

**6. Cost Price in Alligation**: When mixing items at different costs to achieve a target selling price or profit margin, use cost prices as C₁ and C₂.

**7. Concentration Problems**: Pure substance has 100% concentration; pure dilutant (water, oil) has 0% of the active ingredient.

**8. Ratio Addition**: When mixing A:B = m:n with C:D = p:q in quantities x and y, the combined mixture requires finding the total of each component.

Worked Examples

**Example 1: Basic Alligation** **Problem**: A merchant mixes rice at ₹30/kg with rice at ₹40/kg to sell the mixture at ₹35/kg. In what ratio should he mix them?

**Solution**: Using alligation rule: C₁ = 30 (cheaper), C₂ = 40 (dearer), M = 35 (mean) Ratio = (M − C₂):(C₁ − M) = (35 − 40):(30 − 35) = (−5):(−5) Taking absolute values: 5:5 = 1:1

**Answer**: Mix in ratio 1:1.

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**Example 2: Percentage Concentration** **Problem**: How much water must be added to 60 litres of milk containing 8% water to make it 20% water?

**Solution**: Initial: 8% water means 92% milk. Final: 20% water means 80% milk. Milk quantity remains constant = 60 × 0.92 = 55.2 litres. If final mixture has 80% milk: Total mixture = 55.2/0.80 = 69 litres. Water to be added = 69 − 60 = 9 litres.

**Answer**: 9 litres.

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**Example 3: Replacement Problem** **Problem**: A container has 40 litres of milk. 4 litres are removed and replaced with water. This is done twice. What is the final quantity of milk?

**Solution**: Fraction removed each time = 4/40 = 1/10. After first replacement: Milk = 40 × (1 − 1/10) = 40 × 9/10 = 36 litres. After second replacement: Milk = 36 × 9/10 = 32.4 litres. Or directly: Milk = 40 × (9/10)² = 40 × 81/100 = 32.4 litres.

**Answer**: 32.4 litres.

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**Example 4: Three-Component Mixture (Iterative Alligation)** **Problem**: Mix liquids at ₹20, ₹30, and ₹40 per litre to get a mixture at ₹28 per litre. If the first two are in ratio 2:3, find the ratio of all three.

**Solution**: Step 1: Mix ₹20 and ₹30 in ratio 2:3. Mean of this mix = (2×20 + 3×30)/5 = (40+90)/5 = 26. Step 2: Now mix this ₹26 mixture with ₹40 to get ₹28. Using alligation: (28−40):(26−28) = (−12):(−2) = 12:2 = 6:1. So ₹26 mixture : ₹40 = 6:1. Since ₹26 mixture contains ₹20:₹30 = 2:3, scale this by 6 parts. Final ratio ₹20:₹30:₹40 = 12:18:5.

**Answer**: 12:18:5.

Common Mistakes

**Mistake 1**: Confusing which value to subtract in alligation → **Fix**: Always subtract diagonally in the cross-method: (Mean − Lower value):(Higher value − Mean). Both differences should be positive.

**Mistake 2**: Not converting percentages to actual quantities → **Fix**: When dealing with percentages, first find the actual quantity of each component. A 25% solution of 100L has 25L solute, not 25% of mixture.

**Mistake 3**: Using simple average instead of weighted average → **Fix**: If 2L of ₹30 liquid mixes with 3L of ₹40 liquid, mean ≠ (30+40)/2. Use (2×30 + 3×40)/(2+3) = 36.

**Mistake 4**: Incorrect replacement formula application → **Fix**: The formula Final = Initial × (1−f)ⁿ applies only to the component NOT being added. The added component follows a different pattern.

**Mistake 5**: Forgetting to simplify ratios → **Fix**: Always reduce ratios to simplest form. The ratio 10:15:20 should be written as 2:3:4 to avoid calculation errors and match answer options.

Quick Reference

• **Alligation shortcut**: Ratio = (M − C₂):(C₁ − M) where C₁ < M < C₂.
• **Replacement formula**: After n replacements of fraction f, remaining = Initial × (1−f)ⁿ.
• **Water addition to reduce concentration**: Water needed = (Original volume × Original %) / (Final %) − Original volume.
• **Pure substance mixing**: Pure has 100% concentration; pure dilutant has 0%.
• **Ratio-to-percentage**: If A:B = 3:7, then A = 30% and B = 70% of mixture.
• **Cross-verify with weighted average**: Your answer ratio should satisfy (Q₁V₁ + Q₂V₂)/(Q₁+Q₂) = Mean value.