Mixtures and Alligations: Complete Aptitude Guide

The alligation rule solves mixture problems in two arithmetic steps: subtract the mean price from each ingredient’s price diagonally, and the differences give the required mixing ratio.

That sentence covers the full method. Everything below is worked applications of it.

What Is the Alligation Rule?

Two ingredients are mixed: a cheaper one at value p1 and a costlier one at value p2, to produce a mixture at mean value p (where p1 < p < p2). The quantity ratio is:

• q1 : q2 = (p2 - p) : (p - p1)

where q1 is the quantity of the cheaper ingredient and q2 is the quantity of the costlier ingredient.

Derivation

The weighted average equation is:

• q1 × p1 + q2 × p2 = (q1 + q2) × p

Rearranging:

• q1 × (p - p1) = q2 × (p2 - p)
• q1 / q2 = (p2 - p) / (p - p1)

The “alligation cross” is a visual shortcut for this: write p1 on the left and p2 on the right, p in the middle, then take diagonal differences. The difference to the right of p belongs to the left ingredient (cheaper) and vice versa.

The rule applies to any measurable property: price per kg, alcohol concentration, speed, or any other quantity that averages linearly.

Example 1: Copper alloy blending

• Problem: Alloy A contains 30% copper. Alloy B contains 60% copper. In what ratio must they be mixed to produce a 40% copper alloy?
• Step 1: Apply the alligation cross. p1 = 30, p2 = 60, p = 40.
• Step 2: q1 : q2 = (60 - 40) : (40 - 30) = 20 : 10 = 2 : 1
• Verify: (30 × 2 + 60 × 1) / 3 = 120 / 3 = 40% ✓

Example 2: Tea blend at equal ratio

• Problem: A shopkeeper blends tea at Rs. 60/kg with tea at Rs. 90/kg to produce a mixture at Rs. 75/kg. Find the ratio.
• Step 1: p1 = 60, p2 = 90, p = 75.
• Step 2: q1 : q2 = (90 - 75) : (75 - 60) = 15 : 15 = 1 : 1
• Verify: (60 × 1 + 90 × 1) / 2 = 75 ✓

Worked Examples: Prices and Concentrations

Example 3: Profit built into the mean price

• Problem: Mix tea at Rs. 60/kg with tea at Rs. 65/kg so the mixture can be sold at Rs. 68.20/kg with a 10% profit. Find the mixing ratio.
• Step 1: Find the cost price of the mixture. CP = SP × 100 / 110 = 68.20 × 100 / 110 = Rs. 62/kg
• Step 2: Apply alligation. p1 = 60, p2 = 65, p = 62.
• Step 3: q1 : q2 = (65 - 62) : (62 - 60) = 3 : 2
• Verify: (60 × 3 + 65 × 2) / 5 = (180 + 130) / 5 = 62 ✓. SP = 62 × 1.1 = 68.20 ✓
• Answer: Mix cheaper tea and costlier tea in ratio 3 : 2

Example 4: Milk adulteration with water

• Problem: A milkman sells adulterated milk at Rs. 36/litre, claiming it is pure milk worth Rs. 48/litre. What fraction of the mixture is water?
• Step 1: Water costs Rs. 0/litre. Apply alligation. p1 = 0 (water), p2 = 48 (milk), p = 36.
• Step 2: q_water : q_milk = (48 - 36) : (36 - 0) = 12 : 36 = 1 : 3
• Step 3: Water fraction = 1 / (1 + 3) = 1/4 = 25%
• Verify: (0 × 1 + 48 × 3) / 4 = 144 / 4 = 36 ✓

Example 5: Acid solution blending

• Problem: Two acid solutions with concentrations 20% and 50% are blended to produce a 30% solution. Find the ratio.
• Step 1: p1 = 20, p2 = 50, p = 30.
• Step 2: q1 : q2 = (50 - 30) : (30 - 20) = 20 : 10 = 2 : 1
• Verify: (20 × 2 + 50 × 1) / 3 = 90 / 3 = 30% ✓

Example 6: Finding the quantity of each component

• Problem: A merchant blends coffee at Rs. 180/kg with coffee at Rs. 270/kg to produce 60 kg at Rs. 210/kg. How many kg of each type?
• Step 1: Apply alligation. p1 = 180, p2 = 270, p = 210.
• Step 2: q1 : q2 = (270 - 210) : (210 - 180) = 60 : 30 = 2 : 1
• Step 3: Cheaper coffee (Rs. 180) = (2/3) × 60 = 40 kg. Costlier coffee (Rs. 270) = 20 kg.
• Verify: (180 × 40 + 270 × 20) / 60 = (7200 + 5400) / 60 = 210 ✓

Replacement and Dilution Scenarios

Replacement problems introduce a new mechanic: a fixed volume of the existing mixture is drawn out and replaced with a different liquid. The alligation cross still applies to the concentrations, but it is often faster to work directly with the volume fractions.

Example 7: One-time replacement from a pure liquid

• Problem: A 25-litre vessel contains pure milk. 5 litres are removed and replaced with water. Find the ratio of milk to water.
• After removal: 25 - 5 = 20 litres of milk remain. Empty space = 5 litres.
• After adding water: Milk = 20 L, water = 5 L.
• Ratio milk : water = 20 : 5 = 4 : 1

Example 8: One-time replacement from a mixture

• Problem: A 30-litre vessel holds milk and water in ratio 7 : 3. 6 litres of the mixture are removed and replaced with water. Find the new milk-to-water ratio.
• Initial quantities: Milk = 30 × 7/10 = 21 L, water = 30 × 3/10 = 9 L.
• Milk removed in 6L draw: 6 × (7/10) = 4.2 L. Water removed: 1.8 L.
• After removal: Milk = 21 - 4.2 = 16.8 L. Water = 9 - 1.8 = 7.2 L.
• After adding 6 L water: Milk = 16.8 L. Water = 7.2 + 6 = 13.2 L.
• New ratio: 16.8 : 13.2 = 168 : 132 = 14 : 11
• Shortcut: Milk remaining = Initial milk × (1 - draw/total) = 21 × (24/30) = 21 × 4/5 = 16.8 L ✓

Repeated Replacement: Formula and Verification

When the same draw-and-replace operation repeats k times on a vessel of total volume n, each draw removes x litres:

Final original liquid = Initial × ((n - x) / n)^k

Derivation

Let A_k = amount of original liquid after k steps.

• After step 1 (vessel was pure original): A_1 = n - x = n × (n-x)/n
• After step 2 (mixture fraction of original = A_1/n): original removed = x × A_1/n. So A_2 = A_1 - x × A_1/n = A_1 × (1 - x/n) = n × ((n-x)/n)²
• By induction: A_k = n × ((n-x)/n)^k

The common mis-statement

Some prep guides write the formula as Initial × (1 - x/n)^k but plug in “n = initial amount of liquid A” instead of “n = total vessel volume.” When the vessel contains a pure liquid from the start, these are equal and the error is invisible. When the vessel holds a mixture (liquid A plus liquid B), plugging in the volume of A alone gives a wrong answer. Always use the full vessel volume for n.

Example 9: Three-step milk replacement (verified manually)

• Problem: A 125-litre container holds pure milk. In each step, 25 litres are drawn out and replaced with water. How much milk remains after 3 steps?
• Formula: Milk remaining = 125 × ((125 - 25) / 125)³ = 125 × (4/5)³ = 125 × 64/125 = 64 litres
• Manual check (step by step):
  • Step 1: Remove 25 L pure milk. Add 25 L water. Milk = 100 L, water = 25 L.
  • Step 2: Milk concentration = 100/125. Milk in 25 L draw = 25 × 100/125 = 20 L. Milk left = 80 L. Add 25 L water. Milk = 80, water = 45.
  • Step 3: Milk concentration = 80/125. Milk in 25 L draw = 25 × 80/125 = 16 L. Milk left = 64 L. ✓

Example 10: Finding the fraction replaced

• Problem: A jar of whisky contains 50% alcohol. Some is replaced with a blend containing 21% alcohol. The final alcohol concentration is 32%. What fraction of the original whisky was replaced?
• Let fraction replaced = f. Final concentration equation:
  • 0.50 × (1 - f) + 0.21 × f = 0.32
  • 0.50 - 0.29f = 0.32
  • f = 0.18 / 0.29 = 18/29 ≈ 62%
• Alligation cross verification: original (50%) : replacement (21%), mean 32%.
  • Ratio original : replacement = (32 - 21) : (50 - 32) = 11 : 18
  • Fraction replaced = 18 / (11 + 18) = 18/29 ✓

Where These Problems Appear in Tests

Mixtures and alligations problems are a consistent fixture in campus aptitude tests. The TCS NQT Numerical Ability section routinely includes price-blend and dilution questions. The AMCAT Quantitative module covers ratio-and-proportion topics that frequently map to the alligation cross.

Mu Sigma’s MuApt tests numerical reasoning at speed and mixture problems are a standard item type there. DE Shaw’s preliminary quant round is harder: expect both the basic alligation and the repeated-replacement formula, sometimes combined in a single two-part question.

For general placement aptitude preparation, mixtures and alligations sits alongside time and work as one of the more formula-light topic areas, as the alligation cross itself requires no memorised formula beyond the weighted average equation. A student who can derive the repeated-replacement formula from the weighted average, rather than simply recite it, is better placed for problem variants that deviate from the standard template.

The campus placement evaluation test is another common context where mixture problems appear alongside percentages, ratios, and profit-and-loss questions.

Deriving the repeated-replacement formula from scratch, rather than trusting the textbook without checking, is the same instinct that distinguishes a systematic engineer from one who runs code they don’t understand. TinkerLLM is a ₹299 entry point where that same first-principles instinct meets live model APIs. If working through Example 9’s step-by-step derivation above left you curious what that rigour produces on actual LLM tasks, start there.