Mixture and Alligation: Complete Method, Formula and Tricks

Mixture and alligation is one of those topics that looks intimidating until you understand the one core rule that governs every single question in this chapter.

That rule is the alligation rule — and once you internalize it, problems that appear complex on the surface become two-step calculations. Whether the question involves mixing two liquids, combining two groups of people with different averages, blending products at different prices, or diluting a solution — the alligation rule applies every time.

In SSC CGL, mixture and alligation typically contributes 2–3 questions. In IBPS PO and SBI PO, it appears as part of arithmetic DI sets and standalone questions in Prelims. In CAT, alligation shortcuts dramatically speed up weighted average calculations across arithmetic and data interpretation. Across all three exams, this is a topic where full marks are achievable with a clear method and consistent practice.

This guide covers every question type — mean price problems, ratio of mixing, replacement in vessels, successive dilution, and alligation applied to averages and percentages — with complete worked examples for each.

Part 1: The Alligation Rule — The One Formula You Need

What Is Alligation?

Alligation is a method to find the ratio in which two ingredients at different prices (or concentrations, or averages) must be mixed to produce a mixture at a desired mean price (or concentration, or average).

The Alligation Rule

When two ingredients A and B are mixed:

• Ingredient A has value d₁ (cheaper/lower)
• Ingredient B has value d₂ (dearer/higher)
• The mixture has mean value m

The ratio in which A and B are mixed:

Quantity of A : Quantity of B = (d₂ − m) : (m − d₁)

The Cross-Subtraction Diagram

This visual makes the rule impossible to forget:

       d₁              d₂
        \              /
                m
        /              \
 (d₂ − m)        (m − d₁)

• Left diagonal gives the quantity of the cheaper ingredient: (d₂ − m)
• Right diagonal gives the quantity of the dearer ingredient: (m − d₁)

Always subtract toward the center (mean). Never subtract away from it.

Worked Example 1 — Basic Alligation

Tea costing Rs. 60/kg is mixed with tea costing Rs. 90/kg to produce a mixture costing Rs. 75/kg. Find the ratio of mixing.

Using the diagram:

• d₁ = 60, d₂ = 90, m = 75
• Quantity of cheaper (Rs. 60) = 90 − 75 = 15
• Quantity of dearer (Rs. 90) = 75 − 60 = 15
• Ratio = 15 : 15 = 1 : 1

Verification: (60×1 + 90×1) ÷ 2 = 150/2 = 75 ✓

Worked Example 2 — Unequal Prices

Rice at Rs. 40/kg is mixed with rice at Rs. 65/kg to get a mixture at Rs. 50/kg. Find the ratio.

• d₁ = 40, d₂ = 65, m = 50
• Quantity of cheaper = 65 − 50 = 15
• Quantity of dearer = 50 − 40 = 10
• Ratio = 15 : 10 = 3 : 2

Verification: (40×3 + 65×2) ÷ 5 = (120 + 130)/5 = 250/5 = 50 ✓

Part 2: Mean Price Method — Finding the Mixture Value

When quantities are already given and the question asks for the mixture price or average — use the weighted mean formula directly.

Weighted Mean Formula

Mean = (Q₁ × V₁ + Q₂ × V₂) ÷ (Q₁ + Q₂)

Where Q₁, Q₂ are quantities and V₁, V₂ are values (price, concentration, or average).

Worked Example 3

20 litres of milk worth Rs. 18/litre is mixed with 30 litres of milk worth Rs. 23/litre. What is the price of the mixture per litre?

Mean = (20×18 + 30×23) ÷ (20+30)
= (360 + 690) ÷ 50
= 1050 ÷ 50
= Rs. 21/litre

Alligation Shortcut for Weighted Mean

When the ratio of quantities is given — use alligation in reverse to find the mean.

40 litres of solution A (concentration 30%) is mixed with 60 litres of solution B (concentration 50%). Find the concentration of the mixture.

Mean = (40×30 + 60×50) ÷ 100
= (1200 + 3000) ÷ 100
= 42%

Part 3: Finding Quantities When Total Is Given

A very common exam question gives you the total mixture quantity and asks how much of each ingredient was used.

Method

Step 1: Find the ratio using alligation rule → a : b
Step 2: Total parts = a + b
Step 3: Quantity of A = a/(a+b) × Total, Quantity of B = b/(a+b) × Total

Worked Example 4

In what ratio must water (cost = 0) be mixed with milk at Rs. 30/litre to reduce the cost to Rs. 20/litre? If 50 litres of the mixture is needed, how many litres of each?

Step 1 — Alligation:

• d₁ = 0 (water), d₂ = 30 (milk), m = 20
• Water quantity = 30 − 20 = 10
• Milk quantity = 20 − 0 = 20
• Ratio = 10 : 20 = 1 : 2

Step 2 — Actual quantities for 50 litres:

• Water = 1/3 × 50 = 16.67 litres
• Milk = 2/3 × 50 = 33.33 litres

Worked Example 5 — Three Ingredient Mixing

A grocer mixes 10 kg of sugar at Rs. 15/kg, 15 kg at Rs. 20/kg, and 25 kg at Rs. 25/kg. Find the average cost per kg.

Mean = (10×15 + 15×20 + 25×25) ÷ (10+15+25)
= (150 + 300 + 625) ÷ 50
= 1075 ÷ 50
= Rs. 21.50/kg

For three ingredients — use weighted mean directly. Alligation rule applies only for two ingredients at a time.

Part 4: Replacement in Vessels — The Most Tested Subtopic

Replacement problems are the most frequently tested type of mixture question in SSC CGL, IBPS PO, and SBI PO. A vessel contains a mixture; some quantity is removed and replaced with a pure ingredient. This repeats one or more times.

The Replacement Formula

After n replacements, the quantity of original ingredient remaining:

Final quantity = Initial quantity × (1 − r/V)ⁿ

Where:

• V = total volume of vessel
• r = quantity removed (and replaced) each time
• n = number of replacements

Fraction remaining after n replacements = (1 − r/V)ⁿ

Worked Example 6 — Single Replacement

A vessel contains 40 litres of milk. 8 litres is removed and replaced with water. Find the amount of milk remaining.

Milk remaining = 40 × (1 − 8/40)
= 40 × (32/40)
= 40 × 4/5
= 32 litres

Worked Example 7 — Multiple Replacements

A vessel contains 50 litres of milk. 10 litres is removed and replaced with water three times. How much milk remains after the third replacement?

Milk remaining = 50 × (1 − 10/50)³
= 50 × (4/5)³
= 50 × 64/125
= 3200/125
= 25.6 litres

Worked Example 8 — Finding Number of Replacements

A vessel has 64 litres of milk. After n replacements (each removing 16 litres and replacing with water), only 27 litres of milk remain. Find n.

27 = 64 × (1 − 16/64)ⁿ
27/64 = (3/4)ⁿ
(3/4)³ = 27/64
n = 3

Shortcut: Express both sides as fractions with the same base — the power is your answer.

Part 5: Successive Dilution

Successive dilution is replacement applied to concentration — used when the question asks about percentage concentration rather than absolute quantity.

Formula

Final concentration = Initial concentration × (1 − r/V)ⁿ

This is the same as the replacement formula — just applied to concentration percentage instead of volume.

Worked Example 9

A vessel contains 80 litres of a 90% alcohol solution. 20 litres is removed and replaced with water twice. Find the final concentration of alcohol.

Final concentration = 90% × (1 − 20/80)²
= 90% × (3/4)²
= 90% × 9/16
= 810/16
= 50.625%

Worked Example 10 — Concentration Given as Ratio

A 60-litre vessel contains milk and water in ratio 5:1. 12 litres is removed and replaced with water once. Find the new ratio of milk to water.

Initial milk = 5/6 × 60 = 50 litres
Initial water = 10 litres

After removing 12 litres (which contains milk and water in ratio 5:1):

• Milk removed = 5/6 × 12 = 10 litres
• Water removed = 1/6 × 12 = 2 litres

After replacement:

• Milk = 50 − 10 = 40 litres
• Water = 10 − 2 + 12 = 20 litres
• New ratio = 40 : 20 = 2 : 1

Part 6: Alligation Applied to Averages and Percentages

Alligation is not limited to price problems. Anytime two groups with different averages or percentages are combined — alligation gives the mixing ratio instantly.

Alligation for Averages

The average age of Group A (30 people) is 25 years. The average age of Group B is 35 years. The combined average is 28 years. Find the number of people in Group B.

Using alligation:

• d₁ = 25, d₂ = 35, m = 28
• Ratio of Group A to Group B = (35−28) : (28−25) = 7 : 3

Group A = 30 people → 7 parts = 30 → 1 part = 30/7
Group B = 3 parts = 3 × 30/7 = ~12.86 ≈ 13 people

Alligation for Pass Percentages

In Section A, 60% students passed. In Section B, 80% students passed. Overall 65% passed. Find the ratio of students in A to B.

• d₁ = 60, d₂ = 80, m = 65
• Ratio A : B = (80−65) : (65−60) = 15 : 5 = 3 : 1

Alligation for Profit Percentages

A merchant mixes two varieties of goods. Variety 1 gives 20% profit, Variety 2 gives 35% profit. To achieve 30% overall profit, find the mixing ratio.

• d₁ = 20, d₂ = 35, m = 30
• Ratio = (35−30) : (30−20) = 5 : 10 = 1 : 2

Part 7: Special Cases and Exam Traps

Case 1 — When Water Is Free (Cost = 0)

Water adulteration problems always set d₁ = 0 for water. The alligation rule still applies normally.

Milk at Rs. 48/litre is adulterated with water to sell at Rs. 48/litre and gain 20% profit. Find the ratio of water to milk.

Cost price must be = 48/1.20 = Rs. 40/litre (this is the required mean price)

• d₁ = 0 (water), d₂ = 48 (milk), m = 40
• Water : Milk = (48−40) : (40−0) = 8 : 40 = 1 : 5

Case 2 — Mixing Three Varieties Using Alligation Twice

For three varieties A, B, C — apply alligation to any two pairs and find the common ratio.

Three varieties of rice cost Rs. 20, Rs. 30, and Rs. 40/kg. Mix them to get Rs. 28/kg.

Step 1: Mix Rs. 20 and Rs. 40 → Ratio = (40−28) : (28−20) = 12 : 8 = 3 : 2
Step 2: Mix Rs. 30 and the result at Rs. 28:

• But Rs. 28 already achieved with A and C — Rs. 30 variety can be added in any proportion
• This is a flexible mixing problem — exam will usually constrain one quantity

Case 3 — The "How Much to Add" Trap

How much water must be added to 30 litres of milk worth Rs. 24/litre so that the mixture is worth Rs. 20/litre?

Using alligation (water = Rs. 0):

• Ratio of water to milk = (24−20) : (20−0) = 4 : 20 = 1 : 5
• Milk = 30 litres = 5 parts → 1 part = 6 litres
• Water to add = 6 litres

SSC CGL and IBPS PO — Exam Pattern for This Topic

3-Week Mixture and Alligation Practice Plan