Mixture and Alligation

Mixture and Alligation

Overview

Mixture and Alligation is a high-scoring arithmetic topic in SBI Clerk Prelims that appears in both direct questions and as a sub-technique within Data Interpretation sets. The concept helps you quickly find the ratio in which two ingredients at different prices (or concentrations) must be mixed to obtain a desired average price (or concentration).

The questions are typically straightforward—mixing two types of rice, replacing part of a solution, or finding the ratio of milk to water. Mastering the alligation rule gives you a 30-second shortcut compared to equation-based methods, which is crucial given the time pressure in Prelims.

The key skill to develop: identify the "cheaper," "dearer," and "mean" values, apply the cross-difference rule, and interpret the resulting ratio correctly.

Key Concepts

• Mixture: Combining two or more ingredients of different values (price, concentration, percentage) to form a new product with an intermediate value.
• Alligation Rule: When two ingredients are mixed, the ratio of their quantities equals the inverse ratio of their deviations from the mean value. This is the foundation of all mixture problems.
• Cheaper and Dearer: The ingredient with value below the mean is "cheaper"; the one above is "dearer." The mean must always lie between the two.
• Mean Price/Concentration: The resultant value of the mixture—this is what you're trying to achieve or what's given in the problem.
• Replacement Problems: When part of a mixture is removed and replaced with another substance, use the formula for repeated replacement to find the final concentration.
• The alligation cross method: Draw an X connecting cheaper-to-mean and dearer-to-mean, then take differences to get the ratio directly.

Formulas / Key Facts

The Alligation Rule

Quantity of Cheaper   Dearer Value − Mean Value
─────────────────── = ─────────────────────────
Quantity of Dearer    Mean Value − Cheaper Value

Visual Cross Method

Cheaper (C)                    Dearer (D)
         \                    /
          \                  /
           \                /
            Mean (M)
           /                \
          /                  \
    (D − M)                (M − C)

Ratio of Cheaper : Dearer = (D − M) : (M − C)

Repeated Replacement Formula When a vessel contains A litres of liquid and B litres is taken out and replaced with another liquid n times:

Final quantity of original liquid = A × (1 − B/A)ⁿ = A × ((A − B)/A)ⁿ

Key Facts to Remember

• The ratio gives Cheaper : Dearer (not the other way around)
• For percentage problems, treat percentages as values directly (e.g., 20% and 50% solutions)
• In replacement problems, the concentration decreases geometrically with each replacement
• If mixing in ratio a : b, the mean lies at (a × C + b × D)/(a + b)

Worked Examples

Example 1: Basic Price Mixing

Problem: In what ratio must rice at ₹40/kg be mixed with rice at ₹60/kg to get a mixture worth ₹45/kg?

Solution:

• Cheaper = 40, Dearer = 60, Mean = 45
• Cheaper : Dearer = (60 − 45) : (45 − 40) = 15 : 5 = 3 : 1

Answer: 3 : 1

Example 2: Concentration Problem

Problem: A vessel contains 80 litres of milk. 8 litres of milk is taken out and replaced with water. This process is repeated once more. Find the quantity of milk remaining.

Solution:

• Initial quantity (A) = 80 litres
• Quantity replaced each time (B) = 8 litres
• Number of operations (n) = 2

Using formula: Final milk = 80 × (1 − 8/80)² = 80 × (72/80)² = 80 × (9/10)² = 80 × 81/100 = 64.8 litres

Answer: 64.8 litres

Example 3: Finding the Mean

Problem: Two varieties of tea costing ₹80/kg and ₹100/kg are mixed in the ratio 3 : 2. Find the cost of the mixture per kg.

Solution:

• Cost of mixture = (3 × 80 + 2 × 100)/(3 + 2)
• = (240 + 200)/5
• = 440/5
• = ₹88/kg

Answer: ₹88 per kg

Example 4: Milk-Water Problem

Problem: A mixture contains milk and water in the ratio 5 : 3. If 16 litres of water is added, the ratio becomes 5 : 5. Find the initial quantity of milk.

Solution:

• Let initial milk = 5x, initial water = 3x
• After adding 16 litres water: milk = 5x, water = 3x + 16
• New ratio: 5x : (3x + 16) = 5 : 5 = 1 : 1
• Therefore: 5x = 3x + 16
• 2x = 16, so x = 8
• Initial milk = 5 × 8 = 40 litres

Answer: 40 litres

Common Mistakes

• Reversing the ratio: Students often write (M − C) : (D − M) instead of (D − M) : (M − C). Remember: the difference from the dearer gives the quantity of cheaper, not the other way around.
• Confusing what the ratio represents: The alligation ratio gives Cheaper quantity : Dearer quantity. Some students assign it to the wrong ingredients and get the inverse answer.
• Forgetting repeated replacement is multiplicative: In replacement problems, students sometimes subtract twice instead of using the exponential formula. Each replacement reduces the concentration by the same fraction, not the same amount.
• Not converting units properly: When mixing percentages with actual quantities, convert everything to the same unit (either all percentages or all absolute values) before applying alligation.
• Applying alligation when mean is outside the range: If the desired mean is less than both values or greater than both, mixing cannot achieve it. The mean must lie strictly between cheaper and dearer values.

Quick Reference

• Cheaper : Dearer = (Dearer − Mean) : (Mean − Cheaper)
• For replacement n times: Final = Initial × ((Total − Replaced)/Total)ⁿ
• The ratio from alligation is always Cheaper : Dearer—label carefully
• Mean value always lies between the two extremes
• For mixing in ratio a : b, Mean = (aC + bD)/(a + b)
• In percentage mixtures, treat percentages as direct values in the formula