Mixture and Alligation
Overview
The topic deals with combining two or more ingredients at different prices, concentrations, or ratios to form a mixture with an average or resultant value. At the Clerk level, you'll encounter straightforward two-component problems—mixing two types of rice at different prices, diluting milk with water, or combining solutions of different concentrations.
The alligation rule provides a shortcut that eliminates lengthy algebraic calculations, making it perfect for time-bound exams. Mastering this single technique lets you solve what would otherwise require setting up and solving linear equations. The concept also interconnects with weighted averages, profit-loss (mixing to achieve a target selling price), and ratio problems, giving you versatile problem-solving ability.
Focus your preparation on understanding when to apply alligation versus when simple ratio calculations suffice. The Clerk exam favors direct application questions without multi-step complexity.
Key Concepts
- Mixture: Combining two or more ingredients to form a new product. Each ingredient has a characteristic value (price, concentration, strength).
- Alligation Rule: A technique to find the ratio in which two ingredients at different values must be mixed to produce a mixture at a desired mean value.
- Cheaper and Dearer: The ingredient with the lower value is "cheaper"; the one with the higher value is "dearer." The mean value always lies between these two.
- Cross-Difference Method: The ratio of quantities equals the ratio of differences between the mean and each ingredient's value, taken crosswise.
- Replacement Problems: When part of a mixture is removed and replaced with a pure ingredient (often water), a specific formula applies for repeated operations.
- Concentration/Strength: In liquid mixtures, concentration = (quantity of pure substance / total quantity) × 100%.
- The mean value must always lie between the two given values—if it doesn't, the problem setup is impossible or you've misread the question.
Formulas / Key Facts
Alligation Formula (The Core Rule)
When mixing two ingredients:
- Cheaper quantity : Dearer quantity = (Dearer value − Mean value) : (Mean value − Cheaper value)
Visually represented as:
Cheaper Value (C) Dearer Value (D)
\ /
\ /
Mean Value (M)
/ \
/ \
(D − M) (M − C)Ratio of Cheaper : Dearer = (D − M) : (M − C)
Repeated Replacement Formula
If a container has Q units of pure liquid, and R units are removed and replaced with another liquid (like water), repeated n times:
Final quantity of original liquid = Q × (1 − R/Q)ⁿ
Concentration After Mixing
When mixing two solutions of concentrations C₁ and C₂ with quantities Q₁ and Q₂:
Resultant concentration = (C₁ × Q₁ + C₂ × Q₂) / (Q₁ + Q₂)
Key Facts to Remember
- Water has zero cost/concentration in most problems
- Adding water increases quantity but decreases concentration
- Removing mixture and adding water: use replacement formula
- In cost problems, alligation gives the mixing ratio by quantity (weight/volume)
Worked Examples
Example 1: Basic Price Mixing
Rice at ₹40/kg is mixed with rice at ₹60/kg to get a mixture worth ₹45/kg. Find the ratio of mixing.
Solution:
- Cheaper = ₹40, Dearer = ₹60, Mean = ₹45
Using alligation:
- Cheaper : Dearer = (60 − 45) : (45 − 40)
- = 15 : 5
- = 3 : 1
Example 2: Milk and Water
A mixture contains milk and water in ratio 5:3. How much water must be added to 40 litres of this mixture to make the ratio 1:1?
Solution: In 40 litres of mixture:
- Milk = 40 × (5/8) = 25 litres
- Water = 40 × (3/8) = 15 litres
For ratio 1:1, water must equal milk = 25 litres
Water to be added = 25 − 15 = 10 litres
Example 3: Repeated Replacement
A container has 80 litres of milk. 8 litres are drawn out and replaced with water. This is done 3 times. Find the quantity of milk remaining.
Solution: Using formula: Final milk = Q × (1 − R/Q)ⁿ
- Q = 80 litres, R = 8 litres, n = 3
- Final milk = 80 × (1 − 8/80)³
- = 80 × (1 − 1/10)³
- = 80 × (9/10)³
- = 80 × 729/1000
- = 58.32 litres
Example 4: Concentration Problem
How many litres of a 30% acid solution must be mixed with a 70% acid solution to get 20 litres of a 40% acid solution?
Solution: Using alligation:
- 30% : 70%, Mean = 40%
- Ratio = (70 − 40) : (40 − 30) = 30 : 10 = 3 : 1
Total parts = 3 + 1 = 4 parts = 20 litres
- 30% solution = 20 × (3/4) = 15 litres
- 70% solution = 20 × (1/4) = 5 litres
Common Mistakes
- Reversing the ratio: Students often write (M − C) : (D − M) instead of (D − M) : (M − C). Remember: cheaper quantity gets the difference from dearer value (cross-difference).
- Forgetting water has zero value: When water is added to milk or any solution, treat water's cost or concentration as 0, not omitting it from calculation.
- Confusing ratio of quantities with ratio of prices: Alligation gives the ratio of quantities to be mixed, not prices. If asked "in what ratio by cost," you need an additional step.
- Using replacement formula for single removal: The formula Q × (1 − R/Q)ⁿ is for repeated replacement. For single replacement, simply calculate directly or use n = 1.
- Mean value outside the range: If your calculated mean doesn't lie between the cheaper and dearer values, recheck—you've likely swapped values or misread the problem.
Quick Reference
- Alligation ratio: Cheaper : Dearer = (D − Mean) : (Mean − C)
- Mean always lies between the two given values—no exceptions
- Water = 0 in cost and concentration problems
- Repeated replacement: Final = Initial × (1 − R/Q)ⁿ
- Cross-difference: Think "opposite subtraction"—cheaper gets dearer's difference
- Verify: Multiply back to check if your ratio produces the stated mean value