Alligation is a method to find the ratio in which two or more ingredients should be mixed to get a mixture at a desired price or value. It helps solve problems on mixtures like milk and water, cost price combinations, or even solutions in chemistry.
- Alligation: A rule to mix two things with different prices to get a mixture at a target price.
- Mean Price: Price of one unit of the final mixture.
- Cheaper Ingredient: Item with the lower price.
- Dearer Ingredient: The item with the higher price.
1. Alligation Rule Formula
To find the ratio in which two ingredients (A and B) are mixed:
Ratio = (CP of B – Mean Price) : (Mean Price – CP of A)
Where:
– CP = Cost Price
– A = Cheaper ingredient
– B = Dearer ingredient
– Mean Price = Price of the mixture
2. Alligation Rule – Cross Method Layout
So, Required Ratio (A : B) = (d – M) : (M – c)
3. Replacement Formula
If a container contains ‘x’ units of a liquid, and ‘y’ units are replaced
with water ‘n’ times, the quantity of pure liquid left is:
Remaining = x * (1 – y/x)^n
Where:
– x = Total quantity of liquid
– y = Quantity removed and replaced each time
– n = Number of operations
4. Examples
Example 1
Question:
Mix sugar costing ₹30/kg and ₹50/kg to get a mixture worth ₹36/kg.
Solution:
Using the formula:
Ratio = (50 – 36) : (36 – 30) = 14 : 6 = 7 : 3
Answer
So, mix them in a ratio of 7:3 (costlier : cheaper)
Example 2: Mixing Two Items
Question:
Milk costs ₹20 per litre and water costs ₹5 per litre. In what ratio should they be mixed to get a mixture worth ₹15 per litre?
Solution:
Use Alligation Rule:
Dearer (20)
\
\ (20 – 15) = 5
\
Mean (15)
/
/ (15 – 5) = 10
/
Cheaper (5)
Ratio = (20 – 15) : (15 – 5) = 5 : 10 = 1 : 2
Answer
Mix milk and water in the ratio 1:2
Example 3: Replacement Problem
Question:
A 20-litre container is full of milk. 5 litres of milk is removed and replaced with water.
This process is repeated 2 times. How much milk is left?
Solution:
Use the Replacement Formula:
Remaining milk = x * (1 – y/x)^n
Here,
x = 20 litres (total),
y = 5 litres (removed each time),
n = 2 (number of times)
Remaining = 20 * (1 – 5/20)^2
= 20 * (3/4)^2
= 20 * (9/16)
= 11.25 litres
Answer
11.25 litres of milk remains in the container
