Alligation - Concept and tricks to solve Alligation and Mixture problems

Alligation & Mixture

In this post we will learn about Alligation and Mixture. We will learn new concept with multiple methods and tricks to solve Alligation and Mixture problems easily within seconds. To understand this chapter Read here about Ratio and proportion because this chapter will use same concepts.

What is Alligation?

Alligation is the rule which enables us to find the ratio in which two or more ingredients need to be mixed to produce a new mixture of desired strength, price etc.

There are two types of Alligation:
 1.  Alligation Medial: It is used to find the average value or quantity of mixture when the prices of two or more ingredients and the proportion in which they are mixed are given.

 2.  Alligation Alternate: It is used to find the amount of each ingredient needed to make a mixture of a given quantity.

This method is applicable for ratio, percentage, rate, speed etc.

MEAN PRICE: It is the cost price of the mixture.

Rule of Alligation & Mixture

 1.  If gradients are mixed in a ratio, then:

(Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)
This is Alligation and mixture formula.

Example: In what proportion must rice at Rs 3.10 per kg be mixed with rice at Rs 3.60 per kg, so that the mixture be worth Rs 3.25 per kg?
Solution: Cheaper rice : Dearer rice = (3.60 – 3.25)/(3.25 – 3.10)= 7/3 or 7:3

 2.  If a container contains 'X' unit of pure liquid from which 'Y' units are taken out and replaced with another liquid. After nth operations:
Quantity of pure liquid left = X (1 – Y/X)n units

Example 1: A container contained 80 Kg of milk. From this container 8 kg of milk was taken out and replaced by water. This process was further repeated twice. How much milk is now contained by the container?
Solution: Putting the values in above formula = 80 (1 – 8/80)3
              ⇒ 80 x 9 x 9 x 9/1000 = 58.32 Kg

Example 2: A can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. Litres of liquid A contained by the can initially was?
Solution:  Method 1:  Basic: Let Liquid A = 7x and Liquid B = 5x. When 9 litre of mixtures are drawn off, liquid A remains = 7x – 9 x 7/12 = 7x – 21/4
Liquid B remains = 5x – 9 x 5/12 = 5x – 15/4

After adding 9 litres of Liquid B new ratio is 7:9.
   ⇒ (7x – 21/4) : (5x – 15/4 + 9) = 7:9
After solving we get x = 3
Quantity of liquid A = 7x = 7 x 3 = 21 litres

 Method 2:  Direct Formula:
Common factor of first ratio = [(Quantity replaced)/sum of terms in first ratio] + [(quantity replaced x antecedent of first ratio)/sum of terms in first ratio]

Putting values, common factor x = (9/12) + (9 x 7/28) = 3
   ⇒ Quantity of A = 7 x 3 = 21 litres

 Method 3:  By Alligation:
Fraction of A in given mixture = 7/12
Fraction of A in the liquid Added = 0 (since pure liquid B is added)
Fraction of A in the new mixture = 7/16

Ratio of original mixture to B = 7/16 : 7/48 = 3:1

It means for 1 litre of liquid B added, quantity of mixture should be 3 litres.
   ⇒ For 9 litres of liquid B, original mixture should be = 9 x 3 = 27 litres.
Therefore initial quantity of mixture in can = 27 + 9 = 36 litres.
   ⇒ A = 7 x 36/12 = 21 litres
   ⇒ B = 5 x 36/12 = 15 litres


These are few methods to solve Alligation and mixture questions. More methods will be discussed in next post. Good Luck 👍