**Mixture:**The new product obtained by mixing two or more ingredients in a certain ratio is called a mixture of those particular ingredients.**Alligation:**It is a method of solving arithmetic problems related to mixtures of ingredients. This rule enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of the desired price.

**Rule of Alligation**

Alligation is a rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of the desired price.

There are two types of methods used in alligation.

**(a) Alligation Method 1**

The rule also helps to find out the **mean or average value of mixture** when the prices of two or more ingredients which may be mixed together and the ratio in which they are mixed are given.

When two ingredients at given prices are known, then the ratio in which these two are mixed to obtain a mixture of

known price is given by:

Hence, Amount of Cheaper: Amount of dearer = (D – M) : (M – C)

Note:

(1) This method can be used when per cent, per hour, per km, per kg etc are being compared

(2) The CP of the unit quantity of the mixture is called the mean price (M)

**Example 1: In what ratio, wheat at Rs 6.20 per kg be mixed with wheat at Rs 7.20 per kg, so that the mixture is worth Rs 6.50 per kg?**

**Sol.**Required ratio is 70 : 30 or 7 : 3

**Example 2:** **In what ratio Rice worth Rs 30 per kg should be mixed with rice worth Rs 32.5 per kg so that on selling the mixture at Rs 34.10 per kg, the profit is 10%.**

**Sol. **Cost price of rice so that the profit is 10% when SP is 34.10 = 34.10 x (100/110) = Rs 31

The ratio in which rice is to be mixed is 1.5:1 Or 3:2

**Example 3:**** The mixture of a certain quantity of milk with 16 litres of water is worth 0.75 per litre. If pure milk is worth 2.25 per litre, how much milk is there in the mixture? **

**Sol. **Cost of water is Rs 0 per litre.

Water : Milk = 1.50 : 0.75 = 2 : 1

⇒ Quantity of milk

**(b) Alligation Method 2: Repeated Dilution**

This is used to calculate pure quantity left after ‘n’ number of processes of repeated replacement is done on the pure quantity. Suppose, a container contains ‘x’ units of a liquid from which ‘y’ units are taken out and replaced by water.

After ‘n’ operations quantity of pure will be:

**Example 4: ****A container contains 50 litres of milk. From this container, 10 litres of milk was taken out and replaced by water. This process is repeated one more time.**

**How much milk is now left in the container?**

**Sol. **Units of milk in the container, x = 50 L

Units of milk to be replaced, y = 10 L

number of process of repeated replacement = 2

Applying the Replacement Method: Amount of milk after 2 operations =