Mixture and Alligation

Alligation And Rule of Alligation

In Alligation & Rule of Alligation, we study mixtures of two or more than two quantities that have different selling process or cost prices. Imagine that you mix a cheap substance with an expensive substance. What should be the value of the resulting mixture? That is what we will study in the Alligation & Rule of Alligation. In the section below, we will see what Alligation is, we will also what mean price and the rule of alligation is. After studying the following section you will be able to answer all the questions based on alligation and the rule of alligation in less than a minute.

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Alligation And Rule of Alligation

To understand the formula of the Alligation and the rule of alligation, we have to understand the concept of weighted average. For example, let us say that we buy 50 packs of chips at the cost of 10 rupees each and 30 packets at the cost of 20 rupees each. What is the average cost? Will the average will be determined by the 20 rupees packs or the 10 rupees packs? This is when we come across the concept of weighted average.

The weighted average here will be: [50×10]+[30×20]/80

In alligation, we will use the same concept. Let us first define the few following terms:

Alligation: It is the 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.

Mean Price: The cost price of a unit quantity of the mixture is called the mean price. Now let us define the rule of alligation.

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Rule Of Alligation

Let us suppose that two ingredients of concentrations ‘a’ and ‘b’ respectively have been mixed in some proportion. Let ‘a’ be the cheaper component and ‘b’ be the dearer or the most costlier component. Then the rule of alligation states that:

[{Quantity of Cheaper substance}/{Quantity of dearer substance}] = [(C.P. of dearer substance) – (Mean price)/(Mean price) – (C.P. of cheaper substance)]

Let ‘c’ be the cost price or C.P. of a unit quantity of a cheaper substance, ‘m’ be the mean price, ‘d’ be the cost price of a unit quantity of the dearer substance, then we can write:

(Quantity of the Cheaper Substance) : (Quantity of the Dearer Substance) = (d – m) : (m – c). Let us get a clearer picture with some an example.

Example 1: In what ratio must rice at Rs 9.30 per kg be mixed with rice sold at Rs. 10.80 per kg, so that the mixture be worth Rs. 10 per kg?

A) 2:1             B) 4:3                      C) 6:5                   D) 8:7

Answer: Using the rule of alligation, we have:

C.P. of 1 kg of rice (in paise) = 1080 paise = d

Also the C.P. of 1 kg rice of 2nd kind (in paise) = 930 paise = c

Also, mean price of the mixture (per kg in paise) or m = 1000 paise. So from the rule of alligation, we have:

(Quantity of Cheaper rice) : (Quantity of dearer rice) = (1080 – 1000)/(1000 – 930)

Therefore the required ratio = 80 : 70 or 8:7 and hence the correct option is D) 8: 7

The Method Of the Repeated Dilutions


There is another way of achieving the results that are given by the rule of alligation. It is known as the method of the repeated dilutions. Suppose a container contains ‘x’ units of liquid from which ‘y’ units are taken out and replaced by water. After n operations, the quantity of pure liquid = [x{1 – (y/n)}n] units. Let us explain this formula with the help of an example:

Example 2: A container contains 40 litres of milk. From this container, 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

A) 33.3 litres                        B) 12.19 litres                    C) 29.16 litres                     D) 30 litres

Answer: The container contains x = 40 litres of milk. The quantity of milk that is taken out and replaced by water = y = 4 litres.

Also, we have been given that number of times the process is repeated or n = 3 (=1 +2 times). Therefore from the method of repeated dilution, substituting the relevant values, we have:

Amount of milk left after three operations = [40{ 1 – (4/40)}3] litres. Therefore we may write:

[40×(9/10)×(9/10)×(9/10)] = 29.16 litres. Therefore the correct option is C) 29.16 litres.

Example 3: A can contains a mixture of two liquids A and B in the ratio 7: 5. When 9 litres of the mixture is drawn off and the can is filled with B, the ratio of A and B becomes 7: 9. How many litres of liquid A was contained by the can initially?

A) 3                 B) 7                    C) 21                      D) 29

Answer: Suppose the can initially contain 7x and 5x litres of mixtures A and B respectively. The quantity of A in the mixture left = [7x – (7/12)×9] litres = [7x – 21/4] litres.

The quantity of B in mixture left = [5x – (5/12)×9] litres = [5x – 15/4] litres.

Therefore we can write: the ratio of the two quantities as [7x – 21/4]/ [5x – 15/4] = 7/9

In other words, we may say that 252x – 189 = 140x + 147

Hence, x = 3. Therefore the can had 7(3) = 21 litres of the quantity A. Thus the correct option is C) 21.

Practice Questions

Q 1: A milk vendor has 2 cans of milk. the first contains 25% water and the rest milk. the second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3: 5?

A) 6 litres, 7 litres                    B) 7 litres, 8 litres                        C) 6 litres, 6 litres                           D) 7 litres, 5 litres.

Ans: C) 6 litres, 6 litres

Q 2: A dishonest milkman professes to sell his milk at cost price but he mixes it with water an thereby gains 25%. The percentage of water in the mixture is:

A) 10%                 B) 15%                         C) 20%                        D) 25%

Ans: C) 20%

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