The rules of Mixture and Alligation Questions help the candidate to find out the proportion between the two-given commodity.
Mixture and Alligation are two different terms. Before solving mixture and alligations questions you should first know what they actually are. So, let’s start with the definition of Mixture.
The mixture is the mixing of two or more group of different types or two or more items of different types.
When two or more than two substances are in any ratio to produce a product is known as Mixture mean price. The cost price of a unit quantity of the mixture is called the mean price.
It deals with how much quantity you want to mix with the two or more items so, the final quantity can be calculated. Alligations are dealing with the quantity.
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.
Before understanding any new topic, it’s very important that you know the basic concepts of the chapter. So, to clear all your doubts we have mentioned Mixture and Alligation tricks and formulas with their examples to solve the question in a competitive exam in minimal time. Lets start with the mixture and allegation examples.
Let’s suppose if the cost price of the cheaper item is ‘Rs. C’ and the cost price of the dearer item be ‘Rs. D’ and the average or the mean price of a mixture be ‘m’ then,
You can also represent the equation as:
Question: In what proportion must tea at Rs. 10 per kilogram be mixed with tea at Rs. 14 per kilogram. So, that the mixture be worth Rs. 12 per kilogram?
Question: There are two containers which are filled with water and milk. The first container is filled in the ratio of 3:1 and the second container is filled with 5:2. If, both the containers mixture is mixed-up then, the ratio of milk to the water in the mixture will be?
Portion of milk in first container = 3/(3+1)=3/4
Portion of water in first container = 1/(3+1)=1/4
Portion of milk in second container = 5/(5+2)=5/7
Portion of water in second container = 2/(5+2)=2/7
Therefore, the ratio of water and the milk in the third container will be = 3/4 + 5/7 : 1/4 + 2/7 = 41/28 : 15/28
Suppose a container contains ‘x’ unit of a liquid from which ‘y’ units are taken out and replaced by water. After an operation, the quantity of pure liquid
Question: 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?
Solution: Amount of milk left after 3 operations will be
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This brings us to the end of our article. Hope you are now aware of the mixture and alligation tricks, formulae and concepts to solve the questions easily.
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