Mixture and Alligation

Today we are going to learn about mixture and alligation. You might be familiar be with the word mixture i.e. a combination of two or more things in a material where each element is distinct from the other. For example, we can have a mixture of milk and water. Most of the time in the market we are not sold the pure substance.

A shopkeeper sells a mixture of two types of sugar in which one is of higher quality than the other. Of course, higher quality sugar will be expensive and sugar of lower quality will be cheaper. If the shopkeeper sells this mixture of sugar at a certain price, how much each type of sugar should be added in the mixture? alligation and Rule of alligation deal with such type of problems. This lesson is very important considering its importance in the market and thus it appears in most of the competitive exams.

Terms used in mixture and alligation problems

Remember the meaning of following terminologies because they will be frequently used while solving mixture and alligation problems.  

Alligation

The process or rule that helps in finding out the ratio in which two or more quantities that have different cost prices must be added to get a mixture of the desired price is known as alligation.

Mean Price

Mean price is defined as the cost price of a unit of the mixture.

What is the rule of Alligation

If the mixture of two quantities having cost prices x and y respectively, x is the cost of expensive quantity and y is the cost of cheaper quantity, are sold at the mean price of m. Then, the rule of alligation is defined as

    \[ \frac {Quantity \enspace of \enspace cheaper \enspace substance}{Quantity \enspace of \enspace dearer \enspace substance}= \]

    \[\frac {C.P. (dearer \enspace substance) - M.P.}{M.P. - C.P. (cheaper \enspace substance)}\]

Where C.P. stands for cost price and M.P. stands for mean price.

OR

    \[\frac{Quantity \enspace of \enspace cheaper}{Quantity \enspace of \enspace dearer} = \frac{x - m}{m - y}\]



The same formula can be more easily represented by cross multiplication method explained in the figure below:

Cross multiplication method for mixture and alligation questions

The formula says that the ratio of the quantities is inversely proportional to the difference between the cost price and the mean price.

Let us now solve questions related to mixture and alligation to apply the concepts learned above.

Also read how to solve unit digit questions using cyclicity of numbers

Mixture and alligation questions

Problem 1
Consider there are two types of rice. The price of one type of rice is Rs. 100 per kg and the price of other rice is Rs. 75 per kg. If both of these rice are added to produce a mixture of price Rs. 80.5 per kg. What should be the ratio of quantities added in the mixture?

Solution
Cost price of dearer substance = x = Rs. 100 per kg
Cost price of cheaper substance = y = Rs. 75 per kg
Mean price = m = Rs. 80.5 per kg

The ratio of quantities added in the mixture can be found by using the rule of alligation.

    \[\frac{Quantity \enspace of \enspace cheaper}{Quantity \enspace of \enspace dearer} = \frac{x - m}{m - y}\]

= (100 – 80.5) / (80.5 – 75)
= 19.5 / 5.5

The ratio of quantities added in the mixture is 19.5:5.5.

The problem can also be found using cross multiplication method.

Shortcut method to solve mixture and alligation Problems

Problem 2
A milkman sells pure milk at the price of Rs. 20 per liter. However, he sells a mixture of milk and a certain quantity of water at the price of Rs. 15 per liter. Find the ratio of the quantity of pure milk and water.

Solution
Cost price of dearer substance, x = Rs. 20 per liter
Cost of price of cheaper substance, y = 0
(The cost price of water is considered zero)
Mean price, m = Rs. 15 per liter

The ratio can be found as follows.

    \[\frac{Quantity \enspace of \enspace cheaper}{Quantity \enspace of \enspace dearer} = \frac{x - m}{m - y}\]

= (20 – 15) / (15 – 0)
= 5 / 15

The ratio of water to milk in the mixture is 5:15 or 1:3.

Problem 3
In what ratio must the water be mixed with milk to get a gain of 50% on selling the mixture at cost price?

Solution
Let the cost price of milk  = x per liter

Cost of price of cheaper quantity, y = 0 (The cost price of water is considered zero)

Selling price of mixture = x per liter, profit = 50%
= { 100 / (100 + P) } × SP
= { 100 / (100 + 50) } × x
= { 100 / (100 + 50) } × x
= (100 / 150) × x = (2/3)x

Cost price of mixture = mean price = (2 / 3)x

The ratio of quantities added in the mixture can be found by using the rule of alligation.

    \[\frac{Quantity \enspace of \enspace cheaper}{Quantity \enspace of \enspace dearer} = \frac{x - m}{m - y}\]

= { x – (2 / 3)x} / (2/3)x
= { x – (2 / 3)x} / (2/3x)x
= (x/3) / (2/3)x

or

(Quantity of water) / (Quantity of milk) = (x / 3) × (3 / 2x)
= (Quantity of water) / (Quantity of milk) = 1 / 2

The ratio of water to milk in the mixture is 1:2.

Whenever we deal with problems in which a certain quantity of pure liquid is replaced by water. For example, a milkman often replaces pure milk by water. Problems regarding finding out the quantity of pure liquid can be found out by using the following method.

Method of Repeated Dilutions

One of the important and commonly asked mixture and alligation question is the problem of repeated dilution. Suppose we have x units of a pure liquid. From which y units are taken out and replaced with water. Moreover, the process is performed n times. Then, the amount of pure liquid can be found as follows.

Amount of pure liquid remaining = [x{ 1 – (y / x) }n] units

Example 1
A container of juice contains 10 liters of juice. 0.5 liters of juice is taken out and replaced by water. The process is repeated 3 times more. Find the quantity of the remaining liquid at the end.

Solution
Amount of pure juice, x = 10

Amount of juice taken out, y = 0.5 = 1/2

Number of times process performed, n = 4 (1 time initially and 3 times repeated)

The amount of juice remaining can be found out using method repeated dilutions.

Amount of pure liquid remaining = [x{ 1 – (y / x) }n] units

Juice remaining = [10{ 1 – (1/2) / (10) }4]
= [10{ 1 – (1/20) }4]
= 10 × (19/20)4= 8.14

Amount of pure juice remaining is 8.14 liters.

Example 2
6 liters of wine is replaced by water from a container full of wine and this process was repeated one time more. Finally, the ratio of wine to water became 9:25. How much wine was there in the container originally?

Solution
Initially, the amount of pure wine = x

Amount of wine replaced = y = 6 liters

Number of times process performed  = n = 2

x can be found as follows.

Amount of pure liquid remaining = [x{ 1 – (y / x) }n] units

= [x{ 1 – (6 / x) }2]
= (9 / 25)x = [x{ 1 – (6 / x) }2]
= (3 / 5)2  = { 1 – (6 / x) }2
= 3 / 5 = 1 – (6 / x)
= 3 / 5 = (x – 6) / x
= 3x = 5x – 30
= 2x = 30
= x = 15

15 liters of wine was present originally.

If you like this post, share this content with your friends and most importantly don’t forget to subscribe to my blog to get this type of amazing post directly into your email id.

 

Share this post