Alligations and Mixture Formulas

July 12, 2023

Formulas For Alligation And Mixture

A mixture, as the name suggests is mixing two or more things together and alligation enables us to find the ratio in which the ingredients/ things have been mixed to form the mixutre.

The most fundamental point to remember while solving mixtures and alligations is that alligation is a way to find the mean value of mixture when the ratio and amount of the ingredients mixed are different and also to find the proportion in which the elements are mixed.

 

allegations and mixtures formula
Mixture and Alligation practice questions

Formulas to Solve Alligations and Mixtures question :-

What is an Alligation?

When two ingredients X and Y of price p and q respectively are mixed together,  such that the price of the resultant mixture is M (mean price), then the ratio (R) in which ingredients are mixed is given by, the rule of the alligation.

(Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

What is a Mixture?

In a mixture,  two or more ingredients are mixed together to get a desired quantity. The quantity can be expressed as ratio or percentage.

For example: When two varieties of sugar are mixed to form a new variety of sugar then it is called as a mixture.

What is a Mean Price?

The cost of a unit quantity of the mixture is called the mean price.

Formulas to Solve Mixture and Alligations

Alligation and Mixture Formulas 1

When two commodities are mixed then ,

 

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Alligation and Mixture Formulas 2

Consider a container contain x unit of liquid A from which y units are taken out and replaced by water. This operation is repeated n number of times, then the quantity of pure liquid will be given by the formula:

Quantity =  \left[x – (1 – \left(\frac{Y}{X}\right)^n \right] units

Therefore,  (Cheaper Commodity) : (Dearer Commodity) =  (d – m)  :  (m-c) 

Alligation and Mixture Formulas 3

Calculate quantity of pure Liquid after ‘n’ successive operations,

If a container contains ‘x’ units of pure liquid , and we replace the liquid with ‘y’  units of water ,

Then after ‘n’ successive operations, the units of pure liquid left is ,

After n operations, the quantity of pure liquid =x1 — yn units.
x

Criss-Cross Method

Some aspirants use the above method in different format, which we call criss-cross method. Below is the format:

The working is, we take the positive difference of mean price and cheaper price and write the difference in the place of Quantity of dearer price. Similarly, take the positive difference of mean price and dearer price and write the difference in the place of Quantity of cheaper price.

Rule of Constant

There are another types of questions on mixtures and alligation where the quantity of one element in the mixture does not change while adding another element to the first mixture. For such questions, I recommend an alternate method of using the rule of constant to get to the answer. The concept uses the simple understanding of percentages. Let us learn it with the help of an example.

Method 1: School textbook approach

We assume that the quantity of water added to be x litres. The quantity of milk in the existing solution is 30% of 40 = 12 litres, with the addition of water, the quantity of new solution becomes (40 + x) litres. As per the problem, the percentage of milk in new solution should be 15 %. we will get x = 40.

Method 2: Rule of Alligation.

We assume that the two solutions of milk and water are added to get the new solution and apply the approach we used in Example 3.

Taking milk as the common element in both the solutions, we have 30% milk in first and 0% milk in the second solution (i.e. pure water). On mixing them, we got 15% milk in the final solution. Therefore,

Or the ratio of the quantity of first and second solution should be 15:15 = 1:1

Hence, 40 litres of pure water should be mixed to get the desired new solution.

Method 3: Rule of Constant

In this rule, we target the element in the mixture whose amount does not change but its percentage changes because of the change in the total amount of the mixture.

As we calculated above, the quantity of milk in the first solution is 12 litres, and it will remain same in the new solution as well. That is,

12 liters = 30% of the first solution =15% of the new solution

Question 1 :

A and B are two alloys of iron and silver prepared by mixing metals in the ratio 4 : 5 and 7 : 5 respectively. If equal quantities of alloys are melted to form a third alloy C, the ratio of iron to silver in C is

Solution :

In alloy A, the ratio of iron to Silver=4:5, 5+4=9

In alloy B, the ratio of iron to Silver=7:5, 7+5=12, as the amount to be mixed.

So take the LCM of 9 and 12. hence, we mix 36 gm of A and 36 gm of B.

(the reason for choosing LCM as the amount is to simplify calculations.)

For a: amount of iron=\frac{4}{9}\times36=16 and amount of silver \frac{5}{9}\times36=20

For B: amount of iron=\frac{7}{12}\times36=21 and amount of silver=\frac{5}{12}\times36=15

Hence, Total amount of iron=16+21=37

Total amount of silver=20+15=35

Hence, Final ratio of iron to Silver in the mix of two alloys= 37:35

Question 2 :

Mixture of alcohol and water contains 35% of alcohol by volume. Then, 40 ml of water is added to such a mixture of 100 ml. The percentage of alcohol in the new mixture is?

Solution : 

 Here percentage alcohol in the mixture is to be found out. So, the quantity of alcoholis kept as numerator
\frac{Alcohol}{Mixture}=\frac{q}{q_{a}+{q_{w}}}=\frac{0.35\times100}{100+40}=\frac{35}{140}
 (since 40 ml water is added
 
Quantity of mixture=100+40).
 
% alcohol = \frac{35}{140}\times100=25%
 
 the mixture contains 20% alcohol.
 

Question 3 : 

Two containers of equal volume contains milk and water in the ratio 3: 5 and 5: 3, respectively. If the contents of both containers are emptied into a third one, what would be the ratio of milk to water in that container?

Solution : 

Let the volume of both the container b x units each. Thus, the first container would contain \frac{3}{8}x units of milk and \frac{5}{8}x units of water, whereas the second container would contain \frac{5}{8}xunits of milk and \frac{3}{8}x units of water. When the content of both are empty into a third one it would have,
 
Milk          \left ( \frac{3}{8}x \right )+\left ( \frac{5}{8}x \right )                    …(1)
 
And water   \left ( \frac{5}{8}x \right )+\left ( \frac{3}{8}x \right )                    ….(2)
 
Thus, the required  ratio. x:x = 1 : 1

Question 4 :

From two ornament Weighing 18 gram and 24 gram containing gold and silver in the ratio of 2: 1 and 5: 1 respectively, a new ornament is made. What is the amount of gold in the new ornament?

Solution : 

from the first ornament, gold = \frac{2}{3}\times18=12

From the second ornament gold= \frac{5}{6}\times24=20

Thus, total gold=32 gm.

Question 5 :

A sum of money is sufficient to pay Sachin’s salary of 45 days and kale’s salary for 60 days. For how many days can the sum pay the salary of both?

Solution: 

Sachin’s salary for one day = 1/45th of the sum of money.

For kale’s salary for one day = 1/60 of the sum of money.

=> (Sachin + kale)’s salary for one day= 1/45+1/60=7/180th of the sum.

Hence, when they both work together, the sun will last for 180/7 days.

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