Mixtures
When two or more than two pure substances/mixtures are mixed in a certain ratio, they create a mixture. Here we shall confine ourselves to mostly homogenous mixtures in view of the questions commonly asked in CAT.
Mixing without Replacement
In this particular type of mixing, two or more than two substances are mixed without any part of any mixture being replaced.Example1
In a mixture of 420 litres the ratio of milk and water is 6:1. Now, 120 litres of the water is added to the mixture. What is the ratio of milk and water in the final mixture?
Solution
Volume of milk = 360 litres and volume of water = 60 litres.
When 120 litres of water is added, volume of water = 180 litres
So, the ratio of milk water = 2:1
Example2
A milkman mixes 20 litres of water with 80 litres of milk. After selling onefourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
Solution
Ratio of milk and water = 20:80
When 1/4th of this mixture is sold, total volume of mixture will be reduced by 25%, so 25% of milk and water both will reduce. So, volume of milk and water after selling out 1/4th of mixture = 60 litres and 15 litres respectively. Addition of 25 litres of water will finally give us the following: volume of milk = 60 litres and volume of water = 40 litres. Hence, the ratio of water and milk = 40:60 = 2:3.
Example3
How many litres of fresh water should be mixed with 30 litres of 50% milk solution so that resultant solution is a 10% milk solution?
Solution
Method 1 Using Alligation
So, the ratio of fresh water added: milk solution = 4:1
Hence, 120 litres of fresh water should be added.
Method 2 Principle of constant volume of one component
Since we add fresh water, the volume of milk will be constant.
Now volume of milk = 15 litres = 10% of the new mixture.
So, 100% of the new mixture = 150 litres
So, volume of fresh water added = 150 âˆ’ 30 = 120 litres.
Method 3 Principle of inverse proportion
We know that concentration is inversely proportional to the volume of solute added.
So, in this case 30 Ã— 50% = 10% Ã— (30 + x), where x is the volume of water added.
So, x = 120 litres
Method 4 Using equation
In the final mixture,
So, x = 120 litres
Mixing with replacement
In this particular type of mixing, two or more than two substances are mixed by replacing some part of a mixture. In these types of questions, total volume may or may not be the same and information regarding the same can be obtained from the question.Case 1
When the quantity withdrawn and the quantity replaced are of the same volume.

1st operation 
2nd operation 
3rd operation 

Taken out 
Left 
Taken out 
Left 
Taken out 
Left 

Milk 
4 
36 
3.6 
32.4 
3.24 
29.16 
Water 
0 
4 
0.4 
7.6 
0.76 
10.84 
It can be seen that the quantity of water or milk withdrawn is 10% of the existing volume of milk or water because only 10% of the total volume of 40 litres taken out.
Case 2
When the quantity withdrawn and the quantity replaced are of the same volume, but the total volume before replacement does not remain the same.
Milk left =
Milk left =
Case 3
When the quantity withdrawn and the quantity replaced are not of the same volume.
Obviously, there will be 36 litres of milk and 5 litres of water.
Milk left =
=
Example
Two vessels A and B of equal capacities contain mixtures of milk and water in the ratio 4:1 and 3:1, respectively. 25% of the mixture from A is taken out and added to B. After mixing it thoroughly, an equal amount is taken out from B and added back to A. The ratio of milk to water in vessel A after the second operation is
 79:21
 83:17
 77:23
 81:19
Solution
Assume there is 20 litres of the mixture in both the vessels.
In vessel A, milk = 16 litres and water = 4 litres
25% from A to B = milk in B = 15 + 4 = 19 litres
= water in B = 5 + 1 = 6 litres
ratio = 19:6
Equal amount from vessel B to vessel A
= milk in
= water in
Hence, the ratio is 79:21
= milk in
= water in