The topic is very popular among the aptitude tests & IQ tests world wide. It is basically the application of the fundamental arithmetic concepts like percentage, ratio, average etc. The mixtures under consideration will be homogeneous in nature.

TYPES OF MIXTURES:- A mixture is not necessarily made up of liquids or solids. There can be various types such as.

• A liquid and a solid
• Two liquids
• Two solids
• Profit percent on two products
• Boys & girls in a group
• A journey covered with different speeds
• and many more

ALLIGATION RULE: – This is nothing but weighted mean concept to solve a question where we mix two things to make a mixture. Let’s understand it with a simple example.

Cost of liquid A is Rs. 40  per kg while the cost of liquid B is Rs. 50 per kg. What will be the cost of the mixture if A & B are mixed in the ratio 3:1?
Sol:- It is obvious that the price of the mixture will be somewhere between 40 & 50. As the ratio is given, we can take 3 kgs of A which costs 3×40 = 120 and 1 kg of B which costs 1×50=50. So total cost of the mixture becomes 120+50 = 170. This 170 is the cost of 4 kgs of mixture. Thus, cost of the mixture per kg is 170/4 = Rs. 42.5 .

Alligation method solution:-

50-M=3M-120
4M=170
M=42.5

Thus, cost of the mixture per kg is Rs. 42.5 .

Larger value can be called as dearer value and smaller value can be called as Cheaper value. Value of the mixture can be called mean value. The ratio in which we are mixing the Dearer value and the Cheaper value is X: Y. In general we can write:

(D-M)/(M-C) = Y/X

Let’s try one more case before we proceed to the examples:-
Cost of good quality sugar is Rs 60 per kg and the cost of poor quality sugar is Rs 40 per kg in what ratio these two should be mixed so that the cost of the mixture is Rs 48 per Kg

Sol:-

X/Y = (48-40)/(60-48) = 8/12 = ( 2)/( 3)

Let’s now move on to try few good examples:-

Example 1:- 60% of the boys passed in the exam while 80 % of the girla passed in the exams and 64% of the students passed in the exam, what is the number of boys in the class of 300 students.

Sol:-

X/Y = (80-64)/(64-60) = 16/4 = ( 4)/( 1)

As the ratio of boys to girls is 4:1, number of boys = (4/(4+1)) 300 = 240

Example 2:- Metal A is 7 times as heavy as silver where as metal B is 17 times as heavy as silver. In what ratio A & B should be mixed two make an alloy which is 10 times as heavy as silver?

Sol:-

X/Y = (17-10)/(10-7) = 7/( 3)

So A & B should be mixed in the ratio 7:3.

Example 3:- A car covers a distance of 400 kms in 8 hrs by covering some distance at the speed of 45 km/hr and the remaining distance at the speed of 65 km/hr. For how much time it travelled with speed of 45 km/hr?

Sol:-

X/Y = (65-50)/(50-45) = 15/5 = ( 3)/( 1)

So ans = (3/(3+1)) 8 = 6 hrs

Example 4:- Fresh Grapes contain 80 water whereas dry grapes contain 40% water. How many kgs of dry grapes can be obtained from 48 kgs of fresh grapes ?
Sol:- In this problem, if you look the whole things in the reverse order, you will be able to see that if we mix water & dry grapes, we will get fresh grapes.
Let’s take water percentage as reference and apply allegation rule.

X/Y = (100-80)/(80-40) = 20/40 = ( 1)/( 2)

This means if we mix dry fruits & water in the ratio 1:2, we will get fresh grapes. So, weight of dry grapes =(1/(1+2))48=16kgs

That means if equal quantities of the given solution & water are mixed, we will get the required solution so, we will replace 30 letres of the solution with water.

Removal and Replacement :-
The only thing that we need to understand in this type of questions is that if we take out some part of a solution, the concentrate of the original solution.

Example 5:- From 60L of 80% milk solutions.How many litres shoulb be replaced with water so that the resultant is 40% milk solution?

Sol.

X/Y = (40-0)/(80-40) = 1/1

Percentage of milk in water is zero. Using allegation, we can easily find out in what ratio the given solution & pure water should be mixed to get the desired solution.

Example 6 :- From 60L of 60% milk solution 10L is replaced with water. After that from the new mixture formed 20L is replaced with water. What is the percentage of milk in the final mixture?
Sol. As we know that if one sixth of the mixture is going out. Here the original quantity of milk is 60/100×60=36 L
So.fianl quantity of milk = 36((60-10)/60)((60-20)/60)
= 36 x (5 )/6 x 2 = 20 L
Percentage of milk = (20/60) x 100 =33.33%

Example 7 :- From a container which has pure milk in it 30% is taken out and replaced with water. Then, from the solution formed, 30% is taken out and replaced with water.Then one more time the process is repeated. What is the percentage of water in the find solution.
Sol. As we did in the previous example milk remaining
((100-30)/100)((100-30)/100)((100-30)/100)
= (7/10)^3 = 343/1000

So, % of milk remaining = 34.3
Hence,% of water = 100-34.3 =65.7%