Dear Readers,

Today we are going to discuss Mixture and Allegation topic of Quantitative section for Bank and Insurance Exams. Students feel that Mixture and Allegation is the hardest topic but after studying these notes you will say it’s a myth. Before Mixture and Allegation, we suggest you to go through Ratio and Proportion, this will help you to clear your doubts.

We will help you understand this topic by examples which are already discussed in Video session conducted by Meritshine.

Example1: Milk and water are mixed in a vessel in the ratio 7 : 22 and in another vessel in the ratio

21 : 37. In what ratio should the two be mixed to get milk and water in the ratio 25 : 62 in the resultant mixture?

Solution: Milk : Water

Mixture1 7 : 22 29

Mixture2 21 : 37 58

Final Mix. 25 : 62 87

Now, we try to always make the same quantity in all mixtures. We can see that 29, 58 and 87 are multiples of 29, if we take LCM to make quantity similar it will be 174.

Mixture 1 42 : 132

Mixture 2 63 : 111

Final Mix. 50 : 124

Allegation can be apply on any one subsatnces of the solution. We can apply it on either Milk or water, but we will apply it on milk as its calculation seems easier.

Mixture 1 Final Mix. Mixture2

42 50 63

Difference: 8 13

So ratio will be reciprocal of difference i.e. 13:8.

In exams you don’t need to write everything. First you must chose those data are which are less complicate. Then, apply allegation directly.

mix1 = 7+22 = 29,

mix2= 21+ 37 = 58

fin.mix = 25+62 = 87

Now apply allegation on milk after equating quantity of all mixtures.

42 50 63

8 13

So, answer is 13:8.

Example 2: The ratio of land and water on our planet is 1 : 2. If this ratio 2 : 3 in the northern hemisphere, what is it in the southern hemisphere?

Solution: In this question, if you understand the language of question then it is the easiest one.

In a planet, there are two parts: Northern hemisphere and southern hemisphere.

Let total area of planet is 60. (Why we assume 60? Because ratios given, 1+2 = 3 and 2+3 = 5, it should be multiple of this to make calculation easy.)

Ratio of land and water on planet = 1:2

So land on planet = (1/3)×60 = 20

Water on planet = (2/3)× 60 = 40

Area of northern hemisphere = area of southern hemisphere = 30

Given, ratio of land and water on north hemi. = 2 : 3

So, land on north.hemi = (2/5)×30 = 12

water on north. Hemi = (3/5)×30 = 18

Ramaining land and water of planet will be on southern hemi.

land on south hemi = land on planet – land on north hemi

= 20- 12 = 8

water on south hemi = water on planet – water on north hemi

= 40 – 18 = 22

Hence ratio is 8:22 i.e. 4:11

Example3: A vessel contains milk and water in the ratio 8 : 3 while another contains them in the ratio 5 : 1. A 35 litre vessel is to filled with the two such that it contains milk and water in the ratio 4 : 1. What quantity of the mixture should be picked from the first vessel?

Solution: Quantity of final mixture is 35 litre and ratio of milk and water is 4:1.

So, it means that milk = (4/5)×35 = 28 litres

water = (1/5)× 35 = 7 litres

Now if we analyse the given data, we can see that if we take 8 litres of milk form mixture 1 and 20 litres of milk form mixture 2, then our resultant mixture will contain 28 litres of milk.

Similarly, water from mix.1 = 3 litres and from mix.2 = 4 litres, then resultant mix. Will have 7 litres of water.

You must realise that milk and water taken from the mixture 2 has the same ratio as it is in question

20 : 4 i.e. 5:1

So, quantity of mixture 1 taken is 8+3 = 11

Example 4: The incomes of A, B, C are in the ratio 4 : 5 : 6, and their expenditures are in the ratio 8 : 9 : 10. If A saves 1/5th of his income, in what ratio do they save?

Solution: This question seems to be tough but it is not. If you solve these questions practically, it will be very easy for you.

Last line of question says A saves 1/5th of his income it means, if he has 5 Rs. he will save only 1 Rs.

So, ratio of income and expenditure is 5:4

A B C

Income 4 : 5 : 6

Expen. 8 : 9 : 10

Now, try to make income and expenditure of A in ratio 5:4 , for this we have to multiply income ratio by 5 and expenditure ratio by 2.

Income 20 : 25 : 30

Expen. 16 : 18 : 20

Savings 4 : 7 : 10

Example 5: A mixture contains wine and water in the ratio 8 : 5. What part of the mixture must be removed and replaced with water for the resultant mixture to have equal quantities of water and wine?

Solution: In this type, we try to make the quantity same which is not replacing. As in this mixture is not getting replaced by wine, so we make wine ratio same.

Wine : Water

8 : 5 = 13

to make it equal in water we have to add 3 units of water So

8 : 8 = 16

So, the part which is removed = (16-13)/16 = 3/16 part is removed and replaced with water.

Example 6: A mixture contains wine and water in the ratio 7 : 12. What part of the mixture must be removed and replaced with wine for the resultant mixture to have wine and water in the ratio

5 : 6?

Solution: Applying the same method as previous one.

first, make 5: 6 near to 10:12

So, Initially Wine : water

7 : 12 = 19

To get desired ratio 10:12, we have to add 3 litrs of wine

and 10 +12 = 22

Hence the mixture to be replaced is (22-19)/22 = 3/22

Example 7: A mixture contains wine and water in the ratio 8 : 13. What part of the mixture must be removed and replaced with wine for the resultant mixture to have wine and water in the ratio

4 : 5?

Solution: As in this mixture is replaced with wine, so we will try to keep the water same.

wine : water

mixture1 8 : 13

mixture2 4 : 5

To make the water ratio same, we will multiply mixture1 by 5 and mixture2 by 13.

So, mix1 40 : 65

mix2 52 : 65

Hence, the part of mixture to be replaced is (52-40)/(52+65) = 12/117

Example 8: From a solution containing milk and water in the ratio 3 : 4, 7 litres is drawn off and replaced by water. If the resultant mixture contains milk and water in the ratio 1 : 2, then what was the volume of the original solution?

Solution: milk is not getting replaced so we will try to make the ratio of milk same.

Milk : water

Mixture1 3 : 4

Mixture2 1 : 2

multiplying mix2 by 3

mixture2 3 : 6

According to our method, mixture to be replaced = (6-4)/(6+3) = 2/9

and 2 parts of 9 is equal to 7 litres

i.e. 2 = 7

and 9 = 7×9/2 = 63/2 = 31.5

Hence, volume of original solution is 31.5 litres.

### Another approach

This approach is opposite to the previous one discussed above.

Method is applied on which is not changing.

In this case milk is not replaced, so we will apply on it.

So, initial milk Conc. × change in mixture = final milk conc.

(3/7) × X = (1/3)

X = 7/9

We can say that initially there are total 9 parts and (9-7)=2 parts were removed.

2 parts = 7 litres

So, 9 parts = 63/2 = 31.5 litres.

Example 9: From a 30 litre solution containing milk and water in the ratio 4 : 5, 7.5 litres is drawn off and replaced by water. Then 9 litres is drawn off and replaced by water. Finally, 12 litres is drawn off and replaced by water. What is the volume of milk in the resultant mixture?

Solution: As mixture is getting replaced by water so we will apply it for milk.

Final volume of milk = Final total volume × (Conc. Of not changing substances)× (quantity remained after drawn / net quantity after replaced)× n times the process is going…..

You can see that in each replacing process, the same quantity removed and same quantity replaced So, after replacing the quantity will remain same as 30 litres.

Final volume of milk = 30× (4/9)× (22.5/30)× (21/30)× (18/30)

= 30× (4/9)× (3/4)× (7/10)× (3/5)

= 4.2 litres

Example 10: From a 18 litre solution containing milk and water in the ratio 3 : 4, 3 litres

is drawn off and replaced by 5 litres of water. Then 4 litres is drawn off and replaced by 5 litres of water. Finally, 7 litres is drawn off and replaced by 2 litres of water. What is the volume of milk in the resultant mixture?

Solution: At the first replacing process:

Initial solution = 18 litres

solution after withdrawal = 18-3 = 15 litres

Solution after replacing = 15+5 = 20

At the second replacing process:

Initial solution = 20

Solution after withdrawal = 20-4 = 16

Solution after replacing = 16+5 =21

At the third replacing process:

Initial solution = 21

Solution after withdrawal = 21-7 = 14

Solution after replacing = 14+2 = 16

So, Volume of milk in resultant mixture = 16 × (3/7)× (15/20)× (16/21)× (14/16) = 24/7 = 3.43

Example 11: A 10 litre vessel contains 25% milk and the rest is water. Some of it is thrown and replaced with water. Then this operation is done a second time. If the resultant mixture contains just 9% milk, what was the quantity of the mixture removed each time?

Solution: The volume which is replaced is always same. So, ratio of solution after withdrawal to after replacing will be same every time. Let’s say it will be X

Initial conc. of milk × X × X = final conc. of milk

(25/100)× X2 = 9/100

X2 = 9/25

X = 3/5

X = (vol. after withdrawal/vol. after replaced) = 3/5

Vol.after replaced is 10

So, 5= 10

1 = 2

and 3 = 6

Volume after withdrawal = 6

withdrawal volume = 10 – 6 = 4 litres

Thanks,

Team

**AIM**SUCCESS
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