Knowing that alligation is used to determine the mean value of a mixture when the ratio and quantity of the materials blended are different as well as the proportion in which the elements are mixed is necessary to solve mixture and alligation problems.
It could sound a little difficult at first, but as the candidate answers questions based on it, the idea becomes apparent.
Candidates must be familiar with a set of formulas for each topic in order to answer any question requiring numerical abilities. In order to assist candidates with this and make it easier for them to answer the mixing and alignment problems, the following formulas are provided:
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
The allegation can be applied to anyone substance 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 the ratio will be reciprocal of the difference i.e. 13:8.
In exams, you don’t need to write everything. First, you must choose those data are which are less complicated. Then, apply the allegation directly.
mix1 = 7+22 = 29,
mix2= 21+ 37 = 58
fin. mix = 25+62 = 87
Now apply allegation on milk after equating the quantity of all mixtures.
42 50 63
8 13
So, the answer is 13:8.
You can practice important questions from the mixture and allegation chapter in the BYJU'S Exam Prep Bank PO Test Series.
Example 2: The ratio of land and water on our planet is 1: 2. If this ratio is 2 : 3 in the northern hemisphere, what is it in the southern hemisphere?
Solution: In this question, if you understand the language of the question then it is the easiest one.
In a planet, there are two parts: The northern hemisphere and the southern hemisphere.
Let the total area of the planet be 60. (Why do we assume 60? Because ratios are given, 1+2 = 3 and 2+3 = 5, it should be multiple of this to make calculation easy.)
The ratio of land and water on the 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, the ratio of land and water in the northern hemisphere. = 2 : 3
So, land on north. hemisphere = (2/5)×30 = 12
water on the north. Hemi = (3/5)×30 = 18
The remaining land and water of the planet will be in the southern hemisphere.
land on south hemisphere = land on the planet – land on north hemisphere
= 20- 12 = 8
water in the southern hemisphere = water on the planet – water in the northern hemisphere
= 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 fill with the two such that it contains milk and water in the ratio of 4: 1. What quantity of the mixture should be picked from the first vessel?
Solution: The quantity of the final mixture is 35 litres and the 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 from mixture 1 and 20 litres of milk from 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 mixture 2 has the same ratio as it is in question
20: 4 i.e. 5:1
So, the 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 1/5^{th} 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.
The last line of the question says A saves 1/5^{th} of his income which means if he has 5 Rs. he will save only 1 Rs.
So, the ratio of income and expenditure is 5:4
A B C
Income 4: 5: 6
Expense. 8: 9: 10
Now, try to make income and expenditure of A in ratio 5: 4, for this we have to multiply the income ratio by 5 and the expenditure ratio by 2.
Income 20: 25: 30
Expense. 16: 18: 20
Savings 4 : 7: 10
Attend the quality sessions by exam-qualified experts and clear your concepts in the mixture and allegation chapter at BYJU'S Exam Prep Online Classroom Program.
Example 5: A mixture contains wine and water in the ratio of 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 the same which is not replaced. As in this mixture is not getting replaced by wine, so we make the wine ratio the same.
Wine: Water
8 : 5 = 13
to make it equal in the 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 of 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 the previous one.
first, make 5: 6 near to 10:12
So, Initially Wine: water
7 : 12 = 19
To get the desired ratio of 10:12, we have to add 3 litres 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 of 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 this mixture is replaced with wine, so we will try to keep the water the same.
wine: water
mixture1 8: 13
mixture2 4: 5
To make the water ratio the same, we will multiply mixture1 by 5 and mixture2 by 13.
So, mix1 40: 65
mix2 52: 65
Hence, the part of the 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 of 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 the 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.
The method is applied on which is not changing.
In this case, milk is not replaced, so we will apply it.
So, initial milk Conc. × change in mixture = final milk concentration.
(3/7) × X = (1/3)
X = 7/9
We can say that initially there is a total of 9 parts and (9-7)=2 parts were removed.
2 parts = 7 litres
So, 9 parts = 63/2 = 31.5 litres.
You can check other quantitative Aptitude study materials here:
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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 the 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 the 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 an 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 the same. So, the ratio of solution after withdrawal to after replacement will be the same every time. Let’s say it will be X
Initial concentration. of milk × X × X = final concentration. of milk
(25/100)× X^{2} = 9/100
X^{2} = 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
The chapter can be useful for the following exams:
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