#### Mixtures and Alligations

Mixtures and alligations is a concept that is covered in the Arithmetic topic. Each year in the competitive exams held for admission in the PGDM/MBA programs this concept forms an important part of the Quants section. Out of the total questions from arithmetic concepts, mixtures and alligations form around 10% of the questions. The CAT questions on this concept are asked repeatedly, and from the trends of previous years CAT papers, it is evident that one or two questions are asked every year. Other than the Quant section, this topic becomes of use in solving the Data interpretation and Logical reasoning questions also at times.

CAT :

 Mixtures and Alligations Year No of Questions Good attempt Difficulty 2019 Slot 1 1 1 Easy to Moderate Slot 2 1 1 Easy to Moderate 2018 Slot 1 2 1 to 2 Moderate Slot 2 2 1 Moderate to Difficult 2017 Slot 1 1 1 Moderate Slot 2 1 1 Moderate

SNAP :

 Mixtures and Alligations Year Number of Questions Good attempt Difficulty 2020 1 1 Easy to Moderate 2019 1 1 Moderate 2018 1 1 Easy to Moderate

CAT 2021 is expected to follow a similar difficulty level and paper pattern like previous years.

### List of Concepts in that Chapter

Mixtures
and Alligations

Being thorough with this concept can fetch a student marks as the CAT questions from this topic are usually easy to moderate and follow a similar pattern.

If one mixes two different qualities of things, it gives rise to a mixture. The quality of the resulting mixture lies between the qualities of the original constituents.
The quantified value of the resultant mixture will be more than the value of the lowest quality component and less than the value of the highest quality component mixed.

This concept is useful when one wants to increase the profit, increase discounts without compromising on the margins etc.

The basic concept of weighted averages gives rise to the concept of mixtures.

In this case, the weighted average of the mixture will be the resulting value of the mixture.
It is given by:

If qitems of p1 price are mixed with q2 items of pprice, the resulting mixture will be priced at:

P = (q1p1 + q2p2) / (q1+q2)

The generalized formula for the concept explained above is:

P = (q1p1 + q2p+ …… qnpn) / (q1+q2+… qn)

Example: If 10 litres of 80% alcohol is mixed with 15 litres of 60% alcohol, the concentration of the solution is given by:

C = (10*0.80 + 15*0.60) / (10+15)

= 68%

There can be two methods to solve the alligations questions. Being thorough with this concept can fetch a student marks as the CAT questions from this topic are usually easy to moderate and follow a similar pattern.

If one mixes two different qualities of things, it gives rise to a mixture. The quality of the resulting mixture lies between the qualities of the original constituents. The quantified value of the resultant mixture will be more than the value of the lowest quality component and less than the value of the highest quality component mixed.

This concept is useful when one wants to increase the profit, increase discounts without compromising on the margins etc.

The basic concept of weighted averages gives rise to the concept of mixtures.

In this case, the weighted average of the mixture will be the resulting value of the mixture. It is given by:

If q1 items of p1 price are mixed with q2 items of p2 price, the resulting mixture will be priced at:

P = (q1p1 + q2p2) / (q1+q2)

The generalized formula for the concept explained above is:

P = (q1p1 + q2p2 + …… qnpn) / (q1+q2+… qn)

Example: If 10 litres of 80% alcohol is mixed with 15 litres of 60% alcohol, the concentration of the solution is given by:

C = (10*0.80 + 15*0.60) / (10+15)

= 68%

There can be two methods to solve the alligations questions

Method 1

The formula which is also known as the rule of alligation and used to find the ratio in which the ingredients are mixed is Method 2
This concept comes in handy when candidates come across the question of repeated dilution and they need to find the pure quantity left after ‘n’ number of processes of repeated replacement is done on the pure quantity. Suppose a container contains ‘x’ units of a liquid from which ‘y’ units are taken out and replaced by water. Repeating this experiment ‘n’ times, you have to calculate the pure liquid left in the container. The formula for the same is:

Quantity of pure liquid left = x(1-y/x) n

### Some Questions from Previous Papers

Some important CAT questions on mixtures and alligations from previous year papers are given below which can help in the preparation of CAT 2021

1. A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half a litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is? (CAT 2019-Slot 1)

Weight of half litre of the liquid 1 and liquid 2 will be 500 and 400 gm respectively.
So required ratio of liquid 1 and 2 in mixture =(480n-400)/ (500 -480) =80/20 = 4:1
Therefore, the percentage of liquid 1 in the mixture= 4/ (4+1) *100 = 80

2. The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of the salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A, is? (CAT 2019-Slot 2)

Volume of solution in each vessel = 500ml
Salt in each vessel, in A = 10% of 500 = 50 ml
In B = 22 % of 500 = 110 ml
In C = 32% of 500 = 160 ml
When 100 ml of the solution in vessel A is transferred to vessel B, salt in B = 110 + 10% of 100 = 110 + 10 = 120 ml % of salt in B = 120/600 = 20%
So, when 100 ml of the solution in vessel B is transferred to vessel C = 160 + 20% of 100 = 160 + 20 = 180 % of salt in C = 180/600 *100 = 30 %
Now when 100 ml of the solution in vessel C is transferred to vessel A, volume of solution in A = 400+100 = 500
Salt in A = 10% of 400 + 30% of 100 = 40 + 30 = 70
% of salt in A = 70/500 *100 = 14 %

3. A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is? (CAT 2018-Slot 1)

Let the quantities of the paints A and B in the mixture sold be a litres and b litres respectively.
Value at which the entire mixture is sold=264 Profit percent made=10%
Value at which the entire mixture is bought = 264*100/110 = 240
Price at which the entire mixture is bought=24 per litre
Let the cost of B be x per litre.
Cost of A=(x+8) per litre
((x+8) *a+ xb)/10 = 24
Maximum cost of B will occur when a is minimum. b<=a. So, minimum a is 5.
Corresponding b is 5. Then (x+8) * (5) +x*(5) =240 x=20

4. A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts? (CAT 2018-Slot 1)

Let the cost price of peanuts for the wholesaler be x per kg.
Cost price of walnuts for the wholesaler is 3x per kg.
The wholesaler sold 8 kg of peanuts at 10% profit and 16 kg of walnuts at 20% profit to a shopkeeper.
Total cost price to the shopkeeper = (8) *(x)*(1.1) + 16*(3x) *(1.2) = 66.4x
The shopkeeper lost 5 kg walnuts and 3 kg peanuts.
The shopkeeper sold the mixture of 11 kg walnuts and 5 kg peanuts.
His total selling price=166*(16) = 2656
His total cost price= 2656 (100/125) =2124.8
Hence, 66.4x= 2124.8
Price at which the wholesaler bought walnuts = 3x = 96/- per kg

5.A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is? (CAT 2018-Slot 2)

Let the volume of the first and the second Solution be 100 and 300. When they are mixed, quantity of ethanol in the mixture
= (20 + 300S)
Let this Solution be mixed with equal volume i.e. 400 of third Solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final Solution
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or, S=50%
Hence, 50 is the correct answer.

6.A jar contains a mixture of 175 ml water and 700 ml of alcohol. Gopal takes out 10% of the mixture and substitutes it by the water of the same amount. The process is repeated once again. The percentage of water in the mixture is now? (CAT 2018-Slot 2)

Final quantity of alcohol in the mixture = (700/ (700+175)) *((90)2/100) * (700+175) = 567 ml
Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml
Hence, the percentage of water in the mixture = (308/875) *100 = 35.2%

### How to deal with that topic preparation

For the understanding of the candidate, we have divided the preparation into 3 levels for the upcoming CAT 2021.

Level 1
If a candidate is in the initial stages of preparation, he should follow the following steps:

• Try and understand the basic concepts and formulas coming under mixtures and allegations and understand the difference between the two methods to solve the alligation questions
• Do not go to shortcut methods for solving the questions because that might hamper the understanding of the concept
• Avoid using calculators and practice calculations which will strengthen the basics

Level 2

This level is for the students who have cleared their concepts and are looking for faster methods to solve questions

•  Try to understand the techniques always applicable in the questions of this concept and short cuts that can be used
• Since the ratios of the mixtures used in the questions remain more or less the same, try and memorize a few basic ratio values to speed up the calculations
• At this stage, the student should be comfortable in dilution and profit concepts
Level 3
After the student has a good grasp of the basics of the concept and understands under which scenarios the shortcuts can be applied, he reaches this level
• Try to solve questions from previous years CAT, papers and look for solutions available
• Look at the different approaches used to solve the same question and how can the time be effectively reduced on using a particular approach
• CAT judges time management and the questions from this concept are usually easy to moderate and can be cracked in less time if the approach is right. Hence, try and optimize the approach basis the type of question