- Sequential Output Tracing (Input Output)
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- Critical Reasoning: Assumptions & Conclusions
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- Complex Arrangements
- Critical Reasoning: Course of Action
- Critical Path
- Direction and Word Puzzle
- Games And Tournaments
- Linear Arrangement
- Team Selection
- Venn Diagram
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Following Mixtures and Alligation Questions have been asked recently in past CAT papers:

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)

Answer

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)

Answer

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 %

Answer

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

Answer

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.

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 method for 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 alligation 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 alligation 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 alligation method to solve the 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

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 alligation and understand the difference between the two methods to solve the alligation questions. You can use MBAP CAT E Book (Concept Theory). You can also listen to Live Lecture Recording (Basic) on basic concept and MBAP lecture Assignment.

•Do not go to mixture and alligation shortcut tricks for solving the questions because that might hamper the understanding of the concept.

• Avoid using calculators and practice calculations for mixtures and alligation 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 mixtures and alligation problems and learn the concept and practice questions using MBAP CAT E Book (Practice Questions) and MBAP CAT Advance E books for higher level questions.

• 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. One should practice mixtures and alligation previous year questions from MBAP Topic wise Previous Year CAT Questions.

**Level 3**

After the student has a good grasp of the basics of the mixtures and alligation concept and understands under which scenarios the shortcuts can be applied, he reaches this level

• Try to solve mixtures and alligation questions from previous years CAT papers and look for solutions available. For this you can use MBAP Previous Year CAT paper.

• 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 questions are meant to judge time management and the questions from mixtures and alligation 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

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