*Tuesday, April 3rd, 2018*

The concepts of Set Theory are applicable not only in Quant / DI / LR but they can be used to solve syllogism questions as well. Let us first understand the basics of the Venn Diagram before we move on to the concept of maximum and minimum. A large number of students get confused in this so I have listed out each area separately.

**A venn diagram is used to visually represent the relationship between various sets.**

What do each of the areas in the figure represent?

*I – only A; *

*II – A and B but not C; *

*III – Only B; *

*IV – A and C but not B; *

*V – A and B and C; *

*VI – B and C but not A; *

*VII – Only C*

**n(A** **U** **B** **U** **C) = n(A) + n(B) + n(C) — n(A** **n** **B) — n(A** **n** **C) – n(B** **n** **C) + n(A** **n** **B** **n** **C)**

As per the above diagram,

n(A) = I + II + IV + V

n(B) = II + III + V + VI

n(C) = IV + V + VI + VII

n(A n B) = II + V

n(B n C) = V + VI

n(C n A) = IV + V

n(A n B n C) = V

n(A U B U C) = I + II + III + IV + V + VI + VII

Note: While doing such questions, it is advisable that you take the least no. of variables to fill up the empty space. As a practice if n(A n B n C) is missing, take that as ‘x’ and proceed.

**For Maximum and Minimum of values**, the key point to note is:

If you allot a value to the intersection, it will get added to all the individual sets but will bring down the total.

**Example: **In a survey it was found that 80% like tea whereas 70% like coffee. What is the maximum and minimum number of those who like both?

Ans: First thing to note is that no information is mentioned about the people who don’t like either of them. So that value is flexible and can change.

n(tea) = 80

n(coffee) = 70

n(total) = 100 {This includes those who like neither.}

n(tea n coffee) = ??? {We don’t know this value and it is flexible}

If we want to **maximize those who like both**, we have to maximize the value in the intersection. So, we have to minimize the value of the union.

n(tea n coffee)_{max} = 70 {It is limited by the higher of the two values}

In this case, our venn diagram will look something like this (Red is tea & Purple is coffee):

In this case 20% of people like neither tea nor coffee.

If we want to **minimize those who like both**, we have to minimize the value in the intersection. So, we have to maximize the value of the union. We know that the maximum possible value of the union ie n(tea U coffee) = 100

So, we need to figure out the surplus : n(tea) + n(coffee) = 80 + 70 = 150.

The surplus is = 150 – 100 = 50

So, the value of the intersection = value of the surplus = 50

This could have also been obtained by the formula

**n(a U b) = n(a) + n(b) – n (a n**** b)**

In this case, our venn diagram will look something like this (Red is tea & Purple is coffee):

In this case, there is no one who likes neither coffee nor tea.

Let us look at a slightly more complicated problem when we have to deal with three sets and the value of union of the sets is fixed.

**Example:** In a survey it was found that 40 % like tea, 50 % like coffee and 60 % like milk. Every person likes at least one of the three items – tea / coffee / milk. What are the maximum and minimum possible values of those who like all three?

Solution: Currently our Venn Diagram looks like this:

First of all, we need to figure out the surplus.

Surplus = 40 +50 + 60 – 100 = 50

The surplus should be taken care of by adding to the intersection of all three or any of the two.

If we want to **maximize those who like all three**, we need to maximize the intersection of all three. Adding ‘1’ to the intersection of all three takes care of a surplus of ‘2’. To take care of a surplus of 50, we need to make

n (tea n coffee n milk) = 25

*Note: If the union of the sets was not fixed i.e. the line ‘Every person likes at least one of the three items – tea / coffee / milk’ was not given in the question then the answer would have been 40.*

Our venn diagram will now look like this:

*If we want to minimize those who like all three, we need to minimize the intersection of all three. But we have to take care of the surplus of 50. We can do that adding them to the intersection of any two of them. Adding ‘1’ to the intersection of two sets, takes care of a surplus of ‘1’. So,*

n (tea n coffee n milk)_{min }= 0

We can take care of the surplus 50 in many ways by adding them in any order to the intersection of two sets. Three of those many ways are given below in Venn Diagrams.

I hope that this will help you in solving problems related to Set Theory.

How to solve questions based on At least n in Set Theory for CAT Exam?

Permutation and Combination – Fundamental Principle of Counting

Permutation and Combination – Distribution of Objects

How to find Rank of a Word in Dictionary (With or Without Repetition)

Basic Probability Concepts for CAT Preparation

Sequence and Series Problems and Concepts for CAT 2017 Exam Preparation

All questions from CAT 2017 Quantitative Aptitude – Modern Maths

Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is

Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is

Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.

Quantitative Aptitude – Modern Maths – Progressions – Q4: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?

Quantitative Aptitude – Modern Maths – Progressions – Q5: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?

Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?

Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?

Quantitative Aptitude – Modern Maths – P&C – Q5

**A question on Set Theory from CAT 2017 – Forenoon Slot**

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