Saturday, July 4th, 2020
Set Theory is a very important branch of mathematics that has wide ranging applications. It is a very basic and easy concept but has significant use in Algebra, Logic, Combinatorics, probability etc. Thus, for the same reason understanding of this topic is of grave importance to solve various questions in competitive exams such as Banking and MBA entrance exams.
A set is nothing but a collection of distinct objects. The objects could be anything numbers, letters, vectors, etc. For ex, all the students in a class can be grouped into a set or integers such as 2, 3, 4 can form a set etc.
A set is generally denoted with a capital alphabet S or A or and its elements are generally enclosed in curly brackets like this
S = {2, 3 ,4}
Here, 2, 3 and 4 are elements of set S and an element is generally denoted using the symbol Є (belongs to) like this: 2 Є S.
When a set is very large and cannot be contained it can also be denoted in the following way:
A = {x: x is an integer} or,
B = {x: 0<x<10, x Є R}, where R is a set of real numbers
CARDINALITY OF A SET
Cardinality of a set means number of elements in any set. For ex, Consider the set above S = {2, 3 ,4}. Now the set contains 3 elements in it and thus its cardinality is 3. It is denoted by n(S) = 3 or |S| = 3.
TYPES OF SETS
EMPTY SET or NULL SET: A empty set is the one which does not contain any element in it. The set is denoted by the symbol
 and n() = 0, as the set is empty.
B = {1,2} and S = {1, 2, 3, 4, 5}
Then. Both 1 and 2 are present in the set S. Hence, B is subset of S.
Note: A null set is subset of every set.
Every set is subset of itself.
It is denoted by symbol ‘⊂’. Thus, if B is contained in S then it can be written as B ⊂ S.
A = {x: x<5, x Є Z+} and B = {1, 2, 3, 4}
Notice that both the sets are equal as A = {1, 2, 3, 4}
Hence, A = B
Also, notice that n(A) = n(B) in equal sets.
S = {1, 2, 3}, then
P(S) = {,{1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}
If a set is a finite set, then the number of subsets of that sets of that set is |P(S)| = 2n.
A = {1, 2} and B = {2, 3, 4}, then
A Â B = {1, 2, 3, 4}
A = {1, 2} and B = {2, 3, 4}, then
A ∩ B = {2}
 A – B = {x: x Є A and x   B} and B – A = {x: x Є B and x   A}. For ex,
A = {1, 2} and B = {2, 3, 4}, then
A – B = {1}, and B – A = {3,4}
SOME IMPORTANT LAWS OF SET ALGEBRA
Also, if A Â B then,
Since we have now covered all the important formulas and concepts related to sets except one thing i.e. Venn Diagrams, they are pictorial representation of sets and help in easily solving questions
A Venn diagram is a diagram that shows all possible logical relations between a finite collection of sets. They can be used to depict two sets, three sets, 4 sets or more. But for our purpose and level we only need their understanding up till four sets.
A two set Venn diagram is given below
The above diagram is called Venn Diagram where there are two sets A and B and U is the universal set.
Various Relationship between sets using Venn Diagrams
Also, notice that through the Venn diagram that A -B = A ∩ B’
Some basic formulas for Venn Diagram
n(A B) = n(A) + n(B) – n(A ∩ B)
n(A  B  C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
Let A, B and C be three sets and U be the universal set then the Venn diagram for 3 sets can be constructed as following
Solved Examples:
Example 1: In a group of 100 persons, 72 people can speak English and 43 can speak French. How many can speak English only? How many can speak French only and how many can speak both English and French?
Let A denote people who can speak English
Let B denote people who can speak French
Given that: n(A Â B) = 100
To find:
Sol: Since, n(A B) = n(A) + n(B) – n(A ∩ B)
n(A ∩ B) = 72 + 43 – 100
= 15
Now only A can be written as = n(A) – n(A ∩ B)
= 72 – 15 = 57
Similarly, only B = n(B) – n(A ∩ B) = 43 -15 = 28
Example 2: Of the 200 candidates who were interviewed for a position at a call center, 100 had a two-wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both, a two-wheeler and a credit card, 30 had both, a credit card and a mobile phone and 60 had both, a two wheeler and mobile phone and 10 had all three. How many candidates had none of the three?
Sol: Let A denote candidates who have 2-wheelers.
Let B denote candidates with credit card.
Let C denote candidates with mobile phones.
Now we need to find n(Ac∩ Bc∩ Cc)
This can be easily be calculated using D’morgans Law, i.e.
n (A  B  C)c = n(Ac∩ Bc∩ Cc)
Since, n(A  B  C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 100 +70 +140 – 40 – 30 – 60 + 10 = 190
Total Candidates are 200, therefore,
n(A Â B Â C) + n (A Â B Â C)c =200
n (A Â B Â C)c = 10
Example 3: Study the following figure and answer the questions given below.
Find out:
1.) How many doctors are neither artist nor players?
2.) How many doctors are both players and artists?
3.) How many artists are players?
4.) How many players are neither artist nor doctors?
Sol: To solve such questions. Let’s construct the table like we did earlier in 3 set Venn diagram case corresponding to the value of the region. But first,
Let A be Artists
Let P be players
Let D be doctors
Now let’s see what we need to find out
1.) Doctors that are neither Artist nor Players => Only Doctors i.e. Only D = 17
2.) Doctors that are both Artist and Players => D ∩ A ∩ P = 3
3.) Artist that are players => (A ∩ P) + (D ∩ A ∩ P) = 22 +3 = 25
4.) Players that are neither doctors nor artist => Only Players = 25
There can be many questions formed on these kinds of sets. But you can easily solve them using little manipulation and basic addition and subtraction. With a little practice, you can easily and quickly solve questions of sets either using Venn diagrams.
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All questions from CAT Exam 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
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A union (B intersection C) – VENN DIAGRAM