Sets MCQ Quiz - Objective Question with Answer for Sets - Download Free PDF

Concept:

Principle of Inclusion-Exclusion (PIE):

• This principle is used to count the number of elements in the union of multiple sets when there is overlap among the sets.
• It corrects the overcounting by subtracting the sizes of pairwise intersections, adding triple-wise intersections, and so on.

Calculation:

Given,

Total number of students in a class = 100

n(A): Number who like Maths = 60

n(B): Number who like Physics = 50

n(C): Number who like Chemistry = 40

n(A ∩ B) = 30, n(B ∩ C) = 20, n(C ∩ A) = 25

n(A ∩ B ∩ C) = 15

We need to find the number of students who like at least one of the three subjects:

⇒ n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 60 + 50 + 40 − 30 − 20 − 25 + 15

⇒ 150 − 75 + 15 = 90

∴ The number of students who like at least one subject = 90

Concept:

Equivalence Relation and Partition of Sets:

• An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive.
• Each equivalence relation on a set corresponds to a partition of that set.
• A partition divides the set into non-empty, disjoint subsets whose union is the entire set.
• In this question, we count only those partitions where each subset has at least two elements.

Important Terms:

• Partition: A way of dividing the set into disjoint non-empty subsets. Total number of partitions of a set of size n is given by the Bell number.
• Bell Number (Bₙ): The number of partitions of a set with n elements. Notation: Bₙ

Calculation:

Given,

Set S = {1, 2, 3, 4}

⇒ n = 4

We need the number of equivalence relations where no subset (block) has size 1

⇒ That means we need partitions where each block has size ≥ 2

⇒ We analyze possible valid partitions of 4 elements:

⇒ Case 1: One block of size 4 → Partition: {1,2,3,4} ⇒ Only 1 way

⇒ Case 2: Two blocks of size 2 each → Partitions like: { {1,2}, {3,4} } ⇒ Count = 3 ways

⇒ Case 3: One block of size 3 and one singleton → Rejected (singleton not allowed)

⇒ Case 4: Other combinations like {2,1,1}, {1,1,1,1} etc. → Rejected (contains singleton)

∴ The total number of such equivalence relations = 1 + 3 = 4

Given:

Number of elements in the set = 5

Formula Used:

Number of elements in a power set = 2ⁿ, where n is the number of elements in the set.

Calculation:

n = 5

⇒ Number of elements in the power set = 2⁵

⇒ Number of elements in the power set = 2 × 2 × 2 × 2 × 2

⇒ Number of elements in the power set = 32

The power set of a set with 5 elements has 32 elements.

Calculation: 

Elements of the type 3k = 3 

Elements of the type 3k + 1 = 1, 7, 9

Elements of the type 3k + 2 = 2, 5, 11

Subsets containing one element S₁ = 1

Subsets containing two elements

⇒ S₂ = ³C₁ × ³C₁ = 9

Subsets containing three elements

⇒ S₃ = ³C₁ × ³C₁ + 1 + 1 = 11

Subsets containing four elements

⇒ S₄ = ³C₃ + ³C₃ + ³C₂ × ³C₂ = 11

Subsets containing five elements

⇒ S₅ = ³C₂ × ³C₂ × 1 = 9

Subsets containing six elements S₆ = 1

Subsets containing seven elements S₇ = 1 

⇒ sum = 43 

Hence, the correct answer is 43. 

Given:

n(A – B) = 20 + x

n(B – A) = 3x

n(A ∩ B) = x + 1

n(A) = n(B)

Formula used:

Total elements in set A: n(A) = n(A – B) + n(A ∩ B)

Total elements in set B: n(B) = n(B – A) + n(A ∩ B)

Since n(A) = n(B), we equate the expressions.

Calculation:

n(A) = n(A – B) + n(A ∩ B)

⇒ n(A) = (20 + x) + (x + 1)

⇒ n(A) = 21 + 2x

n(B) = n(B – A) + n(A ∩ B)

⇒ n(B) = (3x) + (x + 1)

⇒ n(B) = 4x + 1

Since n(A) = n(B):

⇒ 21 + 2x = 4x + 1

⇒ 21 – 1 = 4x – 2x

⇒ 20 = 2x

⇒ x = 10

We need to find the value of (2x – 5):

⇒ (2 × 10 – 5) = 20 – 5

⇒ 15

∴ The correct answer is option (4).

Given:

Percent of readers who read all the three = 10% 

Percent of readers who read C = 45% 

Percent of readers who read at least two = 35% 

Percent of readers who read A and B = 15% 

Calculations:

Let the percent of readers who read only A and C = a%

Let the percent of readers who read only B and C = b%

Percent of readers who read only C  = c%

According to the question

Percent of readers who read C = 45%

⇒ a% + 10% + b% + c% = 45%

⇒ a% + b% + c% = 35%      ----(1)

Percent of readers who read at least two = 35% 

⇒ a% + 10% + 5% + b% = 35% 

⇒ a% + b% = 20%      ----(2)

Put the value of a% + b% from equation (2) to equation (1)

⇒ 20% + c% = 35% 

⇒ c% = 15% 

∴ Percent of readers who read only C is 15%

CONCEPT:

Let A and B be two sets then A is said to be proper subset of B, if A is a subset of B and A is not equal to B. It is denoted as A ⊂ B.

CALCULATION:

Given: A = {1, 3, 4} and B = {x : x ∈ R and x2 - 7x + 12 = 0}

First let's find the roaster form of set B

In order to do so we need to find the roots of the equation x2 - 7x + 12 = 0

⇒ x2 - 3x - 4x + 12 = 0

⇒ x(x - 3) - 4(x - 3) = 0

⇒ (x - 4) × (x - 3) = 0

⇒ x = 3, 4

⇒ B = {3, 4}

As we can clearly see that, all the elements of B are there in set A but A ≠ B i.e B ⊂ A

Hence, the correct option is 3.

CONCEPT:

Union:

Let A and B be two sets. The union of A and B is the set of all those elements which belong to either A or B or both A and B i.e A ∪ B = {x : x ∈ A or x ∈ B}

Note: If A is a non-empty set such that n(A) = m then numbers of proper subsets of A is given by 2m - 1.

CALCULATION:

Given: A = {1, 2, 5, 7} and B = {2, 4, 6}

As we know that,  A ∪ B = {x : x ∈ A or x ∈ B}.

⇒ A ∪ B = {1, 2, 4, 5, 6, 7}

As we can see that, the number of elements present in A U B = 6 i.e n(A U B) = 6

As we know that, if A is a non-empty set such that n(A) = m then numbers of proper subsets of A is given by 2m - 1.

So, the number of proper subsets of A Δ B = 2⁶ - 1 = 63

Hence, the correct option is 3.

Explanation:

(A - B) ∪ (B - A) =  A Δ B

Symmetric Difference of two Sets: Let A and B be two sets. The symmetric difference of sets A and B is the set (A - B) ∪ (B - A) and is denoted as A Δ B.

i.e A Δ B = (A - B) ∪ (B - A)

The Venn diagram representation of symmetric difference of two sets is shown below

Concept:

Combination: Selecting r objects from given n objects.

• The number of selections of r objects from the given n objects is denoted by \({{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{\;}}\)
•  \({{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{}} = {\rm{}}\frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n\;}} - {\rm{\;r}}} \right)!}}\)

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

Number of elements in A = 10

Number of subsets of A containing exactly two elements = Number of ways we can select 2 elements from 10 elements

⇒ Number of ways we can select 2 elements from 10 elements = ¹⁰C₂ = 45

∴ Number of subsets of A containing exactly two elements = 45

CONCEPT:

Intersection:

Let A and B be two sets. The intersection of A and B is the set of all those elements which are present in both sets A and B.

The intersection of A and B is denoted by A ∩ B i.e A ∩ B = {x : x ∈ A and x ∈ B}

Note: If A is a non-empty set such that n(A) = m then numbers of proper subsets of A is given by 2m - 1.

CALCULATION:

Given: A = {1, 2, 3, 4, 5, 7, 8, 9} and B = {2, 4, 6, 7, 9}

As we know that, A ∩ B = {x : x ∈ A and x ∈ B}

⇒ A ∩ B = {2, 4, 7, 9}

As we can see that,

The number of elements present in A ∩ B = 4

i.e n(A ∩ B) = 4

As we know that;

If A is a non-empty set such that n(A) = m then

The numbers of proper subsets of A are given by 2m - 1.

So, The number of proper subsets of A ∩  B = 2⁴ - 1 = 15

Hence, the correct option is 2

Concept:

• If a set contains n elements, then the total number of subsets will be 2ⁿ

Calculation:

Total number of subsets of the first is 2^m while total number of subsets of the second is 2ⁿ.

Now it is given that

2^m = 2ⁿ + 56

⇒ 2^m - 2ⁿ = 56

⇒ 2³(2^m-3 - 2^n-3) = 56

⇒ (2^m-3 - 2^n-3) = 7

⇒ (2^m-3 - 2^n-3) = 8 - 1

⇒ (2^m-3 - 2^n-3) = 2³ - 2⁰

Hence compare the terms,

m - 3 = 3 thus m = 6 and

n - 3 = 0 giving n = 3.

Hence, (m, n) = (6, 3)

Concept:

Union of the sets: 

Union of two given sets is the set that contains those elements that are either in A or in B, or in both.

The  union of the sets A and B, denoted by A U B

Intersection of Sets:

The intersection of two given sets is the largest set which contains all the elements that are common to both the sets.

The intersection of Sets A and B, denoted by A ∩ B

Formula:  A ∪ B = A + B - A ∩ B

Calculation:

Given: A ∩ B = A

To Find: A ∪ B

As we know, 

A ∪ B = A + B - A ∩ B

⇒ A ∪ B = A + B - A

∴ A ∪ B = B

Concept:

De morgan's law:

(A ∪ B)^c = A^c ∩ B^c

(A ∩ B)^c = A^c ∪ B^c

Distributive law in sets:

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Calculation:

Given: A and B are two sets

To Find: A ∩ (B ∪ A)^c

As we know, (A ∪ B)^c = A^c ∩ B^c

Therefore (B ∪ A)^c = B^c ∩ A^c

Now A ∩ (B ∪ A)^c = A ∩ (B^c ∩ A^c)

= (A ∩ B^c) ∩ (A ∩ A^c)             (Using distributive law)

= (A ∩ B^c) ∩ ϕ                        (∵ x ∩ ϕ = ϕ)

= ϕ 

Calculation:

Given: The four sets have 150, 180, 210 and 240 elements respectively

n(A) = 150

n(B) = 180

n(C) = 210

n(D) = 240

Each pair of sets has 15 elements

n(A ∩ B) = 15

n(A ∩ C) = 15

n(A ∩ D) = 15

n(B ∩ C) = 15

n(B ∩ D) = 15

n(C ∩ D) = 15

Each triple of sets has 3 elements

n(A ∩ B ∩ C) = 3

n(A ∩ B ∩ D) = 3

n(A ∩ C ∩ D) = 3

n(B ∩ C ∩ D) = 3

A ∩ B ∩ C ∩ D = ϕ 

n(A ∩ B ∩ C ∩ D) = 0

Now, number of elements in the union of 4 sets A, B, C and D

n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D) - n(A ∩ B) - n(A ∩ C) - n(A ∩ D) - n(B ∩ C) - n(B ∩ D) - n(C ∩ D) + n(A ∩ B ∩ C) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D) + n(B ∩ C ∩ D) - n(A ∩ B ∩ C ∩ D)

⇒ 150 + 180 + 210 + 240 - 6 × 15 + 4 × 3 - 0

∴ The required number of elements is 702.