Set Theory and types of Sets MCQ Quiz - Objective Question with Answer for Set Theory and types of Sets - Download Free PDF

Concept:

Properties of Irrational Numbers:

• Between any two irrational numbers, there are an infinite number of irrational numbers.
• The set of all irrational numbers between \( \sqrt{12} \) and \( \sqrt{15} \) is an infinite set.

Calculation:

Between any two irrational numbers, there are an infinite number of irrational numbers.

The set of all irrational numbers between \( \sqrt{12} \) and \( \sqrt{15} \) is an infinite set.

Also, the set of all odd integers less than 1000 is \( \{-5, -3, 1, 3, \dots, 999\} \), which is an infinite set.

Hence, Option (a) is correct.

Concept:

Properties of Relations:

• Reflexive Relation: A relation R on a set A is reflexive if (a, a) ∈ R for every element a ∈ A.
• Symmetric Relation: A relation R on a set A is symmetric if (a, b) ∈ R implies (b, a) ∈ R.
• Transitive Relation: A relation R on a set A is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.

Calculation:

Given,

Set A = {1, 2, 3}

Relation R = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}

Check reflexivity:

⇒ (1,1), (2,2), (3,3) ∈ R

⇒ All (a ,a) pairs for a ∈ A are in R

⇒ R is reflexive

Check symmetry:

⇒ (1,2) ∈ R but (2,1) ∉ R

⇒ (2,3) ∈ R but (3,2) ∉ R

⇒ R is not symmetric

Check transitivity:

⇒ (1,2) ∈ R and (2,3) ∈ R

⇒ (1,3) ∈ R

⇒ Other pairs do not form further transitive chains

⇒ R is transitive

∴ R is reflexive and transitive.

Step 1 : Find the elements in A

The correct answer is option 4.

Key Points:

A. 0∈∅" id="MathJax-Element-91-Frame" role="presentation" style="position: relative;" tabindex="0">0∈∅0∈∅ 0∈∅

• Explanation: The empty set (∅" id="MathJax-Element-92-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅ ) contains no elements.

• Conclusion: False, because 0" id="MathJax-Element-93-Frame" role="presentation" style="position: relative;" tabindex="0">00 0  is not an element of ∅" id="MathJax-Element-94-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅ .

B. ∅∈{0}" id="MathJax-Element-95-Frame" role="presentation" style="position: relative;" tabindex="0">∅∈{0}∅∈{0} ∅∈{0}

• Explanation: The set {0}" id="MathJax-Element-96-Frame" role="presentation" style="position: relative;" tabindex="0">{0}{0} {0}  has only one element: 0" id="MathJax-Element-97-Frame" role="presentation" style="position: relative;" tabindex="0">00 0 . The empty set is not listed as an element.

• Conclusion: False, because ∅" id="MathJax-Element-98-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅  is not in {0}" id="MathJax-Element-99-Frame" role="presentation" style="position: relative;" tabindex="0">{0}{0} {0} .

C. ∅∈{∅}" id="MathJax-Element-100-Frame" role="presentation" style="position: relative;" tabindex="0">∅∈{∅}∅∈{∅} ∅∈{∅}

• Explanation: The set {∅}" id="MathJax-Element-101-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅}  has one element: the empty set itself.

• Conclusion: True, because ∅" id="MathJax-Element-102-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅  is explicitly an element of {∅}" id="MathJax-Element-103-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅} .

D. {∅}∈{∅}" id="MathJax-Element-104-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}∈{∅}{∅}∈{∅} {∅}∈{∅}

• Explanation: The set {∅}" id="MathJax-Element-105-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅}  contains only ∅" id="MathJax-Element-106-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅ , not {∅}" id="MathJax-Element-107-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅}  itself.

• Conclusion: False, because {∅}" id="MathJax-Element-108-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅}  is not an element of itself.

E. {∅}⊂{∅,{∅}}" id="MathJax-Element-109-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}⊂{∅,{∅}}{∅}⊂{∅,{∅}} {∅}⊂{∅,{∅}}

• Explanation:

  • {∅,{∅}}" id="MathJax-Element-110-Frame" role="presentation" style="position: relative;" tabindex="0">{∅,{∅}}{∅,{∅}} {∅,{∅}}  has two elements: ∅" id="MathJax-Element-111-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅  and {∅}" id="MathJax-Element-112-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅} .

  • The subset {∅}" id="MathJax-Element-113-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅}  is contained within it.

• Conclusion: True, because every element of {∅}" id="MathJax-Element-114-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅}  (which is just ∅" id="MathJax-Element-115-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅ ) is also in {∅,{∅}}" id="MathJax-Element-116-Frame" role="presentation" style="position: relative;" tabindex="0">{∅,{∅}}{∅,{∅}} {∅,{∅}}

Correct Option:

Only C and E are true.
Answer: Option 4 (C and E only). 

Key Concepts:

1. ∈" id="MathJax-Element-117-Frame" role="presentation" style="position: relative;" tabindex="0">∈∈ ∈  (Element of): Checks if an item is directly inside a set.

   • Example: ∅∈{∅}" id="MathJax-Element-118-Frame" role="presentation" style="position: relative;" tabindex="0">∅∈{∅}∅∈{∅} ∅∈{∅}  is true.

2. ⊂" id="MathJax-Element-119-Frame" role="presentation" style="position: relative;" tabindex="0">⊂⊂ ⊂  (Subset of): Checks if all elements of one set are in another.

   • Example: {∅}⊂{∅,{∅}}" id="MathJax-Element-120-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}⊂{∅,{∅}}{∅}⊂{∅,{∅}} {∅}⊂{∅,{∅}}  is true.

3. The empty set (∅" id="MathJax-Element-121-Frame" role="presentation" style="position: relative;" tabindex="0">∅∅ ∅ ) is not an element of every set—only when explicitly included (e.g., {∅}" id="MathJax-Element-122-Frame" role="presentation" style="position: relative;" tabindex="0">{∅}{∅} {∅} ).

This aligns perfectly with Option 4.

20, Only football 190 etc.

Given \(N = 500, n(F) = 285,\)

\(n(H) = 195, n(B) = 115, n(F \cap B) = 45,\)

\(n(F \cap H) = 70, n(H \cap B) = 50,\)

\(n(F' \cap H' \cap B') = 50.\)

To find \(n(F \cap H \cap B), n(F \cap H' \cap B'), n(F' \cap H \cap B'), n(F' \cap H' \cap B)\), we have

\(50 = n(F' \cap H' \cap B') = n((F \cup H \cup B)')\)

By De-Morgan law

\(= N - n(F \cup H \cup B) = N - \{S_1 - S_2 + S_3\}\)

\(= N - \{n(F) + n(H) + n(B) - n(F \cap H) - n(H \cap B) - n(B \cap F) + n(F \cap H \cap B)\}\)

\(= 500 - 285 - 195 - 115 + 70 + 50 + 45 - n(F \cap H \cap B)\)

\(= 665 - 595 - n(F \cap H \cap B)\)

\(= 70 - n(F \cap H \cap B)\)

\(\therefore n(F \cap H \cap B) = 20\)

Again, \(n(F \cap H' \cap B') = \{F \cap (H \cup B)'\}\)

By De-Morgan law

\(= n(F) - \{F \cap (H \cup B)\}\)

\(= n(F) - n((F \cap H) \cup (F \cap B))\)

\(= n(F) - n(n(F \cap H) + n(F \cap B) - n(F \cap H \cap B))\)

\(= 285 - 70 - 45 + 20 = 190\)

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.

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:

Power Set:

Let A be a set, then the set of all the possible subsets of A is called the power set of A and is denoted by P(A).

Note: If A is a finite set with m elements. Then the number of elements (cardinality) of the power set of A is given by: n (P(A)) = 2^m.

CALCULATION:

Given: A = {1, 3, 5}

Here, we have to find the cardinality of the power set of A i.e n (P(A))

As we know that if A is a finite set with m elements. Then the number of elements (cardinality) of the power set of A is given by: n (P(A)) = 2m.

Here, we can see that, the given A has 3 elements i.e n(A) = 3

So, the cardinality of the given set is n(P(A)) = 2³ = 8

Hence, the correct option is 4.

CONCEPT:

• A set which contains finite number of elements is called a finite set.
• A set which has infinite number of elements is called an infinite set
• A set which does not contain any element is called an empty set.

CALCULATION:

Given: B = {x : x ∈ N such that x2 + 11x + 30 = 0}

First let's solve the quadratic equation x2 + 11x + 30 = 0

⇒ x² + 5x + 6x + 30 = 0

⇒ x(x + 5) + 6(x + 5) = 0

⇒ (x + 6) (x + 5) = 0

⇒ x = - 5 or - 6

According to the definition of the given set x is a natural number but we know that neither x = - 5 nor x = - 6 is a natural number

So, the given set is an empty set i.e B = ϕ

Hence, the correct option is 1.

Concept:

• The power set (or powerset) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set.

Calculations:

Given, A = {x, y, z}. 

The power set (or powerset) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set.

Powerset of A = {ϕ,{x}, {y}, {z}, {x, y}, {y, z}, {x, z},{x, y, z}}.

Hence, the number of subsets in the powerset of A is 8.

Concept:

The Pigeonhole Principle: Let there be n boxes and (n + 1) objects. Then, under any assignment of objects to the boxes, there will always be a box with more than one object in it. This can be reworded as, if m pigeons occupy n pigeonholes, where m > n, then there is at least one pigeonhole with two or more pigeons in it. 

Calculation:        

We divide the set into n classes {1, 2}, {3, 4},......{2n - 1, 2n}.

By the pigeonhole principle, given n +1 elements at least two of them will be in the same case {2k - 1, 2k} (1 ≤ k ≤ n). But 2k - 1 and 2k are relatively prime because their difference is 1. 

Concept:

Equivalent set: Two sets A and B are said to be equivalent if they have the same number of elements (cardinality), regardless of what the elements are. , i.e n(A) = n(B).

Equal set: Two sets A and B are said to be equal if they have exactly the same elements. The repetition of elements does not matter in sets because a set is defined as a collection of distinct objects, so repeated elements are generally ignored.

If A = B, then n(A) = n(B) and for any x ∈ A, x ∈ B too.

Calculation:

1. A = {1, 3, 5} and B = {2, 4, 7} are equivalent sets. → True

Set A has three elements: 1, 3, and 5. Set B also has three elements: 2, 4, and 7. Since both sets have three elements, they are indeed equivalent sets.

2. A = {1, 5, 9} and B = {1, 5, 5, 9, 9} are equal sets. → True

Set A contains the elements 1, 5, and 9. Set B also contains the elements 1, 5, and 9 (despite some elements being repeated, duplicates are not considered in set theory). Since both sets have the same elements, they are equal.

Therefore, both statements are correct:

Statement 1 is correct: A and B are equivalent sets.
Statement 2 is correct: A and B are equal sets.

Hence, The correct answer is "Both 1 and 2".

Source:- https://ncert.nic.in/textbook/pdf/kemh101.pdf, Page No. 7-8.

Concept:

The number of elements in the power set of any set A is 2n where n is the number of elements of the set A.

Calculation:

Let A = {0, 1, 2, …, 9}

Number of elements in set A = 10

∴ The number of element in the power set P(A) = 2¹⁰ = 1024

Hence, option (3) is correct.