Mathematical Sets and Relations – Multiple Choice Questions -

Last Updated on: 17th February 2025, 03:26 pm

Sets and Relations – Multiple Choice Questions

In Multiple Choice Questions Tests, each question is given with multiple alternative options (normally 4), out of which only one option is correct. The examinee has to quickly select the correct option within specified time.

Multiple Choice Questions are commonplace for Academic Courses Examinations, Competitive Tests, Entrance Tests, etc.

Examinees should well understand the pattern, rules and tricks to quickly find the correct answer to Multiple Choice Questions. Click the link Multiple Choice Questions answering guidelines to play Video explaining various useful and quick tricks and tips.

Click the link to get Centralised Links of Videos of Multiple Choice Questions Bank, organised as per Subjects.

1. If every element of a set P is also an element of set Q then set P is a subset of Q then the set Q is known as

(a)        Set of P
(b)        Sub set of P
(c)        Super set of P
(d)        Any of these.

Intent: To test the understanding of the relationship between subsets and supersets.

Options Analysis:

(a) Set of P – Incorrect. A “set of P” is not a formal mathematical term.
(b) Subset of P – Incorrect. The given statement defines P as a subset of Q, but Q itself is not a subset of P.
(c) Super set of P – Correct. If P is a subset of Q, then Q is the superset of P.
(d) Any of these – Incorrect. Only option (c) is correct.

Correct Answer: (c) Super set of P

Logical Deliberation:
A superset contains all elements of another set. Since P is a subset of Q, it follows that Q is the superset of P.

Every subset is a subset of a super set
Hence Option (c) is the correct answer

2. The empty set is also known as

(a)        Zero set
(b)        Null set
(c)        Void set
(d)        (b) or (c)

Intent: To test knowledge of different terms used to describe an empty set.

Options Analysis:

(a) Zero set – Incorrect. While “zero set” may be used informally, it is not a widely accepted term.
(b) Null set – Correct. The empty set is commonly referred to as the “null set.”
(c) Void set – Correct. “Void set” is another term for the empty set.
(d) (b) or (c) – Correct. Since both (b) and (c) are correct, this option is also correct.

Correct Answer: (d) (b) or (c)

Logical Deliberation:
The empty set is the set with no elements and is denoted by ∅. It is also called a null set or void set.

Empty set is also known as null set or void set.
Hence Option (d) is the correct answer

3. Inter section of two set A & B is known as

$\displaystyle \begin{array}{*{20}{l}} {(a)A\subset B} \\ {(b)A\cap B} \\ {(c)A\cup B} \\ {(d)A\not\subset B} \end{array}$

Intent: To test the knowledge of set operations, particularly intersection.

Options Analysis:

(a) A ∪ B (Union) – Incorrect. This represents the union of sets, not their intersection.
(b) A ⊂ B (Subset) – Incorrect. This denotes a subset relation, not an intersection.
(c) A ∩ B (Intersection) – Correct. The intersection contains only the common elements of sets A and B.
(d) A ÷ B (Not a standard set operation) – Incorrect. Division is not defined for sets in this manner.

Correct Answer: (b) A ∩ B

Logical Deliberation:
The intersection of two sets A and B includes only the elements that are present in both A and B.

Intersection of two set A & B is known as ( $\displaystyle {A\cap B}$ )
Hence Option (b) is the correct answer

4. If n(A) = n(B), for two finite set A & B is known

(a)        Singleton set
(b)        Equal sets
(c)        Equivalent sets
(d)        None of these

Intent: To test the knowledge of equivalent and equal sets.

Options Analysis:

(a) Singleton set – Incorrect. A singleton set has only one element, unrelated to this question.
(b) Equal sets – Incorrect. Equal sets must have the same elements, not just the same number of elements.
(c) Equivalent sets – Correct. If two sets have the same number of elements, they are equivalent sets.
(d) None of these – Incorrect. The correct answer is given in option (c).

Correct Answer: (c) Equivalent sets

Logical Deliberation:
Two sets are equivalent if they have the same number of elements, even if the elements themselves differ.

Two finite sets are called equivalent if they have same number of element. i.e. n (A) = n (B)
Hence Option (c) is the correct answer

5. The objects of a set are called its
(a)        Universal set
(b)        Sub set
(c)        Unit
(d)        Elements of set

Intent: To test knowledge of basic set terminology.

Options Analysis:

(a) Universal set – Incorrect. A universal set contains all elements under discussion, not just the objects of one set.
(b) Subset – Incorrect. A subset is a set derived from another set, not individual objects.
(c) Unit – Incorrect. The term “unit” is not used for set elements.
(d) Elements of set – Correct. The objects in a set are called elements.

Correct Answer: (d) Elements of set

Logical Deliberation:
Each item in a set is called an element (or member) of that set.

Objects of a set are called its element
Hence Option (d) is the correct answer

6. The methods of describing a set
(a)        Roster from
(b)        Set Builder from
(c)        Either (a) or (b)
(d)        None of the above

There are two method of describing a set
(1) Roster from
(2) Set builder from
Hence Option (c) is the correct answer

7. Universal set is denoted by _____________ .
(a) { }
(b) $\displaystyle \phi$
(c) U
(d) None of these

Intent: To test knowledge of the notation used for the universal set.

Options Analysis:

(a) U – Correct. The universal set is commonly denoted by U.
(b) {} (Empty set) – Incorrect. An empty set contains no elements, unlike the universal set.
(c) None of these – Incorrect. The universal set has a standard notation, which is U.

Correct Answer: (c) U

Logical Deliberation:
The universal set (U) contains all elements under discussion.

Universal set is denoted by $\displaystyle U$
Hence Option (c) is the correct answer

8. _____________is also known as Tabular from.
(a)        Rule method
(b)        Roster form
(c)        Set-Builder form
(d)        Any of the above

Intent: To determine the equivalent term for a method of set representation.

Options Analysis:

(a) Rule method – Incorrect. This refers to the set-builder method.
(b) Roster form – Correct. The roster method (listing elements) is also called tabular form.
(c) Set-builder form – Incorrect. This is a different method using a defining property.
(d) Any of the above – Incorrect. Only the roster form is the tabular form.

Correct Answer: (b) Roster form

Logical Deliberation:
The roster method is also called tabular form, as it lists elements explicitly.

Another name for Roster from is Tabular form. In such form, the elements are being separated by commas and are enclosed within braces { }.
Hence Option (b) is the correct answer

9.    ____________ is also known as Rule Method.
(a)        Rule method
(b)        Roster from
(c)        Set-Builder from
(d)        None

Intent: To test knowledge of different methods of defining sets.

Options Analysis:

(a) Rule method – Correct. The rule method describes set elements using a logical condition.
(b) Roster form – Incorrect. The roster method lists elements explicitly rather than using a rule.
(c) Set-Builder form – Correct. The set-builder method is also known as the rule method.
(d) None – Incorrect. A correct answer exists among the choices.

Correct Answer: (c) Set-Builder form

Logical Deliberation:
The set-builder method (also called the rule method) defines a set by specifying a condition that its members satisfy.

Another name for Set Builder form is Rule Method. It is a statement or an expression to represent all the elements of a set.
Hence Option (c) is the correct answer

10. Under  _______________ , we just make a list of all elements of the set and put them under { }
(a)        Roster method
(b)        Set-Builder form
(c)        Both (a) & (b)
(d)        None of the above

Intent: To check understanding of the roster method of defining sets.

Options Analysis:

(a) Roster method – Correct. The roster method explicitly lists all elements of a set within curly brackets { }.
(b) Set-Builder form – Incorrect. The set-builder method defines a set using a property instead of listing elements.
(c) Both (a) & (b) – Incorrect. Only the roster method lists elements explicitly.
(d) None of the above – Incorrect. The correct method exists in option (a).

Correct Answer: (a) Roster method

Logical Deliberation:
The roster method represents a set by listing all its elements within { }.

Under Roster method we make a list of elements separated by comma and we put it under { }
Hence Option (a) is the correct answer

11. $\displaystyle \mathbf{If}\text{ }\mathbf{A}\text{ }\subset \text{ }\mathbf{B},\text{ }\mathbf{B}\text{ }\subset \text{ }\mathbf{A}\text{ }\mathbf{then}\text{ }\mathbf{two}\text{ }\mathbf{sets}\text{ }\mathbf{A}\text{ }\And \text{ }\mathbf{B}\text{ }\mathbf{are}~$                       .
(a)        Proper sub-set
(b)        Equal set
(c)        Equivalent sets
(d)        Power set

Intent: To determine when two sets are considered equal.

Options Analysis:

(a) Proper subset – Incorrect. A proper subset means one set is strictly contained within another, but this condition states equality.
(b) Equal set – Correct. If A ⊆ B and B ⊆ A, then A and B must contain the same elements, making them equal.
(c) Equivalent sets – Incorrect. Equivalent sets only require the same number of elements, not necessarily the same elements.
(d) Power set – Incorrect. A power set is the set of all subsets of a given set.

Correct Answer: (b) Equal set

Logical Deliberation:
Two sets A and B are equal if and only if A ⊆ B and B ⊆ A, meaning they contain exactly the same elements.

( $\displaystyle If\text{ }A\subset B\text{ }and\text{ }B\subset A\text{ }then\text{ }two\text{ }setsA\text{ }\And \text{ }B\text{ }are\text{ }equal\text{ }set$ )
Hence Option (b) is the correct answer

12. Set A and set B be two sets, the  ______________of A and B is $\displaystyle \mathbf{A}\text{ }\cap \text{ }\mathbf{B}~~$ .
(a)        Intersection
(b)        Equivalent
(c)        Union
(d)        Void

Intersection of set A & set B = ( $\displaystyle A\cap B$ )
Hence Option (a) is the correct answer

13. Set A = {1, 2, 3} & set B = {2, 1, 3}.
(a)        A & B are Equal
(b)        A & B is disjoint
(c)        A and B is NOT Equal
(d)        A is proper sub set of B

Intent: To test understanding of set equality.

Options Analysis:

(a) A & B are equal – Correct. Since A and B contain the same elements, they are equal.
(b) A & B are disjoint – Incorrect. Disjoint sets have no common elements, but here A and B are identical.
(c) A & B are not equal – Incorrect. The sets have the same elements, so they are equal.
(d) A is a proper subset of B – Incorrect. A proper subset must contain fewer elements than the set it belongs to.

Correct Answer: (a) A & B are equal

Logical Deliberation:
Sets are equal if they contain the same elements, regardless of order. Since A = {1, 2, 3} and B = {2, 1, 3}, they are equal.

Set A and set B are equal set because these sets have same elements.
Hence Option (a) is the correct answer

14. If A = {2, 5, 7} and B = {3, 5, 7, 8}, the value of A $\displaystyle \cup$ B is
(a)        {5, 7, 8}
(b)        {3, 8}
(c)        {2, 3, 5, 7, 8}
(d)        {0}

Set A = {2, 5, 7} B = {3, 5, 7, 8}
A $\displaystyle \cup$ B (A union B)
= {2, 3, 5, 7, 8}
Hence Option (c) is the correct answer

15. If A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, the value of A – {B $\displaystyle \cup$ C } is
(a)        {1, 2, 3}
(b)        {2, 3, 4, 5}
(c)        {1}
(d)        {0}.

A = { 1, 2, 3, 4} B = {2, 4, 6, 8} C = {3, 4, 5, 6}
B $\displaystyle \cup$ C = {2, 3, 4, 5, 6, 8}
A – {B $\displaystyle \cup$ C } = {1, 2, 3, 4} – {2, 3, 4, 5, 6, 8} = {1}
Hence Option (c) is the correct answer

16. If A = {a, b, c, d, e, f} B= {a, e, i, o, u} and C = {m, n, o, p, q, r, s, t, u} then A $\displaystyle \cup$ B is
(a) {a, b, c, d, e, f, i, o, u}
(b) {a, b, c}
(c) {d, e, f}
(d) $\displaystyle \phi$

Intent: To test the correct representation of an empty (null) set.

Options Analysis:

(a) {0} – Incorrect. This set contains one element (0), so it is not empty.
(b) {∅} – Correct. This is the standard notation for an empty set.
(c) 0 – Incorrect. This does not represent a set.
(d) – Correct. Since both (b) and (c) represent an empty set, this is also a valid answer.

Correct Answer: (d) Both (b) and (c)

Logical Deliberation:
The empty set (null set) is represented as ∅ or {}, meaning it has no elements.

$\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}} \\ {B\text{ }=\text{ }\left\{ {a,\text{ }e,\text{ }i,\text{ }o,\text{ }u,} \right\}} \\ {C\text{ }=\text{ }\left\{ {m,\text{ }n,\text{ }o,\text{ }p,\text{ }q,\text{ }r,\text{ }s,\text{ }t,\text{ }u} \right\}} \\ {A\cup B\text{ }=\text{ }\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f,\text{ }i,\text{ }o,\text{ }u} \right\}} \end{array}$

Hence Option (a) is the correct answer

17. The subsets of {0} are
(a) 1
(b) {0}
(c) $\displaystyle \phi$
(d) $\displaystyle \phi$ and {0} .

$\displaystyle \phi \text{ and }\left\{ 0 \right\}\text{ }are\text{ }subset\text{ }of\text{ }every\text{ }set$
Hence Option (d) is the correct answer

18. If A = {1, 2, 3} and B = {2, 4}, the value of A $\displaystyle \cap ~~$ B is
(a) {1, 2, 4}
(b) {2}
(c) {1, 2}
(d) $\displaystyle \phi$

$\displaystyle \begin{array}{*{20}{l}} {A=\left\{ {1,2,3} \right\}\text{ }and\text{ }B=\left\{ {2,4} \right\},} \\ {\therefore A\cap B=\left\{ 2 \right\}~\left( {only\text{ }common\text{ }elements.} \right)} \end{array}$
Hence Option (b) is the correct answer

19. $\displaystyle \mathbf{A}\text{ }=\phi ~~~~$ , which one of the following statements is correct ?

$\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\phi \in A} \\ {\left( b \right)} & {A\in \phi } \\ {\left( c \right)} & {A\not\subset \phi } \\ {\left( d \right)} & {\phi \subset A} \end{array}$

If A = $\displaystyle \phi$ then $\displaystyle \phi$ $\displaystyle \subset$ A . Because f is a subset of every set.
Hence Option (d) is the correct answer

20. If A ={1, 2, 3}, B = {3, 4} and C = {4, 5, 6} then $\displaystyle \mathbf{A}\cup (\mathbf{B}\cap \mathbf{C})$ is equal to
(a)        {1, 2, 3,4}
(b)        {3}
(c)        {1, 2, 3, 4, 5, 6}
(d)        {1, 2} .

$\displaystyle \begin{array}{*{20}{l}} {A=\left\{ {1,\text{ }2,\text{ }3} \right\},\text{ }B\text{ }=\text{ }\left\{ {3,\text{ }4} \right\},~~~~~~C\text{ }=\text{ }\left\{ {4,\text{ }5,\text{ }6} \right\}} \\ {(B\cap C)\text{ }\left\{ 4 \right\}.\text{ }\left[ {\text{ }only\text{ }common\text{ }elements\text{ }of\text{ }set\text{ }B\text{ }and\text{ }set\text{ }C} \right]} \\ {A\cup (B\cap C)=\left\{ {1,\text{ }2,\text{ }3,\text{ }4} \right\}} \end{array}$
Hence Option (a) is the correct answer

21. Of the following sets, the one having exactly eight subsets is
(a)        {1, 2}
(b)        {0, 1, 2}
(c)        {8}
(d)        {1, 2, 3, 4, 5, 6, 7, 8} .

The number of subsets (called Power Set) of a set of n elements contains 2ⁿ-1 subsets. Here {0, 1, 2} has 3 elements. It contains 2³-1 = 8 subsets :

$\displaystyle {\left\{ 0 \right\},\text{ }\left\{ 1 \right\},\text{ }\left\{ 2 \right\},\text{ }\left\{ {0,\text{ }1} \right\},\text{ }\left\{ {0,\text{ }2} \right\}\text{ }\left\{ {1,\text{ }2} \right\}\text{ }\left\{ {0,\text{ }1,\text{ }2} \right\},\phi }$
Hence Option (b) is the correct answer

22. If A = {1, 2, 3, 4} and B = {5, 6, 7}, then $\displaystyle \mathbf{A}\cap \mathbf{B}$ is

$\displaystyle \begin{array}{l}(a)\left\{ {1,2} \right\}\\(b)\left\{ {6,7} \right\}\\(c)\left\{ 3 \right\}\\(d)\phi \end{array}$

A = {1, 2, 3, 4} B = {5, 6, 7}
$\displaystyle A\cap B\text{ }=\phi$ (as there is no common elements.)
Hence Option (d) is the correct answer

23. The number of subsets of the set {a , b, c} is
(a)        12
(b)        6
(c)        8
(d)        9.

The number of subsets (called Power Set) of a set of n elements contains 2n-1 subsets.. Here {0, 1, 2} contains 23-1 = 8 subsets

$\displaystyle Subsets\text{ }of\text{ }set\text{ }\left\{ {a,\text{ }b,\text{ }c} \right\}\text{ }are\text{ }\left\{ a \right\},\text{ }\left\{ b \right\},\text{ }\left\{ c \right\},\text{ }\left\{ {a,\text{ }c} \right\},\text{ }\left\{ {a,\text{ }b} \right\},\text{ }\left\{ {a,\text{ }b,\text{ }c} \right\},\text{ }\left\{ {b,\text{ }c} \right\},\text{ }\phi$
Hence Option (c) is the correct answer

24. {1, 2, 3} $\displaystyle \cup$ {2, 3, 4} is equal to
(a) {1, 2}
(b) $\displaystyle \phi$
(c) {1, 2}
(d) {1, 2, 3, 4} .

= {1, 2, 3} $\displaystyle \cup$ {2, 3, 4}
= {1, 2, 3, 4}
Hence Option (d) is the correct answer

25. If A = {x : x² – 3x + 2 = 0} and B = {x : x² + 4x – 12 = 0}, then A – B is equal to

$\displaystyle \begin{array}{l}(a)\left\{ 3 \right\}\\(b)\left\{ {1,6} \right\}\\(c)\left\{ 1 \right\}\\(d)\phi \end{array}$

Eq A : x2 – 3x + 2 = 0 Eq B : x2 + 4x – 12 = 0

Or, x2 – x – 2x + 2 = 0 Or, x2 + 6x – 2x – 12 = 0

Or, x (x – 1) – 2 (x – 1) = 0 Or, x (x + 6) – 2 (x + 6) = 0

Or, (x – 1) (x – 2) = 0 Or, (x – 2) (x + 6) = 0

$\displaystyle \begin{array}{*{20}{l}} {\therefore x=1\text{ }Or,\text{ }2} & {\therefore x\text{ }=\text{ }2\text{ }Or,-6} \\ {\therefore A=\left\{ {1,2} \right\}~B=\{-6,2\}} & ~ \\ {\therefore A-B\text{ }=\text{ }\left\{ 1 \right\}~\left[ {items\text{ }of\text{ }set\text{ }A\text{ }not\text{ }available\text{ }in\text{ }set\text{ }B} \right]} & {} \end{array}$

Hence Option (c) is the correct answer

26. Write down the power set of the set {0}
(a) $\displaystyle \phi$
(b) {0}
(c) { $\displaystyle \phi$ }
(d) { $\displaystyle \phi$ , {0} }.

$\displaystyle Power\text{ }set\text{ }is\text{ }the\text{ }set\text{ }of\text{ }all\text{ }subsets.\text{ }\phi \text{ }is\text{ }subset\text{ }of\text{ }each\text{ }set.\text{ }So,\text{ }Power\text{ }set\text{ }of\text{ }the\text{ }set\text{ }\left\{ 0 \right\}\text{ }is\text{ }\{\phi ,\text{ }\left\{ 0 \right\}\text{ }\}.~$

Hence Option (d) is the correct answer

27. If U = { 2, 3, 4, 5, 6, 7, 8, 9,10, 11}, A = {2, 4, 7}, B = {3, 5, 7, 9, 11} and C = {7, 8, 9, 10, 11}, compute : $\displaystyle (\mathbf{A}\cap \mathbf{U})\cap (\mathbf{B}\cup \mathbf{C})$
(a) {7}
(b) $\displaystyle \phi$
(c) {2}
(d) {10}

U = { 2, 3, 4, 5, 6, 7, 8, 9,10, 11},
A = {2, 4, 7},
B = {3, 5, 7, 9, 11}
C = {7, 8, 9, 10, 11},

$\displaystyle \begin{array}{*{20}{l}} {A\cap U=\left\{ {2,\text{ }4,\text{ }7} \right\},~~~~~~~~~~~~~~~~~~\left( {common\text{ }in\text{ }both\text{ }U\text{ }\And \text{ }A} \right)} \\ {B\cup C=\left\{ {3,\text{ }5,\text{ }7,\text{ }8,\text{ }9,\text{ }11} \right\}} \\ {(A\cap U)\cap (B\cup C)\text{ }=\text{ }\left\{ 7 \right\}} \end{array}$

Hence Option (a) is the correct answer

28. If U = n {a, b, c, d, e, f}, A = {a, b, c}, find (U $\displaystyle \cup$ A)
(a) U
(b) A
(c) $\displaystyle \phi$
(d) {a, b, c}

$\displaystyle \begin{array}{*{20}{l}} {U=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\},} \\ {A=\left\{ {a,\text{ }b,\text{ }c} \right\}} \\ {(U\cup A)=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}=U} \end{array}$

Hence Option (a) is the correct answer

29. If U = {a, b, c, d, e, f}, A = {a, b, c}, B = {c, d, e, f} and C = {c, d, e}, find $\displaystyle (\mathbf{A}\cup \mathbf{B})\cup \mathbf{C}~$
(a)        {a, b, c}
(b)        B
(c)        {c, d, e}
(d)        U .

A = {a, b, c}
B = {c, d, e, f}
C = {c, d, e}

$\displaystyle \begin{array}{*{20}{l}} {(A\cup B)=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}} \\ {(A\cup B)\cup C=\left\{ {a,\text{ }b,\text{ }c,\text{ }d,\text{ }e,\text{ }f} \right\}=U} \end{array}$

Hence Option (d) is the correct answer

30. If U = {a, b, c, d, e, f}, A = {a, b, c}, B = {c, d, e, f}, C = {c, d, e}, find $\displaystyle (\mathbf{A}\cap \mathbf{B})\cup (\mathbf{A}\cap \mathbf{C})$
(a)        {c}
(b)        {a}
(c)        {f}
(d)        {e}

A = {a, b, c}
B = {c, d, e, f}
C = {c, d, e}

$\displaystyle \begin{array}{*{20}{l}} {A\cap B\text{ }=\text{ }\left\{ c \right\}} \\ {A\cap c\text{ }=\text{ }\left\{ c \right\}} \\ {(A\cup B)\cup (A\cup C)\text{ }=\text{ }\left\{ c \right\}} \end{array}$

Hence Option (a) is the correct answer

31. If S and T are two sets such that S has 21 elements, T has 32 elements, and S $\displaystyle \cap$ T has 11 elements, how many elements does S $\displaystyle \cup$ T have ?
(a)        48
(b)        26
(c)        42
(d)        36

S has 21 elements
T has 32 elements
$\displaystyle S\text{ }\cap \text{ }T\text{ }has\text{ }11\text{ }elements$

So, there are 11 common elements in S & T
So, uncommon elements in S = 21 – 11 = 10
Un common elements in T = 32 – 11 = 21
$\displaystyle S\cup T=11+10+21=42$

We may also calculate it as 21 (S) + 32 (T) – 11 common elements = 53-11=42
Hence Option (c) is the correct answer

32. In a group of 1,000 people, there are 750 people who can speak Hindi and 400 who can speak English. How many can speak Hindi only ?
(a)        600
(b)        700
(c)        850
(d)        900

No. of people can speak in Hindi = 750
No. of people can speak in English = 400
No. of people can speak in both Hindi & English
= 750 + 400 – 1000 = 1150 – 1000 = 150
No. of people can speak in only Hindi = 750 – 150 = 600

Hence Option (a) is the correct answer

33. In a class of 50 students, 35 opted for Mathematics and 37 opted for Biology. How many have opted for both Mathematics and Biology ? How many have opted for only Mathematics ? (Assume that each student has to opt for at least one of the subjects).
(a)        20
(b)        21
(c)        13
(d)        18

No. of students opted for Mathematics = 35
No. of students opted for Biology = 37
No. of students opted for Mathematics & Biology
= 35 + 37 – 50 = 72 – 50 = 22
No. of students opted only for Mathematic (35 – 22) = 13

Hence Option (c) is the correct answer

34. In a town with a population of 5,000, 3,200 people are egg-eaters, 2,500 meat-eaters and 1,500 eat both egg and meat. How many are pure vegetarians ?
(a)        400
(b)        800
(c)        700
(d)        750

Both egg-eater & meat-eater = 1500
Only egg-eater = 3200 – 1500 = 1700
Only meat-eater = 2500 – 1500 = 1000
vegetarians = 5000 – (1700 + 1500 + 100) = 5000 – 4200 = 800

Hence Option (b) is the correct answer

35. If S = {x : x² + 1 = 0, x real}, then S is

$\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & 0 \\ {\left( b \right)} & {\left\{ 0 \right\}} \\ {\left( c \right)} & {\{\phi \}} \\ {\left( d \right)} & \phi \end{array}$

$\displaystyle \begin{array}{*{20}{l}} {~{{x}^{2}}+1=0~~~~~~~~{{x}^{2}}=-1~(x\text{ }is\text{ }not\text{ }real\text{ }number,\text{ }imaginary\text{ }number} \\ {As\text{ }per\text{ }problem,\text{ }x\text{ }is\text{ }real.\text{ }There\text{ }is\text{ }no\text{ }such\text{ }real\text{ }number\text{ }(\surd -1\text{ }is\text{ }imaginary\text{ }number)} \\ {\therefore s=\phi } \end{array}$

Hence Option (d) is the correct answer

36. Which of the following is correct statement ?

$\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\phi =\text{ }\left\{ \text{ } \right\}} \\ {\left( b \right)} & {\phi =\text{ }\left\{ 0 \right\}} \\ {\left( c \right)} & {\phi =\text{ }\{\phi \}} \\ {\left( d \right)} & {All\text{ }of\text{ }the\text{ }above} \end{array}$

$\displaystyle \phi \text{ }is\text{ }indicated\text{ }by\text{ }\left\{ \text{ } \right\}$

Hence Option (a) is the correct answer

37. If A = {1, 2, 3}, B = {2, 3, 4}, then A $\displaystyle \cap$ B is
(a)        {2, 3}
(b)        {2}
(c)        {3,2}
(d)        Both a & c

$\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\text{ }\left\{ {1,\text{ }2,\text{ }3} \right\},} \\ {B\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }4} \right\},} \\ {A\cap B\text{ }=\text{ }\left\{ {2,\text{ }3} \right\}~or\text{ }\left\{ {3,2} \right\}\text{ }\left( {common\text{ }elements\text{ }of\text{ }both\text{ }sets} \right)} \end{array}$

Hence Option (d) is the correct answer

38. Let A = {0, 1, 3, 4}, B = {5, 6, 1, 3, 9} and C = {1, 2, 3, 4, 9, 13}, then $\displaystyle (\mathbf{A}\cap \mathbf{B})\cup \mathbf{C}$ is
(a)        C
(b)        {0, 1, 2, 3, 4, 9, 13}
(c)        A
(d)        B

A = {0, 1, 3, 4}
B = {5, 6, 1, 3, 9}
C = {1, 2, 3, 4, 9, 13}

$\displaystyle \begin{array}{*{20}{l}} {A\cap B\text{ }=\text{ }\left\{ 3 \right\}} \\ {A\cap B\cup C\text{ }=~\left\{ {1,\text{ }2,\text{ }3,\text{ }4,\text{ }9,\text{ }13} \right\}\text{ }=\text{ }\left\{ C \right\}} \end{array}$

Hence Option (a) is the correct answer

39. Let A = {1, 2, 3,4,5,6,7,8,9}, B = {2, 4, 6, 7, 8} and C = {3, 4, 5, 8, 9, 10}, then (A – B)
(a)        {1, 3, 4, 5, 8, 9, 10}
(b)        {1, 2, 3, 4, 5, 9}
(c)        {6, 7, 8}
(d)        {1, 3, 4}

A = {1, 2, 3, 4,5,6,7,8,9},
B = {2, 4, 6, 7, 8}
C = {3, 4, 5, 8, 9, 10},

A – B = {1, 3, 5, 9}

$\displaystyle (A-B)\cup C\text{ }=\text{ }\left\{ {1,\text{ }3,\text{ }4,\text{ }5,\text{ }8,\text{ }9,\text{ }10} \right\}$

Hence Option (a) is the correct answer

40. On a survey of 100 boys it was found that 50 used white shirt, 40 used red and 30 used blue. 20 were habituated in using both white and red shirts, while 15 used both red and blue shirts and 10 used blue and white shirts. Find the number of boys using all the colours.
(a)        22
(b)        25
(c)        45
(d)        40.

$\displaystyle \begin{array}{*{20}{l}} {~W=\text{ }50~~~~~~~~~~~~~~~~~~R\text{ }=\text{ }40~~~~~~~~~~~~~~~~~~~~~B\text{ }=\text{ }30} \\ {W\cap R\text{ }=\text{ }20~~~~~~~~~R\cap B\text{ }=\text{ }15~~~~~~~~~~~W\cap B\text{ }=\text{ }10} \\ {W\cup R\cup B\text{ }=\text{ }100} \\ {W\cap R\cap B\text{ }=\text{ }?} \end{array}$

$\displaystyle \begin{array}{*{20}{l}} {We\text{ }know} \\ {W\cup R\cup B\text{ }=\text{ }W\text{ }+\text{ }R\text{ }+\text{ }B-W\cap R-R\cap B~-W\cap B} \\ {+\text{ }W\cap R\cap B} \end{array}$

$\displaystyle \begin{array}{*{20}{l}} {Or,\text{ }100\text{ }=\text{ }50\text{ }+\text{ }40\text{ }+\text{ }30-20-15-10\text{ }+\text{ }W\cap R\cap B} \\ {Or,\text{ }100\text{ }=\text{ }120-45\text{ }+\text{ }W\cap R\cap B} \\ {Or,\text{ }100\text{ }=\text{ }75\text{ }+\text{ }W\cap R\cap B} \\ {Or,\text{ }W\cap R\cap B\text{ }=\text{ }100-75\text{ }=\text{ }25} \end{array}$

Hence Option (b) is the correct answer

41. In a group of 75 persons, 20 take tea but not coffee and 30 take tea. How many take coffee but not tea ?
(a)        45
(b)        32
(c)        50
(d)        60

Only tea = 20
both coffee & tea = 30 – 20 = 10
Only coffee = 75 – (20 + 10) = 45

Hence Option (a) is the correct answer

42. Set the positive integers, less than 3 is
(a)        {-1}
(b)        {0, 1}
(c)        {1, 2, 3}
(d)        {1, 2}

Positive integers less 3 are 1, 2.

So, answer = {1, 2}

Hence Option (d) is the correct answer

43. A = {2, 3, 5, 7}, B = {4, 6, 8, 10} then A $\displaystyle \cap$ B can be written as

$\displaystyle \begin{array}{*{20}{l}} {\left( a \right)} & {\left\{ \text{ } \right\}} \\ {\left( b \right)} & {\{\phi \}} \\ {\left( c \right)} & {A\text{ }\cup \text{ }B} \\ {\left( d \right)} & {\left\{ 0 \right\}} \end{array}$

$\displaystyle \begin{array}{*{20}{l}} {A\text{ = }\left\{ {2,\text{ }3,\text{ }5,\text{ }7} \right\}} & {} \\ {B=\left\{ {4,\text{ }6,\text{ }8,\text{ }10} \right\}} & {(Because\text{ }there\text{ }is\text{ }no\text{ }common\text{ }Element.} \\ {A\cap B=\left\{ \text{ } \right\}} & {So,\text{ }A\cap B\text{ }is\text{ }a\text{ }null\text{ }set.)} \\ {\{\phi \}\text{ }is\text{ }not\text{ }a\text{ }null\text{ }set,\text{ }it\text{ }is\text{ }a\text{ }singleton\text{ }set} & {} \end{array}$

Hence Option (a) is the correct answer

44. By Roster method to express integers greater than 5 and less than 8.
(a) {5, 6, 7}
(b) {5, 6}
(c) $\displaystyle \phi$
(d) {6, 7}

Integers greater than 5 and less than or equal to 8 are 6, 7,

So, required answer {6, 7}

Hence Option (d) is the correct answer

45. Write the set containing all days of the week beginning with S.
(a)        {Sunday, Monday}
(b)        {Saturday, Sunday}
(c)        {Friday, Saturday}
(d)        none

Days $\displaystyle \to$ Monday , Wednesday, Tuesday, Thursday, Friday, Saturday, Sunday
So, all days of the week beginning with S are Saturday, Sunday.
Required set = {Saturday, Sunday}

Hence Option (b) is the correct answer

46. A $\displaystyle \cap$ A is equal to
(a) A
(b) U
(c) $\displaystyle \phi$
(d) A’
$\displaystyle A\cap A\text{ }=\text{ }A~~\left( {Always} \right)$

Hence Option (a) is the correct answer

47. A $\displaystyle \cap$ A’ is equal to
(a) U
(b) $\displaystyle \phi$
(c) A
(d) A $\displaystyle \cup$ A’

$\displaystyle \begin{array}{*{20}{l}} {Set\text{ }A'\text{ }contain\text{ }elements\text{ }not\text{ }available\text{ }in\text{ }set\text{ }A.} \\ {So,\text{ }there\text{ }is\text{ }no\text{ }common\text{ }element\text{ }between\text{ }set\text{ }A\text{ }and\text{ }set\text{ }A'} \\ {\therefore A\cap A'=\phi } \end{array}$

Hence Option (b) is the correct answer

48. A = {1, 2, 3, 4, 5}, B = {4, 6, 7, 5}, C = {2, 3, 8, 9}, then $\displaystyle \mathbf{A}\cap (\mathbf{B}\cup \mathbf{C})$ is
(a)        Set A
(b)        Set B
(c)        Set C
(d)        {2, 3, 4, 5}

$\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\text{ }\left\{ {1,\text{ }2,\text{ }3,\text{ }4,\text{ }5} \right\}} \\ {B\text{ }=\text{ }\left\{ {4,\text{ }6,\text{ }7,\text{ }5} \right\},\text{ }C\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }8,\text{ }9} \right\},} \\ {B\cup C\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }4,\text{ }5,\text{ }6,\text{ }7,\text{ }8,\text{ }9} \right\}} \\ {A\cap (B\cup C)\text{ }=\text{ }\left\{ {2,\text{ }3,\text{ }4,\text{ }5} \right\}} \end{array}$

Hence Option (d) is the correct answer

49. V = {x : x is vowel}, Y = {x : x is ‘a’ or any letter before ‘e’ in the alphabet}, then V $\displaystyle \cap$ Y is
(a) {a, o, u}
(b) $\displaystyle \phi$
(c) {a, e, i, o, u}
(d) {a}

$\displaystyle \begin{array}{*{20}{l}} {V\text{ }=\text{ }\left\{ {a,\text{ }e,\text{ }i,\text{ }o,\text{ }u} \right\}} \\ {Y\text{ }=\text{ }\left\{ {a,\text{ }b,\text{ }c,\text{ }d,} \right\}} \\ {V\cap Y\text{ }=\text{ }\left\{ a \right\}} \end{array}$

Hence Option (d) is the correct answer

50. If A has 32 elements, B has 42 elements and A $\displaystyle \cup$ B has 62 elements. The number of elements in A $\displaystyle \cap$ B is
(a)        12
(b)        70
(c)        20
(d)        50

$\displaystyle \begin{array}{*{20}{l}} {A\text{ }=\text{ }32\text{ }elements} \\ {B\text{ }=\text{ }42\text{ }elements} \\ {A\cup B\text{ }=\text{ }62\text{ }elements} \\ {A\cup B\text{ }=\text{ }A\text{ }+\text{ }B-A\cap B} \\ {62\text{ }=\text{ }32\text{ }+\text{ }42-A\cap B} \\ {Or,\text{ }A\cap B\text{ }=\text{ }74-62\text{ }=\text{ }12} \end{array}$

Hence Option (a) is the correct answer

51. In a group of 20 children, 8 drink tea but not coffee and 13 like tea. The number of children drinking coffee but not tea is
(a)        5
(b)        7
(c)        3
(d)        4.

Total number of children = 20
Drink only tea = 8
Like tea = 13
Drink both tea & coffee (13 – 8) = 5
Drink only coffee = (20 – 8 – 5) = 7

Hence Option (b) is the correct answer

52. The number of subsets of the {6, 8, 11} is
(a)        7
(b)        5
(c)        8
(d)        3

Number of subsets = 2n-1, where n is the number of elements in the given set. The number of elements in the set is 3. So, the number of subsets = 23 -1 =8, having elements as :

$\displaystyle \left\{ 6 \right\},\text{ }\left\{ 8 \right\},\text{ }\left\{ {11} \right\},\text{ }\left\{ {6,\text{ }8} \right\},\text{ }\left\{ {6,\text{ }11} \right\},\text{ }\left\{ {11,\text{ }8} \right\},\text{ }\left\{ {6,\text{ }8,\text{ }11} \right\},\text{ }\phi$

Hence Option (c) is the correct answer

Mathematical Sets and Relations – Multiple Choice Questions

Like this:

Work from Home. Earn Money

Thank you for your response. ✨