Set Theory / Venn Diagrams Questions and Answers

Q: In a class of 50 students, 30 passed in Mathematics, 25 passed in English, and 10 passed in both. How many students passed in at least one subject ?

45 students passed in at least one subject.Adding the two pass lists counts the 10 students who passed both subjects twice, so subtract them once:\[ n(M \cup E) = 30 + 25 - 10 = 45 \]So 45 passed at least one subject and the other 5 of the 50 failed both.55 is the plain sum 30 + 25, which double-counts the 10. 50 is the class size, which would mean no one failed both, and 40 subtracts the overlap twice.

Q: In the given diagram, circles A, B and C represents players who can play Cricket, Hockey and Volleyball, respectivelyRegion of the diagramNumberA only2B only4C only9A and B only8B and C only3A and C only7A, B and C5Then the number of players who can play Cricket and Volleyball but not Hockey is :

The number is 7.Cricket is circle A and Volleyball circle C. 'Cricket and Volleyball but not Hockey' is the part common to A and C that lies outside B, the region marked 7.5 is the centre, where all three circles meet, so those players also play hockey. 8 is A and B only, and 9 is C only.

Q: If A = {x : x is a prime number less than 10}, find the cardinality of A.

We need to list the prime numbers less than 10.Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves.The prime numbers less than 10 are 2, 3, 5, and 7.So, A = {2, 3, 5, 7}.The cardinality of A, denoted n(A), is the number of elements in the set, which is 4.

Important Instructions

21.

In a class of 50 students, 30 passed in Mathematics, 25 passed in English, and 10 passed in both. How many students passed in at least one subject ?

A.40
B.45
C.50
D.55

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

45 students passed in at least one subject.

Adding the two pass lists counts the 10 students who passed both subjects twice, so subtract them once:

\[ n(M \cup E) = 30 + 25 - 10 = 45 \]

So 45 passed at least one subject and the other 5 of the 50 failed both.

55 is the plain sum 30 + 25, which double-counts the 10. 50 is the class size, which would mean no one failed both, and 40 subtracts the overlap twice.

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22.

In the given diagram, circles A, B and C represents players who can play Cricket, Hockey and Volleyball, respectively

Region of the diagram	Number
A only	2
B only	4
C only	9
A and B only	8
B and C only	3
A and C only	7
A, B and C	5

Then the number of players who can play Cricket and Volleyball but not Hockey is :

Exam Year(s)

UKPSC Lower PCS - 2025,

Answer: B

The number is 7.

Cricket is circle A and Volleyball circle C. 'Cricket and Volleyball but not Hockey' is the part common to A and C that lies outside B, the region marked 7.

5 is the centre, where all three circles meet, so those players also play hockey. 8 is A and B only, and 9 is C only.

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23.

If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, what is A ∪ B?

A.{3, 4}
B.{1, 2, 3, 4, 5, 6}
C.{1, 2, 5, 6}
D.{1, 2}

Answer: B

The union collects every element of either set, listed once.

\[ A \cup B = \{1, 2, 3, 4, 5, 6\} \]

3 and 4 belong to both but are not repeated.

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24.

If A = {a, b, c, d} and B = {c, d, e, f}, what is A ∩ B?

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

Answer: C

The intersection keeps only the elements common to both sets.

\[ A \cap B = \{c, d\} \]

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25.

Let U = {1, 2, 3, 4, 5, 6, 7, 8} be the universal set and A = {1, 3, 5, 7}. Find A'.

A.{1, 3, 5, 7}
B.{}
C.{2, 4, 6, 8}
D.{1, 2, 3, 4, 5, 6, 7, 8}

Answer: C

The complement contains everything in the universal set that is not in \( A \).

\[ A' = \{2, 4, 6, 8\} \]

Together \( A \) and \( A' \) make up the whole of \( U \) and share nothing.

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26.

If a set A has 5 elements, what is the number of elements in its power set, n(P(A))?

A.5
B.10
C.25
D.32

Answer: D

The power set contains every subset, and each element is either in or out.

\[ n(P(A)) = 2^{n(A)} = 2^{5} = 32 \]

That count includes the empty set and \( A \) itself.

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27.

If n(A) = 20, n(B) = 30 and n(A ∪ B) = 40, find n(A ∩ B).

A.5
B.10
C.15
D.20

Answer: B

We use the formula for the cardinality of the union of two sets:

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Substituting the given values:

40 = 20 + 30 - n(A ∩ B)

40 = 50 - n(A ∩ B)

n(A ∩ B) = 50 - 40 = 10

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28.

In a class of 50 students, 30 like Math, 25 like Science, and 10 like both. How many students like neither Math nor Science?

A.5
B.10
C.15
D.20

Answer: A

Let M be the set of students who like Math and S be the set of students who like Science.

n(M) = 30, n(S) = 25, n(M ∩ S) = 10.

Number of students who like at least one subject is n(M ∪ S) = n(M) + n(S) - n(M ∩ S) = 30 + 25 - 10 = 45.

Total students = 50.

Number of students who like neither = Total students - n(M ∪ S) = 50 - 45 = 5.

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29.

Which of the following represents the shaded region in the Venn diagram?

A.\( A \cup B \)
B.\( A \cap B \)
C.\( A - B \)
D.\( B - A \)

Answer: C

The shaded region lies inside circle A but outside circle B - the part of \( A \) that \( B \) does not overlap.

\[ A - B = \{\, x \mid x \in A \text{ and } x \notin B \,\} \]

\( A \cap B \) would be the lens where they overlap, and \( B - A \) the mirror image on the other side.

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30.

If A = {x : x is a prime number less than 10}, find the cardinality of A.

Answer: B

We need to list the prime numbers less than 10.

Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves.

The prime numbers less than 10 are 2, 3, 5, and 7.

So, A = {2, 3, 5, 7}.

The cardinality of A, denoted n(A), is the number of elements in the set, which is 4.

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Set Theory / Venn Diagrams Questions and Answers

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Set Theory / Venn Diagrams Questions and Answers

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Tags: Set Theory Questions and Answers || Venn Diagrams MCQ Questions || Set Operations GK || Venn Diagram Quantitative Aptitude GK || Logical Reasoning Venn Diagrams

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