RBSE (Rajasthan Board of Secondary Education) Class 11 Maths Chapter 1, “Sets,” covers fundamental concepts like the definition of sets, representation of sets, empty sets, finite and infinite sets, equal sets, subsets, power sets, universal sets, Venn diagrams, and operations on sets.

Exercise 1.4 specifically focuses on operations on sets, primarily:

Union of Sets (∪): The union of two sets A and B, denoted by A∪B, is the set of all elements that are in A or in B (or in both). A∪B={x:x∈A or x∈B}
Intersection of Sets (∩): The intersection of two sets A and B, denoted by A∩B, is the set of all elements that are common to both A and B. A∩B={x:x∈A and x∈B}
Disjoint Sets: Two sets A and B are said to be disjoint if their intersection is the empty set, i.e., A∩B=Φ.
Difference of Sets ($- $): The difference of two sets A and B, denoted by A−B, is the set of all elements that are in A but not in B. A−B={x:x∈A and x∈/B}

RBSE Class 11 Maths: Chapter 1 (Sets) – Exercise 1.4 Solutions

Question 1: Find the union of each of the following pairs of sets:

(i) X={1,3,5}, Y={1,2,3} * Solution: X∪Y={1,2,3,5} (Elements that are in X or Y or both).

(ii) A={a,e,i,o,u}, B={a,b,c} * Solution: A∪B={a,b,c,e,i,o,u}

(iii) A={x:x is a natural number and multiple of 3}, B={x:x is a natural number less than 6} * Solution: First, list the elements of each set: A={3,6,9,12,…} B={1,2,3,4,5} A∪B={1,2,3,4,5,6,9,12,…} This can also be written in set-builder form: {x:x is 1,2,3,4,5 or a multiple of 3}

(iv) A={x:x is a natural number and 1<x≤6}, B={x:x is a natural number and 6<x<10} * Solution: A={2,3,4,5,6} B={7,8,9} A∪B={2,3,4,5,6,7,8,9}

(v) A={1,2,3}, B=Φ * Solution: A∪B={1,2,3} (Union with the empty set results in the original set).

Question 2: Let A={a,b}, B={a,b,c}. Is A⊂B? What is A∪B?

Solution:
- Is A⊂B? Yes, because every element of A (‘a’ and ‘b’) is also an element of B.
- What is A∪B? A∪B={a,b,c}. (Since A⊂B, A∪B=B).

Question 3: If A and B are two sets such that A⊂B, then what is A∪B?

Solution: If A⊂B, it means all elements of A are already in B. Therefore, when you combine A and B, the resulting set will simply be B, as B already contains all elements of A. So, A∪B=B.

Question 4: If A={1,2,3,4}, B={3,4,5,6}, C={5,6,7,8} and D={7,8,9,10}, find:

(i) A∪B * Solution: A∪B={1,2,3,4,5,6}

(ii) A∪C * Solution: A∪C={1,2,3,4,5,6,7,8}

(iii) B∪C * Solution: B∪C={3,4,5,6,7,8}

(iv) B∪D * Solution: B∪D={3,4,5,6,7,8,9,10}

(v) A∪B∪C * Solution: A∪B∪C=(A∪B)∪C={1,2,3,4,5,6}∪{5,6,7,8}={1,2,3,4,5,6,7,8}

(vi) A∪B∪D * Solution: A∪B∪D=(A∪B)∪D={1,2,3,4,5,6}∪{7,8,9,10}={1,2,3,4,5,6,7,8,9,10}

(vii) B∪C∪D * Solution: B∪C∪D=(B∪C)∪D={3,4,5,6,7,8}∪{7,8,9,10}={3,4,5,6,7,8,9,10}

Question 5: Find the intersection of each of the following pairs of sets of Question 1:

(i) X={1,3,5}, Y={1,2,3} * Solution: X∩Y={1,3} (Elements common to X and Y).

(ii) A={a,e,i,o,u}, B={a,b,c} * Solution: A∩B={a}

(iii) A={3,6,9,12,…}, B={1,2,3,4,5} * Solution: A∩B={3}

(iv) A={2,3,4,5,6}, B={7,8,9} * Solution: A∩B=Φ (These are disjoint sets).

(v) A={1,2,3}, B=Φ * Solution: A∩B=Φ (Intersection with the empty set is always the empty set).

Question 6: If A={3,5,7,9,11}, B={7,9,11,13}, C={11,13,15} and D={15,17}, find:

(i) A∩B * Solution: A∩B={7,9,11}

(ii) B∩C * Solution: B∩C={11,13}

(iii) A∩C∩D (or A∩(C∩D)) * Solution: C∩D={15} A∩(C∩D)={3,5,7,9,11}∩{15}=Φ

(iv) A∩C * Solution: A∩C={11}

(v) B∩D * Solution: B∩D=Φ

(vi) A∩(B∪C) * Solution: First find B∪C={7,9,11,13,15} Then, A∩(B∪C)={3,5,7,9,11}∩{7,9,11,13,15}={7,9,11}

(vii) A∩D * Solution: A∩D=Φ

(viii) A∩(B∪D) * Solution: First find B∪D={7,9,11,13,15,17} Then, A∩(B∪D)={3,5,7,9,11}∩{7,9,11,13,15,17}={7,9,11}

(ix) (A∩B)∩(B∪C) * Solution: First find A∩B={7,9,11} Then find B∪C={7,9,11,13,15} Then, (A∩B)∩(B∪C)={7,9,11}∩{7,9,11,13,15}={7,9,11}

(x) (A∪D)∩(B∪C) * Solution: First find A∪D={3,5,7,9,11,15,17} Then find B∪C={7,9,11,13,15} Then, (A∪D)∩(B∪C)={3,5,7,9,11,15,17}∩{7,9,11,13,15}={7,9,11,15}

Question 7: If A={x:x is a natural number}, B={x:x is an even natural number}, C={x:x is an odd natural number}, D={x:x is a prime number}, find:

(i) A∩B * Solution: A contains all natural numbers {1,2,3,…}. B contains even natural numbers {2,4,6,…}. The intersection will be the elements common to both, which are all even natural numbers. A∩B=B={x:x is an even natural number}

(ii) A∩C * Solution: A contains all natural numbers. C contains odd natural numbers {1,3,5,…}. The intersection will be all odd natural numbers. A∩C=C={x:x is an odd natural number}

(iii) A∩D * Solution: A contains all natural numbers. D contains prime numbers {2,3,5,7,…}. The intersection will be all prime numbers. A∩D=D={x:x is a prime number}

(iv) B∩C * Solution: B contains even natural numbers. C contains odd natural numbers. There is no common element between even and odd numbers. B∩C=Φ

(v) B∩D * Solution: B contains even natural numbers {2,4,6,…}. D contains prime numbers {2,3,5,7,…}. The only even prime number is 2. B∩D={2}

(vi) C∩D * Solution: C contains odd natural numbers {1,3,5,7,…}. D contains prime numbers {2,3,5,7,…}. The intersection will be all odd prime numbers. C∩D={3,5,7,11,…}={x:x is an odd prime number}

Question 8: Which of the following pairs of sets are disjoint?

(i) {1,2,3,4} and {x:x is a natural number and 4≤x≤6} * Solution: Let A={1,2,3,4} and B={4,5,6}. A∩B={4}. Since the intersection is not empty, they are not disjoint.

(ii) {a,e,i,o,u} and {c,d,e,f} * Solution: Let A={a,e,i,o,u} and B={c,d,e,f}. A∩B={e}. Since the intersection is not empty, they are not disjoint.

(iii) {x:x is an even integer} and {x:x is an odd integer} * Solution: Let A={…,−4,−2,0,2,4,…} and B={…,−3,−1,1,3,…}. A∩B=Φ. There is no common element between even and odd integers. So, they are disjoint.

Question 9: If A={3,6,9,12,15,18,21}, B={4,8,12,16,20}, C={2,4,6,8,10,12,14,16}, D={5,10,15,20}, find:

(i) A−B * Solution: Elements in A but not in B: A−B={3,6,9,15,18,21} (12 is in B).

(ii) A−C * Solution: Elements in A but not in C: A−C={3,9,15,18,21} (6 and 12 are in C).

(iii) A−D * Solution: Elements in A but not in D: A−D={3,6,9,12,18,21} (15 is in D).

(iv) B−A * Solution: Elements in B but not in A: B−A={4,8,16,20} (12 is in A).

(v) C−A * Solution: Elements in C but not in A: C−A={2,4,8,10,14,16} (6 and 12 are in A).

(vi) D−A * Solution: Elements in D but not in A: D−A={5,10,20} (15 is in A).

(vii) B−C * Solution: Elements in B but not in C: B−C={20} (4, 8, 12, 16 are in C).

(viii) C−B * Solution: Elements in C but not in B: C−B={2,6,10,14} (4, 8, 12, 16 are in B).

(ix) D−B * Solution: Elements in D but not in B: D−B={5,10,15} (20 is in B).

(x) C−D * Solution: Elements in C but not in D: C−D={2,4,6,8,12,14,16} (10 is in D).

(xi) D−C * Solution: Elements in D but not in C: D−C={5,15,20} (10 is in C).

Question 10: If X={a,b,c,d} and Y={f,b,d,g}, find:

(i) X−Y * Solution: X−Y={a,c} (Elements in X but not in Y).

(ii) Y−X * Solution: Y−X={f,g} (Elements in Y but not in X).

(iii) X∩Y * Solution: X∩Y={b,d} (Elements common to X and Y).

Question 11: If R is the set of real numbers and Q is the set of rational numbers, then what is R−Q?

Solution:
- Real numbers (R) include all rational and irrational numbers.
- Rational numbers (Q) are numbers that can be expressed as a fraction p/q, where p and q are integers and q=0.
- Irrational numbers are real numbers that cannot be expressed as a simple fraction (e.g., 2,π).
- R−Q represents the set of elements that are in R but not in Q. This set comprises all numbers that are real but not rational.
- Therefore, R−Q={x:x is an irrational number}.
- Solution: R−Q is the set of irrational numbers.

Question 12: State whether each of the following statements is true or false. Justify your answer.

(i) {2,3,4,5} and {3,6} are disjoint sets. * Solution: False. The intersection of these two sets is {3}, which is not empty. Therefore, they are not disjoint.

(ii) {a,e,i,o,u} and {a,b,c,d} are disjoint sets. * Solution: False. The intersection is {a}, which is not empty. Therefore, they are not disjoint.

(iii) {2,6,10,14} and {3,7,11,15} are disjoint sets. * Solution: True. The intersection of these two sets is Φ (the empty set), as there are no common elements. Therefore, they are disjoint.

(iv) {2,6,10} and {3,7,11} are disjoint sets. * Solution: True. The intersection of these two sets is Φ (the empty set), as there are no common elements. Therefore, they are disjoint.

Last Updated on July 22, 2025 by Aman Singh

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Aman Singh

Aman Singh | M.Sc. Mathematics, RRBMU University Alwar

A seasoned Mathematics Educator with 7 years of dedicated experience in the field of education. Specializing in simplifying complex mathematical concepts, Aman has a proven track record of helping students master advanced topics. Holds an M.Sc. in Mathematics from RRBMU University, Alwar. Passionate about leveraging conceptual clarity and effective teaching methodologies to drive student success and achievement.

"Transforming mathematical complexity into conceptual clarity."

For the past 7 years, Aman Singh has been on a mission to redefine math education. Armed with an M.Sc. in Mathematics from RRBMU University Alwar, Aman brings a deep well of knowledge and seven years of classroom insight to every lesson. Specializing in turning student struggle into genuine mastery, Aman believes math isn't just about numbers—it's about building confidence and problem-solving muscle.