Class 11 Maths NCERT Solutions Chapter 1 Ex 1.4 – Union and Intersection of Sets | Boundless Maths
Ex 1.4 Class 11 Maths NCERT Solutions · Chapter 1

Class 11 Maths NCERT Solutions Chapter 1 Ex 1.4 – Union and Intersection of Sets

Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 1 Ex 1.4 — all 12 questions solved, with Venn diagrams for union, intersection, set difference, and disjoint sets, for CBSE 2026-27.

This is the longest exercise in the chapter, but it repeats only four operations across many sets: union (combine everything), intersection (keep only the overlap), difference (what's in one set but not the other), and the special case of disjoint sets (no overlap at all). Questions 1–4 build up union, Questions 5–7 do the same for intersection, Question 8 introduces disjoint sets visually, Questions 9–10 cover set difference, and Questions 11–12 close with a classic result (R − Q = irrational numbers) and a set of true/false disjoint checks. Every diagram-worthy idea here is illustrated with a Venn diagram alongside the working.

12Questions
Easy–HardDifficulty Mix
2026-27CBSE Syllabus

Class 11 Maths NCERT Solutions Chapter 1 Ex 1.4 — All 12 Questions

1

Find the union of each of the following pairs of sets:
(i) X=\{1,3,5\}, Y=\{1,2,3\}
(ii) A=\{a,e,i,o,u\}, B=\{a,b,c\}
(iii) A=\{x : x \text{ is a natural number and a multiple of 3}\}, B=\{x : x \text{ is a natural number less than 6}\}
(iv) A=\{x : x \text{ is a natural number and } 1\lt x\le6\}, B=\{x : x \text{ is a natural number and } 6\lt x\lt10\}
(v) A=\{1,2,3\}, B=\phi

Easy +
Solution

The union of two sets combines every element that is in either set, writing each shared element only once.

U A B
A∪B: the shaded region is everything in A, in B, or in both.

(i) Combine the elements of X and Y, listing 1 and 3 (common to both) only once.

X\cup Y=\{1,2,3,5\}

(ii) Combine the elements of A and B; a is common to both and is listed once.

A\cup B=\{a,b,c,e,i,o,u\}

(iii) A=\{3,6,9,12,\ldots\} (multiples of 3), B=\{1,2,3,4,5\}. Combining gives every natural number less than 6, together with every multiple of 3.

A\cup B=\{1,2,3,4,5,6,9,12,15,\ldots\}

(iv) A=\{2,3,4,5,6\}, B=\{7,8,9\}. These sets share no elements, so the union simply lists all of them together.

A\cup B=\{2,3,4,5,6,7,8,9\}

(v) Since B=\phi has no elements, the union is just A itself.

A\cup B=\{1,2,3\}
2

Let A=\{a,b\}, B=\{a,b,c\}. Is A\subset B? What is A\cup B?

Easy +
Solution

Both a and b, the elements of A, are also elements of B.

Yes, A ⊂ B.

Since A is already entirely contained in B, taking the union adds nothing new — every element of A is already listed in B.

A\cup B=\{a,b,c\}=B
3

If A and B are two sets such that A\subset B, then what is A\cup B?

Easy +
Solution

If every element of A already lies in B, then combining A with B introduces no new elements beyond what B already contains.

A ∪ B = B
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\cup B
(ii) A\cup C
(iii) B\cup C
(iv) B\cup D
(v) A\cup B\cup C
(vi) A\cup B\cup D
(vii) B\cup C\cup D

Medium +
Solution

(i) Combine A and B, listing shared elements 3 and 4 once.

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

(ii) Combine A and C — they share no elements.

A\cup C=\{1,2,3,4,5,6,7,8\}

(iii) Combine B and C, listing shared elements 5 and 6 once.

B\cup C=\{3,4,5,6,7,8\}

(iv) Combine B and D — they share no elements.

B\cup D=\{3,4,5,6,7,8,9,10\}

(v) Combine all elements of A, B and C.

A\cup B\cup C=\{1,2,3,4,5,6,7,8\}

(vi) Combine all elements of A, B and D.

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

(vii) Combine all elements of B, C and D.

B\cup C\cup D=\{3,4,5,6,7,8,9,10\}
5

Find the intersection of each pair of sets of Question 1 above.

Easy +
Solution

The intersection of two sets keeps only the elements common to both.

U A B
A∩B: only the overlap of A and B is shaded.

(i) The elements common to X=\{1,3,5\} and Y=\{1,2,3\} are 1 and 3.

X\cap Y=\{1,3\}

(ii) The only letter common to A=\{a,e,i,o,u\} and B=\{a,b,c\} is a.

A\cap B=\{a\}

(iii) A = multiples of 3, B = {1,2,3,4,5}. The only number that is both a multiple of 3 and less than 6 is 3.

A\cap B=\{3\}

(iv) A=\{2,3,4,5,6\} and B=\{7,8,9\} share no elements.

A\cap B=\phi

(v) Since B=\phi has no elements, there can be nothing common to A and B.

A\cap B=\phi
6

If A=\{3,5,7,9,11\}, B=\{7,9,11,13\}, C=\{11,13,15\} and D=\{15,17\}; find:
(i) A\cap B
(ii) B\cap C
(iii) A\cap C\cap D
(iv) A\cap C
(v) B\cap D
(vi) A\cap(B\cup C)
(vii) A\cap D
(viii) A\cap(B\cup D)
(ix) (A\cap B)\cap(B\cup C)
(x) (A\cup D)\cap(B\cup C)

Hard +
Solution

(i) Elements common to A and B: 7, 9, 11.

A\cap B=\{7,9,11\}

(ii) Elements common to B and C: 11, 13.

B\cap C=\{11,13\}

(iii) A\cap C=\{11\} (only 11 is common to A and C). Then \{11\}\cap D=\{11\}\cap\{15,17\}=\phi (11 is not in D).

A\cap C\cap D=\phi

(iv) Elements common to A and C: only 11.

A\cap C=\{11\}

(v) B and D share no elements.

B\cap D=\phi

(vi) First, B\cup C=\{7,9,11,13,15\}. Then intersect with A: elements common to \{3,5,7,9,11\} and \{7,9,11,13,15\} are 7, 9, 11.

A\cap(B\cup C)=\{7,9,11\}

(vii) A and D share no elements.

A\cap D=\phi

(viii) First, B\cup D=\{7,9,11,13,15,17\}. Then intersect with A: common elements are 7, 9, 11.

A\cap(B\cup D)=\{7,9,11\}

(ix) A\cap B=\{7,9,11\} from part (i). B\cup C=\{7,9,11,13,15\} from part (vi). Their intersection: 7, 9, 11 (all three already lie in the smaller set).

(A\cap B)\cap(B\cup C)=\{7,9,11\}

(x) A\cup D=\{3,5,7,9,11,15,17\}. B\cup C=\{7,9,11,13,15\} from part (vi). Elements common to both: 7, 9, 11, 15.

(A\cup D)\cap(B\cup C)=\{7,9,11,15\}
7

If A=\{x : x \text{ is a natural number}\}, B=\{x : x \text{ is an even natural number}\}, C=\{x : x \text{ is an odd natural number}\} and D=\{x : x \text{ is a prime number}\}, find:
(i) A\cap B
(ii) A\cap C
(iii) A\cap D
(iv) B\cap C
(v) B\cap D
(vi) C\cap D

Medium +
Solution

Note that B\subset A, C\subset A and D\subset A, since every even number, odd number, and prime number is itself a natural number.

(i) Since B is entirely contained in A, their intersection is just B.

A\cap B=B=\{2,4,6,8,\ldots\}

(ii) Since C is entirely contained in A, their intersection is just C.

A\cap C=C=\{1,3,5,7,\ldots\}

(iii) Since D is entirely contained in A, their intersection is just D.

A\cap D=D=\{2,3,5,7,11,\ldots\}

(iv) No natural number is both even and odd at once.

B\cap C=\phi

(v) The only even prime number is 2 — every other prime is odd.

B\cap D=\{2\}

(vi) Every prime number except 2 is odd, so the odd primes are all of D except 2.

C\cap D=\{3,5,7,11,13,\ldots\}=D-\{2\}
8

Which of the following pairs of sets are disjoint?
(i) \{1,2,3,4\} and \{x : x \text{ is a natural number and } 4\le x\le6\}
(ii) \{a,e,i,o,u\} and \{c,d,e,f\}
(iii) \{x : x \text{ is an even integer}\} and \{x : x \text{ is an odd integer}\}

Easy +
Solution

Two sets are disjoint only if they share absolutely no elements — their intersection must be empty.

(i) The second set is \{4,5,6\}. It shares the element 4 with \{1,2,3,4\}.

1,2,3 5,6 4
The circles overlap at 4 — NOT disjoint.
Not disjoint (they share the element 4).

(ii) Both sets contain the letter e.

a,i,o,u c,d,f e
The circles overlap at e — NOT disjoint.
Not disjoint (they share the element e).

(iii) No integer can be both even and odd — the two sets share nothing.

even odd
The circles never touch — disjoint sets.
Disjoint (no integer is both even and odd).
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\} and D=\{5,10,15,20\}; find:
(i) A-B
(ii) A-C
(iii) A-D
(iv) B-A
(v) C-A
(vi) D-A
(vii) B-C
(viii) B-D
(ix) C-B
(x) D-B
(xi) C-D
(xii) D-C

Hard +
Solution

A-B keeps only the elements of A that are NOT in B — it removes the overlap but keeps the rest of A.

U A B
Blue-only region = A−B. Teal-only region = B−A. The white overlap is A∩B — it belongs to neither A−B nor B−A.

Working through each part by directly comparing the four given sets:

(i) Remove from A any element also in B (12 is in both).

A-B=\{3,6,9,15,18,21\}

(ii) Remove from A any element also in C (6 and 12 are in both).

A-C=\{3,9,15,18,21\}

(iii) Remove from A any element also in D (15 is in both).

A-D=\{3,6,9,12,18,21\}

(iv) Remove from B any element also in A (12 is in both).

B-A=\{4,8,16,20\}

(v) Remove from C any element also in A (6 and 12 are in both).

C-A=\{2,4,8,10,14,16\}

(vi) Remove from D any element also in A (15 is in both).

D-A=\{5,10,20\}

(vii) Remove from B any element also in C (4, 8, 12, 16 are in both).

B-C=\{20\}

(viii) Remove from B any element also in D (20 is in both).

B-D=\{4,8,12,16\}

(ix) Remove from C any element also in B (4, 8, 12, 16 are in both).

C-B=\{2,6,10,14\}

(x) Remove from D any element also in B (20 is in both).

D-B=\{5,10,15\}

(xi) Remove from C any element also in D (10 is in both).

C-D=\{2,4,6,8,12,14,16\}

(xii) Remove from D any element also in C (10 is in both).

D-C=\{5,15,20\}
10

If X=\{a,b,c,d\} and Y=\{f,b,d,g\}, find:
(i) X-Y
(ii) Y-X
(iii) X\cap Y

Easy +
Solution

X and Y share the elements b and d.

U X Y a, c b, d f, g
X-only = {a,c}, overlap = {b,d}, Y-only = {f,g}.

(i) Remove from X any element also in Y (b and d are in both).

X-Y=\{a,c\}

(ii) Remove from Y any element also in X (b and d are in both).

Y-X=\{f,g\}

(iii) The elements common to both X and Y.

X\cap Y=\{b,d\}
11

If \mathbb{R} is the set of real numbers and \mathbb{Q} is the set of rational numbers, then what is \mathbb{R}-\mathbb{Q}?

Easy +
Solution

\mathbb{R}-\mathbb{Q} consists of every real number that is NOT rational.

\mathbb{R}-\mathbb{Q} = the set of irrational numbers (e.g., \sqrt2,\ \sqrt5,\ \pi).
12

State whether each of the following statement is true or false. Justify your answer.
(i) \{2,3,4,5\} and \{3,6\} are disjoint sets.
(ii) \{a,e,i,o,u\} and \{a,b,c,d\} are disjoint sets.
(iii) \{2,6,10,14\} and \{3,7,11,15\} are disjoint sets.
(iv) \{2,6,10\} and \{3,7,11\} are disjoint sets.

Medium +
Solution

(i) The two sets share the element 3.

2,4,5 6 3
The circles overlap at 3 — NOT disjoint.
False — they share the element 3, so their intersection is not empty.

(ii) The two sets share the element a.

e,i,o,u b,c,d a
The circles overlap at a — NOT disjoint.
False — they share the element a, so their intersection is not empty.

(iii) No number appears in both lists.

2,6,10,14 3,7,11,15
The circles never touch — disjoint sets.
True — the two sets share no elements, so their intersection is empty.

(iv) No number appears in both lists.

2,6,10 3,7,11
The circles never touch — disjoint sets.
True — the two sets share no elements, so their intersection is empty.

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Common Questions

Class 11 Maths NCERT Solutions Chapter 1 Ex 1.4 — FAQs

How many questions are there in Exercise 1.4?
Exercise 1.4 has 12 questions, covering the union of sets, intersection of sets, set difference (A−B), and disjoint sets, with several questions asking for many combinations across the same four sets.
What is the difference between the union and intersection of two sets?
The union A∪B contains every element that is in A, in B, or in both — you combine everything. The intersection A∩B contains only the elements common to both A and B — you keep only the overlap. In a Venn diagram, the union shades both circles entirely, while the intersection shades only the region where they overlap.
Where can I find the official NCERT textbook for this chapter?
Sets is Chapter 1 of the NCERT Class 11 Mathematics textbook, published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the exercise exactly as it appears there.

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