Class 11 Applied Maths | Sets and Relations | Holidays Homework 2025

Q1. Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}.

Find A′, B′, A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′.

Answer: 

i. A′ = U − A = {1,4,5,6}

ii. B' = U − B = {1, 2, 6}

iii. A′ ∩ B′ = {1, 6}

iv. A ∪ B = {2, 3, 4, 5}

v. (A ∪ B)′ = U − (A∪B) = {1,6} 

⇒ ( A ∪ B )′ = A′ ∩ B′ (De Morgan's Law)

Q2.  Let U = {x : x ∈ N, x ≤ 9}; A = {x : x is an even number, x ∈ N, 0 < x < 10};B = {2, 3, 5, 7}. Write the set (A ∪ B)′

Answer: U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {2, 4, 6, 8}

B = {2, 3, 5, 7}

(A ∪ B) = {2, 3, 4, 5, 6, 7, 8}

(A ∪ B)′ = U − (A ∪ B) = {1, 9}

Q3.  Let n (A) = m, and n (B) = n. Find the total number of non-empty relations that can be defined.

Answer: The Cartesian product A×B contains m×n elements.

The number of subsets of A×B (including the empty set) is 2^(m⋅n)

Total non-empty relations=2^(m⋅n) − 1

Q4. Let A = {1, 2, 3}, B = {4} and C = {5}

 (i) Verify that: A × (B – C) = (A × B) – (A × C)

 (ii) Find (A × B) ∩ (A × C)

Answer: 

(i) B – C = {4} - {5} = {4} No common elements

LHS = A × (B – C) = {(1,4), (2, 4), (3,4)}

A × B = {(1,4), (2,4), (3,4)}

A × C = {(1,5), (2,5), (3,5)}

RHS = (A × B) – (A × C) = {(1,4), (2,4), (3,4)}

Thus LHS = RHS

(ii) A × B = {(1,4), (2,4), (3,4)}

A × C = {(1,5), (2,5), (3,5)}

(A × B) ∩ (A × C) = ∅  (empty set)

Q5.  Find x and y if: 

(i) (4x + 3, y) = (3x + 5, – 2) 

(ii) (x – y, x + y) = (6, 10)

Answer: 

(i) (4x + 3, y) = (3x + 5, – 2) 

4x + 3 = 3x + 5

4x - 3x = 5 -3 

x = 2

(ii) (x – y, x + y) = (6, 10)

x – y = 6

x + y = 10

Add both eqns.

2x = 16

x = 8

8 + y = 10

y = 2

Q6. Let A = {1, 2, 3, 5} and B = {4, 6, 9} define a relation from A to B, given by

 R ={(a, b) : a ∈ A, b ∈ B , (a-b) is odd}.

(i) Write R in roster form.

(ii) Find the domain (R) and Range (R).

(iii) Depict R using an arrow diagram

Answer: R = {(1,4), (1,6), (2,9), (2, 9), (3,4), (5,4), (5,6)}

Domain = {1, 2, 3, 5}

Range = {4, 6, 9}

Q7. If A = {1, 2, 3, 4,..., 14} and the relation R is defined from A to A by

 R = {(x,y) : 3x – y = 0, x ,y ∈A}.

 (i) Write R in roster form.

 (ii) Write its domain, codomain and range.

Answer: 

(i) y = 3x and y;  x, y ∈ A

R = {(1,3),(2,6), (3,9), (4,12)}

(ii) Domain of R = {1, 2, 3, 4}

Range = {3, 6, 9, 12}

Codomain = {1, 2, 3, 4,..., 14}

Q8. In a class of 24 students, 16 have taken Biology, 13 have taken Physics, and 12 have taken Chemistry. Six students have taken both Biology and Chemistry, 10 have taken both Biology and Physics, and 5 have taken both Physics and Chemistry. Four students have taken all three subjects. The school needed to gather more information about the various groupings of the students in order to organise classrooms with the appropriate number of lab kits for each subject. Based on the information provided, answer the following questions:

(i) How many students have only Chemistry?

(ii) How many students have only one subject?

(iii) How many students have Biology or Physics but not Chemistry?

Answer: n(U) = 24, n(B) = 16, n(P) = 13, n(C) = 12

n(B ∩ C) = 6, n(B ∩ P) = 10, n(P ∩ C) = 5, n(B ∩ P ∩ C) = 4 

n(B) = n(B_only) + n(B ∩ C) +  n(B ∩ P) - n(B ∩ P ∩ C) 

16 = n(B_only) + 6 + 10 - 4

n(B_only) = 16 - 16 + 4 = 4

n(P) = n(P_only) + n(P ∩ C) +  n(B ∩ P) - n(B ∩ P ∩ C)

13 =  n(P_only) + 5 + 10 - 4

n(P_only) = 17 - 15 = 2

n(C) = n(C_only) + n(P ∩ C) +  n(B ∩ C) - n(B ∩ P ∩ C)

12 = n(C_only) + 5 + 6 - 4

n(C_only) = 12 - 7 = 5 

(i) Chemistry only students = 5

(ii) Students having only one subject = n(P_only) + n(C_only) + n(B_only)

     = 2 + 5 + 4 = 11

(iii) students have Biology or Physics but not Chemistry = n(B_only) + n(P_only) + n(B ∩ P) - n(B ∩ P ∩ C)

 = 4 + 2 + 10 - 4 = 12