BELIEVE EDUCATION
Std.: 9 (English) Maths - I Marks: 30
Date: 26-Jun-2025 Believe Education Time: 1:30minutes
Chapter: 1
Q.1 Multiple Choice Questions 3
1 If not a single element is common in two sets than those sets are
a. Singleton set b. Empty set c. Disjoint set d. Subset
Ans Option c.
2 M∪ N = {1, 2, 3, 4, 5, 6} and M = (1, 2, 4} then which of the following represent set N?
a. {1, 2, 3} b. {3, 4, 5, 6} c. {2, 5, 6} d. {4, 5, 6}
Ans Option b.
3 n (A) + n (A') = ............... .
a. n (B) b. n (U) c. n (AUB) d. n (AnB)
Ans Option b.
Q.2 Solve the following 4
1 Write the following sets in the set builder form.
i. F = {I, N, D, A}
ii. H = {3, 9, 27, 81, 243}
Ans i. F = {a|a is a letter in the word 'INDIA'}
ii. H = {a|a = 3n, n∈ N, n≤ 5}
2 P = { x | x is square root of 25} S = {y | y ∈ I, -5 ≤ y ≤ 5}
Verify P⊆ S
Ans P = {-5, 5}
S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
Hence every element of set P is also an element of set S.
∴ P⊆ S
3 Observe the Venn diangram and write the following sets using listing method.
A
Ans A = {2, 4, 6, 8, 10}
4 Write with reasons, which of the following sets are finite or infinite.
A = {x | x < 10, x is a natural number}
Ans A = {x | x < 10, x is a natural number}
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
i.e, set is finite.
Q.3 Attempt the following (activity) 4
1 A = {1, 3, 2, 7} then write any four subsets of A.
solution: Four subsets of A:
i. B = { _____ }
ii. C = { _____ }
iii. D = { _____ }
iv. E = { _____ }
Ans 1) 3 2) 2, 1 3) 1, 2, 7 4) 1, 7
2 A = {x | x = 2n, n ∈ N, 0 < x ≤ 10}, B = {y | y is an even number, 1 ≤ y ≤ 10}, Are A and B equal sets?
A = {_____}
B = {_____}
∴ A and B are equal sets.
Ans 1) 2, 4, 6, 8, 10 2) 2, 4, 6, 8, 10
Q.4 Answer the following 6
1 Decide which of the following are equal sets and which are not? Justify your answer.
A = {x | 3x - 1 = 2}
B = {x | x is a natural number but x is neither prime nor composite}
C = { x | x ∈ N, x < 2}
Ans A = {x | 3x - 1 = 2}
Here, 3x - 1 = 2
3x = 3
∴ x=1
∴ A={1}
B = {x | x is a natural number but x is neither prime nor composite}
∴ B={1}
C = {x | x ∈ N, x < 2 }
∴ C={1}
Here, A, B and C are equal sets.
i.e., A = B = C
2 Classify the following sets as 'singleton' or 'empty'
i. A = {x|x is a negative natural number}
ii. B = {y|y is an odd prime number y < 4}
Ans i. Eash natural number is positive.
A={ }
∴ It is an empty set.
ii.B = { 3 }
∴ It is a singleton set.
3 U = {1, 2, 3, 7, 8, 9, 10, 11, 12}
P = {1, 3, 7, 10}
then, Verify (P')' = P
Ans U = {1, 2, 3, 7, 8, 9, 10, 11, 12}
P = {1, 3, 7, 10}
∴ P' = {2, 8, 9, 11, 12}
then (P')' is the set of elements which are not in P' but in U
i.e. P
∴ (P')' = P
Q.5 Solve the following 6
1 A.T.V survey says 136 students watch only programme P1, 107 watch only programme P2, 27 watch only
programme P3, 25 students watch P1 and P2 but not P3. 37 watch P2 and P3 but not P1. 53 students watch P1
and P3 but not P2. 40 students watch all three programmes and 80 students do not watch any programme.
Find with the help of Venn diagram.
i. Number of P1 viewers.
ii. Number of P2 or P3 viewers.
iii.Total number of viewers surveyed.
Ans
i. From the Venn diagram,
Number of P1 viewers = n(P1)
= 136 + 25 + 53 + 40
= 254
∴ Number of P1 viewers is 254.
ii. Number of P2 or P3 viewers
= n(P2∪ P3)
= 107 + 25 + 40 + 37 + 53 + 27
= 289
∴ Number of P2 or P3 viewers is 289.
iii. Total Number of viewers surveyed
= Number of only P1 viewers + Number of P2 or P3 viewers + 80
= 136 + 289 + 80
= 505
∴ Total number of viewers surveyed is 505.
2 A = {1, 2, 3, 4, 5} B = {2, 3}
A ∪ B = {1, 2, 3, 4, 5}
Prove that, A ∪ B = A Using Venn diagram
Ans A = {1, 2, 3, 4, 5} B = {2, 3}
A ∪ B = {1, 2, 3, 4, 5}
Hence, if B ⊆ A then A ∪ B = A
Q.6 solve the following (non textual) 4
1 Observe the figure and verify the following equation:
n(A∪ B∪ C) = n(A) + n(B) + n(C) - n(A∩ B) - n(B∩ C) - n(C∩ A) + n(A∩ B∩ C)
Ans L.H.S = n(A∪ B∪ C)
A∪ B∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9}
∴ n(A∪ B∪ C) = 9 ... (i)
R.H.S.
= n(A) + n(B) + n(C) - n(A∩ B) - n(B∩ C) - n(C∩ A) + n(A∩ B∩ C)
A = {1, 2, 3, 4, 5}
∴ n(A) = 5
B = {2, 3, 6, 7, 8}
∴ n(B) = 5
C = {3, 4, 6, 9}
∴ n(C) = 4
A∩ B = {2, 3}
∴ n(A∩ B) = 2
B∩ C = {3, 6}
∴ n(B∩ C) = 2
C∩ A = {3, 4}
∴ n(C∩ A) = 2
A∩ B∩ C = { 3 }
∴ n(A∩ B∩ C) = 1
By putting the above values in R.H.S., we get
n(A) + n(B) + n(C) - n(A∩ B) - n(B∩ C)
- n(C∩ A) + n(A∩ B∩ C)
=5+5+4-2-2-2+1
=9 ... (ii)
∴ From (i) and (ii), we get L.H.S = R.H.S.
∴ n(A∪ B∪ C) = n(A) + n(B) + n(C)
- n(A∩ B) - n(B∩ C) - n(C∩ A) + n(A∩ B∩ C)
Q.7 Answer the following 3
1 In a class, 8 students out of 28 have a dog as their pet animal at home, 6 students have a cat as their pet
animal. 10 students have dog and cat both, then how many students do not have a dog or cat as their pet
animal at home?
Ans
In a class,
Let A be the set of students have a dog as their pet animal at home = n(A) = 8
Let B be the set of students have cat as their pet animal at home = n(B) = 6
Hence, Number of students have dog and cat both = n(A∩ B) = 10
Number of students in class = n(∪ ) = 28
Number of students do not have a dog or cat as their pet animal at home
= n(A∪ B) - n(A) + n(B) + n(A ∩ B)
= 18 + 16 - 10
n(A∪ B) = 24
= 28 - 24 = 4
Total Number of students do not have a dog or cat as their pet animal at home = 4
