ML Aggarwal Operations on sets Exe-6 Class 8 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-6 Questions for Operation on Sets as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.
ML Aggarwal Operation on Sets Exe-6 Class 8 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 8th |
| Chapter-6 | Operation on Sets |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-6Questions |
| Edition | 2023-2024 |
Operation on Sets Exe-6
ML Aggarwal Class 8 ICSE Maths Solutions
Page-102
Question 1. If A=(1,2,3….8), B=(3,5,7,9,11) and C= (0,5,10,20) find
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) A ∩ B
(v) A ∩ C
(vi) B ∩ C
Also find the cardinal number of the sets B ∪ C, A ∪ B, A ∩ C and B ∩ C
Answer :
A = {0, 1, 2, 3, 4, 5, 6, 7, 8}
B = {3, 5, 7, 9, 11}
C = {0, 5, 10, 20}
(i) A ∪ B
A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11}
Cardinal number of set i.e. n(A ∪ B) = 11
(ii) A ∪ C
A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 20}
Cardinal number of set i.e. n(A ∪ C) = 11
(iii) B ∪ C
B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20}
Cardinal number of set i.e. n(B ∪ C) = 8
(iv) A ∩ B
A ∩ B = {3, 5, 7}
Cardinal number of set i.e. n(A ∩ B) = 3
(v) A ∩ C
A ∩ C = {0, 5}
Cardinal number of set i.e. n(A ∩ C) = 2
(vi) B ∩ C
B ∩ C = {5}
Cardinal number of set i.e. n(B ∩ C) = 1
Question 2. Find A’ when
(i) A= {0, 1, 4, 7} and E, = {x | x ϵ W, x ≤ 10}
(ii) A = {consonants} and ξ = {alphabets of English}
(iii) A = boys in class VIII of all schools in Bengaluru} and ξ = {students in class VIII of all schools in Bengaluru}
(iv) A = {letters of KALKA} and ξ = {letters of KOLKATA}
(v) A = {odd natural numbers} and ξ = {whole numbers}.
Answer :
(i) A = {0, 1, 4, 7}
E = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Then, A’ = {2, 3, 5, 6, 8, 9, 10}
(ii) A = {consonants} and ξ = {alphabets of English}
A = {consonants}
ξ = {alphabets of English}
A’ = {Vowels}
(iii) A = {boys in class VIII of all schools in Bengaluru} and ξ = {students in class VIII of all schools in Bengaluru}
A’ = {Girls in class VIII of all schools in Bengaluru}
(iv) A = {letters of KALKA} and ξ = {letters of KOLKATA}
A = {K, A, L, K, A}
ξ = {K, O, L, K, A, T, A}
A’ = {O, T}
(v) A = {odd natural numbers} and ξ = {whole numbers}.
A = {odd natural numbers}
ξ = {whole numbers}
A’ = {0, even whole numbers}
Question 3. If A {x : x ϵ N and 3 < x < 1} and B = {x : x ϵ Wand x ≤ 4}, find
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(iv) B – A
Answer :
A = {x : x ϵ N and 3 < x < 7}
A = {4, 5, 6}
B = {x : x ϵ W and x ≤ 4}
B = {0, 1, 2, 3, 4}
(i) A ∪ B
A ∪ B = {0, 1, 2, 3, 4, 5, 6}
(ii) A ∩ B
A ∩ B = {4}
(iii) A – B
A – B = {5, 6}
(iv) B – A
B – A = {0, 1, 2, 3}
Question 4. If P = {x : x ϵ W and x < 6} and Q = {x : x ϵ N and 4 ≤ x ≤ 9}, find
(i) P ∪ Q
(ii) P ∩ Q
(iii) P – Q
(iv) Q – P
Is P ∪ Q a proper superset of P ∩ Q ?
Answer :
P = {x : x ϵ W and x < 6}
P = {0, 1, 2, 3, 4, 5}
Q = {x : x ϵ N and 4 ≤ x ≤ 9}
Q = {4, 5, 6, 7, 8, 9}
(i) P ∪ Q
P ∪ Q = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(ii) P ∩ Q
P ∩ Q = {4, 5}
(iii) P – Q
P – Q = {0, 1, 2, 3}
(iv) Q – P
Q – P = {6, 7, 8, 9}
By observing the above sets, P ∪ Q is a proper superset of P ∩ Q.
(ML Aggarwal Operations on sets Exe-6 Class 8 ICSE Maths)
Question 5. If A = (letters of word INTEGRITY) and B = (letters of word RECKONING), find
(i) A ∪ B
(ii) A ∩ B
(iii) A – B
(iv) B – A
Also verify that:
(a) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(b) n(A – B) = n(A ∪ B) – n(B)
= n(A) – n(A ∪ B)
(c) n(B – A) = n(A ∪ B) – n(A)
= n(B) – n(A ∩ B)
(d) n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B).
Answer :
A = {I, N, T, E, G, R, Y}
n(A) = 7
B = {R, E, C, K, O, N, I, G}
n(B) = 8
(i) A ∪ B = {I, N, T, E, G, R, Y} ∪ {R, E, C, K, O, N, I, G}
A ∪ B = {I, N, T, E, G, R, Y, C, K, O}
n(A ∪ B) = 10
(ii) A ∩ B = {I, N, T, E, G, R, Y} ∩ {R, E, C, K, O, N, I, G}
A ∩ B = {I, N, E, G, R}
n(A ∩ B) = 5
(iii) A – B = {I, N, T, E, G, R, Y} – {R, E, C, K, O, N, I, G}
A – B = {T, Y}
n(A – B) = 2
(iv) B – A = {I, N, T, E, G, R, Y} – {R, E, C, K, O, N, I, G}
B – A = {C, K, O}
n(B – A) = 3
Now,
(a) n(A ∪ B) = 10
n(A) + n(B) – n(A ∩ B) = 7 + 8 – 5
= 15 – 5
= 10
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(b) n(A – B) = 2
n(A ∪ B) – n(B) = 10 – 8 = 2
n(A) – n(A ∪ B) = 7 – 5 = 2
n(A – B) = n(A ∪ B) – n(B) = n(A) – n(A ∪ B)
(c) n(B – A) = 3
n(A ∪ B) – n(A) = 10 – 7 = 3
n(B) – n(A ∩ B) = 8 – 5 = 3
n(B – A) = n(A ∪ B) – n(A) = n(B) – n(A ∩ B)
(d) n(A ∪ B) = 10
n(A – B) + n(B – A) + n(A ∩ B) = 2 + 3 + 5 = 10
n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B)
Operation on Sets Exe-6
ML Aggarwal Class 8 ICSE Maths Solutions
Page-103
Question- 6. If ξ = {natural numbers between 10 and 40}
A = {multiples of 5} and
B = {multiples of 6}, then
(i) find A ∪ B and A ∩ B
(ii) verify that
n(A ∪ B) = B (A) + n(B) – n(A ∩ B).
Answer :
ξ = {natural numbers between 10 and 40}
ξ = {11,12, 13, 14, 15, …., 39}
ξ is a universal set and A and B are subsets of ξ.
Then, the elements of A and B are to be taken only from ξ.
A = {multiples of 5}
A = {15, 20, 25, 30, 35}
B = {multiples of 6}
B = {12, 18, 24, 30, 36}
(i) A ∪ B = {15, 20, 25, 30, 35, 40} ∪ {12, 18, 24, 30, 36}
A ∪ B = {15, 20, 25, 30, 35, 12, 18, 24, 36}
A ∩ B = {30}
(ii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 5
n(A) = 5
n(B) = 5
n(A ∩ B) = 1
n(A) + n(B) – n(A ∩ B) = 5 + 5 – 1 = 9
By comparing the results,
9 = 9
Hence, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Question 7. If ξ ={1,2, 3, …. 9}, A = {1, 2, 3, 4, 6, 7, 8} and B = {4, 6, 8}, then find.
(i) A’
(ii) B’
(iii) A ∪ B
(iv) A ∩ B
(v) A – B
(vi) B – A
(vii) (A ∩ B)’
(viii) A’ ∪ B’
Also verify that:
(a) (A ∩ B)’ = A’ ∪ B’
(b) n(A) + n(A’) = n(ξ)
(c) n(A ∩ B) + n((A ∩ B)’) = n(ξ)
(d) n(A – B) + n(B – A) + n(A ∩ B)
= n(A ∪ B).
Answer :
ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4, 6, 7, 8}
B = {4, 6, 8}
(i) A’ = ξ – A = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 7, 8}
A’ = {5, 9}
(ii) B’ = ξ – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 6, 8}
B’ = {1, 2, 3, 5, 7, 9}
(iii) A ∪ B = {1, 2, 3, 4, 6, 7, 8} ∪ {4, 6, 8}
A ∪ B = {1, 2, 3, 4, 6, 7, 8}
(iv) A ∩ B = {1, 2, 3, 4, 6, 7, 8} ∩ {4, 6, 8}
A ∩ B = {4, 6, 8}
(v) A – B = {1, 2, 3, 4, 6, 7, 8} – {4, 6, 8}
A – B = {1, 2, 3, 7}
(vi) B – A = {4, 6, 8} – {1, 2, 3, 4, 6, 7, 8}
B – A = { }
(vii) (A ∩ B)’ = ξ – (A ∩ B)
(A ∩ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 6, 8}
(A ∩ B)’ = {1, 2, 3, 5, 7, 9}
(viii) A’ ∪ B’ = {5, 9} ∪ {1, 2, 3, 5, 7, 9}
A’ ∪ B’ = {1, 2, 3, 5, 7, 9}
n(ξ) = 9
n(A) = 7
n(A’) = 2
n(B’) = 6
n(A ∩ B) = 3
n((A ∩ B)’) = 6
n(A’ ∪ B’) = 6
n(A – B) = 4
n(B – A) = 0
n(A ∪ B) = 7
(a) (A ∩ B)’ = A’ ∪ B’
(A ∩ B)’ = {1, 2, 3, 5, 7, 9}
A’ ∪ B’ = {1, 2, 3, 5, 7, 9}
(A ∩ B)’ = A’ ∪ B’
(b) n(A) + n(A’) = n(ξ)
7 + 2 = 9
9 = 9
Hence, by comparing the results, n(A) + n(A’) = n(ξ)
(c) n(A ∩ B) + n((A ∩ B)’) = n(ξ)
3 + 6 = 9
9 = 9
Hence, by comparing the results, n(A ∩ B) + n((A ∩ B)’) = n(ξ)
(d) n(A – B) + n(B – A) + n(A ∩ B) = n(A ∪ B)
4 + 0 + 3 = 7
7 = 7
Hence, by comparing the results, n(A – B) + n(B – A) + n(A ∩ B) = n(A ∪ B)
Question 8. If 4 = {x : x ϵ W, x ≤ 10}, A. = {x : x ≥ 5} and B = {x : 3 ≤ x < 8}, then verify that:
(i) (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’= A’ ∪ B’
(iii) A – B = A ∩ B’
(iv) B – A = B ∩ A
Answer :
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {5, 6, 7, 8, 9, 10}
B = {3, 4, 5, 6, 7}
(i) (A ∪ B)’ = A’ ∩ B’
First consider the Left hand side (LHS) = (A ∪ B)’
(A ∪ B) = {5, 6, 7, 8, 9, 10} ∪ {3, 4, 5, 6, 7}
(A ∪ B) = {3, 4, 5, 6, 7, 8, 9, 10}
(A ∪ B)’ = ξ – A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7, 8, 9, 10}
(A ∪ B)’ = {0, 1, 2}
Hence, LHS (A ∪ B)’ = {0, 1, 2}
Then, Right-hand side (RHS) = A’ ∩ B’
A’ = ξ – A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
A’ = {0, 1, 2, 3, 4}
B’ = ξ – B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
B’ = {0, 1, 2, 8, 9, 10}
A’ ∩ B’ = {0, 1, 2, 3, 4} ∩ {0, 1, 2, 8, 9, 10}
A’ ∩ B’ = {0, 1, 2}
Hence, RHS A’ ∩ B’ = {0, 1, 2}
By comparing LHS and RHS
(A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’= A’ ∪ B’
Consider LHS (A ∩ B)’
(A ∩ B) = {5, 6, 7, 8, 9, 10} ∩ {3, 4, 5, 6, 7}
(A ∩ B) = {5, 6, 7}
(A ∩ B)’ = ξ – (A ∩ B) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7}
(A ∩ B)’ = {0, 1, 2, 3, 4, 8, 9, 10}
Hence, LHS = (A ∩ B)’ = {0, 1, 2, 3, 4, 8, 9, 10}
Now, consider RHS = A’ ∪ B’
A’ ∪ B’ = {0, 1, 2, 3, 4} ∪ {0, 1, 2, 8, 9, 10}
A’ ∪ B’ = {0, 1, 2, 3, 4, 8, 9, 10}
Hence, RHS = A’ ∪ B’ = {0, 1, 2, 3, 4, 8, 9, 10}
By comparing LHS and RHS
(A ∩ B)’= A’ ∪ B’
(iii) A – B = A ∩ B’
Consider the LHS = A – B
A – B = {5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
A – B = {8, 9, 10}
Hence, LHS = A – B = {8, 9, 10}
Now, consider RHS = A ∩ B’
(A ∩ B’) = {5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 8, 9, 10}
(A ∩ B’) = {8, 9, 10}
Hence, RHS = (A ∩ B’) = {8, 9, 10}
By comparing LHS and RHS,
A – B = A ∩ B’
(iv) B – A = B ∩ A’
Consider the LHS = B – A
B – A = {3, 4, 5, 6, 7} – {5, 6, 7, 8, 9, 10}
B – A = {3, 4}
Hence, LHS = B – A = {3, 4}
Now, consider RHS = B ∩ A’
B ∩ A’ = {3, 4, 5, 6, 7} ∩ {5, 6, 7, 8, 9, 10}
B ∩ A’ = {3, 4}
Hence, RHS = B ∩ A’ = {3, 4}
By comparing the LHS and RHS, B – A = B ∩ A’.
Question 9. If n(A) = 20, n(B) = 16 and n(A ∪ B) = 30, find n(A ∩ B).
Answer :
n(A) = 20
n(B) = 16
n(A ∪ B) = 30
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
30 = 20 + 16 – n(A ∩ B)
30 = 36 – n(A ∩ B)
n(A ∩ B) = 36 – 30
n(A ∩ B) = 6
Hence, n(A ∩ B) = 6
(ML Aggarwal Operations on sets Exe-6 Class 8 ICSE Maths)
Question 10. If n ξ = 20 and n(A’) = 7, then find n(A)
Answer :
n(ξ) = 20
n(A’) = 7
n(A’) = n(ξ) – n(A)
7 = 20 – n(A)
n(A) = 20 – 7
n(A) = 13
Hence, n(A) = 13
Question 11. If n(ξ) = 40, n(A) = 20, n(B’) = 16 and n(A ∪ B) = 32, then find n(B) and n(A ∩ B).
Answer :
n(ξ) = 40
n(A) = 20
n(B’) = 16
n(A ∪ B) = 32
n(B’) = n(ξ) – n(B)
16 = 40 – n(B)
n(B) = 40 – 16
n(B) = 24
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
32 = 20 + 24 – n(A ∩ B)
32 = 44 – n(A ∩ B)
n(A ∩ B) = 44 – 32
n(A ∩ B) = 12
Hence, n(A ∩ B) = 12
Question 12. If n(ξ) = 32, n(A) = 20, n(B) = 16 and n((A ∪ B)’) = 4, find :
(i) n(A ∪ B)
(ii) n(A ∩ B)
(iii) n(A – B)
Answer :
n(ξ) = 32
n(A) = 20
n(B) = 16
n((A ∪ B)’) = 4
(i) n(A ∪ B) = n(ξ) – n((A ∪ B)’)
= 32 – 4
n(A ∪ B) = 28
(ii) n(A ∩ B)
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
28 = 20 + 16 – n(A ∩ B)
28 = 36 – n(A ∩ B)
n(A ∩ B) = 36 – 28
n(A ∩ B) = 8
(iii) n(A – B) = n(A) – n(A ∩ B)
= 20 – 8
= 12
Hence, n(A – B) = 12
Question 13. If n(ξ) = 40, n(A’) = 15, n(B) = 12 and n((A ∩ B)’) = 32, find :
(i) n(A)
(ii) n(B’)
(iii) n(A ∩ B)
(iv) n(A ∪ B)
(v) n(A – B)
(vi) n(B – A)
Answer :
n(ξ) = 40
n(A’) = 15
n(B) = 12
n((A ∩ B)’) = 32
(i) n(A)
n(A) = n(ξ) – n(A’)
n(A) = 40 – 15
n(A) = 25
(ii) n(B’)
n(B’) = n(ξ) – n(B)
n(B’) = 40 – 12
n(B’) = 28
(iii) n(A ∩ B) = n(ξ) – n((A ∩ B)’)
= 40 – 32
= 8
(iv) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 25 + 12 – 8
= 37 – 8
= 29
(v) n(A – B) = n(A) – n(A ∩ B)
= 25 – 8
= 17
(vi) n(B – A) = n(B) – n(A ∩ B)
= 12 – 8
= 4
(ML Aggarwal Operations on sets Exe-6 Class 8 ICSE Maths)
Question 14. If n(A – B) = 12, n(B – A) = 16 and n(A ∩ B) = 5, find:
(i) n(A)
(ii) n(B)
(iii) n(A ∪ B)
Answer :
n(A – B) = 12
n(B – A) = 16
n(A ∩ B) = 5
(i) n(A)
n(A) = n(A – B) + n(A ∩ B)
= 12 + 5
= 17
(ii) n(B)
n(B) = n(B – A) + n(A ∩ B)
= 16 + 5
n(B) = 21
(iii) n(A ∪ B)
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 17 + 21 – 5
= 38 – 5
n(A ∪ B) = 33
— : End of ML Aggarwal Operations on sets Exe-6 Class 8 ICSE Maths Solutions :–
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