ML Aggarwal Operation on Sets Exe-6 Class 8 ICSE Ch-6 Maths Solutions

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 :–

Return to –  ML Aggarwal Maths Solutions for ICSE Class -8

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