# ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 6 Operation on sets Venn Diagrams Ex 6.1

## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 6 Operation on sets Venn Diagrams Ex 6.1

Question 1.
If A = {0, 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.
Solution:
A = {0, 1, 2, 3, …, 8}
B = {3, 5, 7, 9, 11}
C = {0, 5, 10, 20}
(i) A ∪ B = {0,1, 2, 3, 4, 5, 6, 7, 8, 9,11},
n(A ∪ B) = 11
(ii) A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 20},
n(A ∪ C) = 11
(iii) B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20},
n(B ∪ C) = 8
(iv) A ∩ B = {3, 5, 7}, n(A ∩ B) = 3
(v) A ∩ C = (0, 5), n(A ∩ C) = 2
(vi) B ∩ C = {5}, 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}.
Solution:
Find A’
(i) A = {0, 1,4,7} and ξ= {x|x ϵ W, x ≤10}
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
∴ A’ = {2, 3, 5, 6, 8, 9, 10}

(ii) A = {consonants} and 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}
ξ = {letters of KOLKATA}
= {KAL} and ξ = {KOLAT}
∴ A’ = {O, T}

(v) 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
Solution:
A = {x : x ϵ N and 3 < x < 7}
= {4, 5, 6}
B = {x: x ϵ W and x ≤ 4}
= {0, 1, 2, 3, 4}
(i) A ∪ B = {0, 1, 2, 3, 4, 5, 6}
or {x : x ϵ W and x ≤ 4}
(ii) A ∩ B = {4}
(iii) A – B = {5, 6}
(iv) 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 ?
Solution:
P = {x: x ϵ W and x < 6}
= {0, 1, 2, 3, 4, 5}
Q = {x : x ϵ N and 4 ≤ x ≤ 9}
= (4, 5, 6, 7, 8}
(i) P ∪ Q = {0, 1, 2, 3, 4, 5, 6, 7, 8}
{x : x ϵ W and x ≤ 8}
(ii) P ∩ Q = {4, 5}
(iii) P – Q = {0, 1, 2, 3}
(iv) Q – P = {6, 7, 8}
Yes, P ∪ Q is the super set of P ∩ Q
because P ∩ Q is subset of P ∪ Q.

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).
Solution:
A = (I, N, T, E, G, R, Y}
B = {R, E, C, K, O, N, I, G}
(i) A ∪ B = {I, N, T, E, G, R, Y} ∪ {R, E, C, K, O, N, I, G}
= {I, N, T, E, G, R, Y, C, K, 0}

(ii) A ∩ B = {I, N, T, E, G, R, Y} ∩ {R, E, C, K, O, N, I, G}
= {I, N, E, R, G}

(iii) A – B = {R, E, C, K, O, N, I, G} – {1, N, T, E, G, R, Y}
= {T, Y}

(iv) B – A = – {R, E, C, K, O, N, I, G} {I, N, T, E, G, R, Y}
= {C, K, O}
Verification:
Here n(A ∪ B) = 10
n(A) = 7
n(B) = 8
n(A ∩ B) = 5
n(A – B) = 2
n(B – A) = 3
(a) n(A ∪ B) = 10
and 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
and 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
and 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
and 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).

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).
Solution:
Here, ξ = {11, 12, 13, ……., 39}
It is understood that A and B are subsets of ξ {Universal set)
So, the elements of A and B are to be taken only from ξ.
A= {multiples of 5}
= multiples of 5 which belong to ξ
= {15, 20, 25, 30, 35}
B = {multiples of 6}
= multiples of 6 which belong to ξ
= {12, 18, 24, 30, 36}
(i) A ∪ B = {15, 20, 25, 30, 35, 12, 18, 24, 36}
A ∩ B = {30}
(ii) n(A) = 5, n(B) = 5
n(A ∪ B) = 9 and n(A ∩ B) = 1
∴ n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
9 = 5 + 5 – 1 = 10 – 1 = 9
Hence, verified.

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).
Solution:
The given sets in the roster form are:
ξ = {1,2, 3, …, 9}
A = {1,2, 3, 4, 6, 7,8} and B = {4, 6, 8}
(i) A’ = ξ – A
= {1,2, 3, 4, 5, 6, 7, 8, 9} – {1,2, 3, 4, 6, 7, 8}
= {5, 8, 9}

(ii) B’ = ξ – B
= {1,2, 3, 4, 5, 6, 7, 8, 9} – {4, 6, 8}
= {1,2,3, 5, 7, 9}

(iii) A ∪ B = {1, 2, 3, 4, 6, 7, 8} ∪ {4, 6, 8}
= {1,2, 3, 4, 6, 7, 8}

(iv) A ∩ B = {1, 2, 3, 4, 6, 7, 8} ∩ {4, 6, 8}
= {4, 6, 8}

(v) A – B = {1, 2, 3, 4, 6, 7, 8} – {4, 6, 8}
= {1, 2, 3, 7}

(vi) B – A = {4, 6, 8} – {1, 2, 3, 4, 6, 7, 8}
= {}

(vii) (A ∩ B)’ = ξ – (A ∩ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 6, 8}
= {1,2, 3, 5, 7, 9}

(viii) A’ ∪ B’ = {5, 8, 9} ∪ {1, 2, 3, 5, 7, 9}
= {1,2, 3, 5, 7, 8, 9}
Verification :
Here n(A) = 6
n(A’) = 3
n(ξ) = 9
n(A ∩ B) = 3
n((A ∩ B)’) = 6
n(A – B) = 4
n(B – A) = 0
n(A ∩ B)’ = 6
n( A’ ∩ B’) = 7
n(A ∪ B) = 8
(a) (A ∩ B)’ = ξ – (A ∩ B)
= {1,2, 3, 4, 5, 6, 7, 8, 9} – (4, 6, 8}
= {1,2, 3, 5, 7, 9}
A’ ∪ B’ = {5, 7, 9} ∪ {1, 2, 3, 5, 7, 9}
= {1, 2, 3, 5, 7, 9}
(A ∩ B)’ = A’ ∪ B’ (Verified)

(b) n(A) + n(A’) = 6 + 3 = 9
n(ξ) = 9
∴ n(A) + n(A’) = n(ξ) (Verified)

(c) n(A ∩ B) + n((A ∩ B)’) = 3 + 6 = 9
n(ξ) = 9
∴ n(A ∩ B) + n((A ∩ B)’) = n(ξ) (Verified)

(d) n(A – B) + n(B – A) + n(A ∩ B)
= 4 + 0 + 3 = 7
n(A ∪ B) = 7
∴ n(A – B) + n(B – A) + n(A ∩ B)
= n(A ∪ B) (Verified)

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’
Solution:
The given sets in the roster form are :
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) L.H.S. = (A ∪ B)’
A ∪ B = {5, 6, 7, 8, 9, 10} ∪ {3, 4, 5, 6, 7}
= {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}
= {0, 1, 2}
R.H.S. = A’ ∩ B’
A’ = ξ – A
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
= {0, 1, 2, 3, 4}
B’ = ξ – B
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
= {0, 1, 2, 8, 9, 10}
A’ ∩ B’ = {0, 1, 2, 3, 4} ∩ {0, 1, 2, 8, 9, 10}
= {0, 1, 2}
Hence (A ∪ B)’ = A’ ∩ B’ (Verified)
(ii) L.H.S. = (A ∩ B)’
A n B = {5, 6,7, 8,9, 10} n {3,4, 5,6, 7} = {5, 6, 7}
(A ∩ B)’ = ξ – (A ∩ B)
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
– {5, 6, 7}
= {0, 1, 2, 3, 4, 8, 9, 10}
R.H.S. = A’ u B’
A’ = ξ – A
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
= {0, 1, 2, 3, 4}
B’ = ξ – B
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
= {0, 1, 2, 8, 9, 10}
∴ A’ ∪ B’ = {0, 1, 2, 3, 4} ∪ {0, 1, 2, 8, 9, 10}
= {0, 1, 2, 3, 4, 8, 9, 10}
Hence (A ∩ B)’ =A’ ∪ B’ (Verified)

(iii) L.H.S. = A – B
∴ A – B = {5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
= {8, 9, 10}
R.H.S. = A ∩ B’
B’ = ξ – B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {3, 4, 5, 6, 7}
= {0, 1, 2, 8, 9, 10}
∴ A ∩ B’ = {5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 8, 9, 10}
= {8, 9, 10}
Hence A – B = A ∩ B’

(iv) L.H.S. = B – A
∴ B – A = {3, 4, 5, 6, 7} – {5, 6, 7, 8, 9, 10}
= {3, 4}
R.H.S. = B ∩ A’
A’ = ξ – A
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {5, 6, 7, 8, 9, 10}
= (0, 1,2, 3, 4}
B ∩ A’ = {3, 4, 5, 6, 7} ∩ {0, 1, 2, 3, 4} = {3, 4}
Hence B – A = B ∩ A’.

Question 9.
If n(A) = 20, n(B) = 16 and n(A ∪ B) = 30, find n(A ∩ B).
Solution:
Given that,
n(A) = 20, n(B) = 16 and
n(A ∪ B) = 30 Then n(A ∩ B) = ?
We know that,
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.

Question 10.
If n ξ = 20 and n(A’) = 7, then find n(A).
Solution:
We know that,
n(A’) = n ξ – n(A)
7 = 20 – n(A)
7 – 20 = – n(A)
-13 = -n(A) ⇒ 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).
Solution:
We know that,
n(B’) = n(ξ) – n(B)
16 = 40 – n(B)
16 – 40 = – n(B)
-24 = -n(B)
n(B) = 24
and also, we know that,
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
32 = 20 + 24 – n(A ∩ B)
32 = 44 – n(A ∩ B)
32 – 44 = – n(A ∩ B)
⇒ -12 = -n(A ∩ B)
⇒ 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)
Solution:
Given that,
n(ξ) = 32, n(A) = 20, n(B) = 16,
n((A ∪ B)’) = 4
(i) n(A ∪ B) = n(ξ) – n((A ∪ B)’)
= 32 – 4 = 28

(ii) We know that,
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
Hence n(A ∩ B) = 8.

(iii) n(A – B) = n(A) – n(A ∩ B)
= 20 – 8 = 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)
Solution:
Given that,
n(ξ) = 40
n(A’)= 15
n(B) = 12
n((A ∩ B)’) = 32
(i) n(A) = n(ξ) – n(A’)
= 40 – 15 = 25

(ii) n(B’) = n(ξ) – n(B) = 40 – 12 = 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 = 25 + 4 = 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

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)
Solution:
Given that n(A – B) = 12,
n(B – A)= 16 and n(A ∩ B) = 5
(i) n(A) = n(A – B) + n(A ∩ B)
{∵ n(A – B) = n(A) – n(A ∩ B)}
= 12 + 5 = 17

(ii) n(B) = n(B – A) + n(A ∩ B)
{∵ n(B – A) = n(B) – n(A ∩ B)}
= 16 + 5 = 21

(iii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 17 + 21 – 5 = 38 – 5 =33