ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 6 Operation on sets Venn Diagrams Check Your Progress
Question 1.
If ξ={x : x ϵ N, r < 25} and A = {x : x is a composite number}, then find A’ in set builder from and also in roster form.
Solution:
ξ = {x : x ϵ N, x < 25}
A = {x : x is a composite number},
then find A’ in set builder form and also in roster form
ξ = {0, 2, 3, 4, ……., 24}
A = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24}
∴ A’ = (1, 2, 3, 5, 7, 11, 13, 17, 19, 23}
= {x : x ϵ N, x is a prime number less than 25}
Question 2.
If ξ = {x | x ϵ I, -5 ≤ x ≤ 5},
A= {-5, -3, 0, 3, 5} and B = {-4, -2, 0, 2, 4}, then
(i) Find A ∪ B and A ∩ B
(ii) Verify that n(A ∪ B) = n(A) + n(B) – n(A ∩ B).
(iii) Find A’ and verify that n(A) + n(A’) = n(ξ).
(iv) Find B’ and verify that n(B’) = n(ξ) – n(B)
(v) Find A ∪ A’, A ∩ A’, B ∪ B’ and B ∩ B’.
(vi) Find (A ∪ B)’ and A’ ∩ B’. Are they equal ?
(vii) Find (A ∩ B)’ and A’ ∪ B’. Are they equal ?
Solution:
ξ = {x | x ϵ 1, – 5 ≤ x ≤ 5}
= {-5, -4, -3, -2, -1,0, 1,2, 3, 4, 5}
n(ξ) = 11
A = {-5, -3, 0, 3, 5} ⇒ n(A) = 5
B = {-4,-2, 0, 2, 4} ⇒ n(B) = 5
(i) A ∪ B = {-5, -4, -3, -2, 0, 2, 3, 4, 5}
⇒ n(A ∪ B) = 9
A ∩ B = {0}
⇒ n(A n B) = 1
(ii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
L.H.S. = n(A ∪ B) = 9
R.H.S. = n(A) + n(B) – n(A ∩ B)
= 5 + 5 – 1 = 9
L.H.S. = R.H.S.
(iii) A’ = ξ – A = {-4, -2, -1, 1, 2, 4}
⇒ n(A’) = 6
Now, n(A) + n(A’) = 5 + 6 = 11 = n(ξ)
Hence verified
(iv) B’ = ξ – B = {-5, -3, -1, 1, 3, 5}
n(B’) = 6
n(ξ) – n(B) = 11 – 5 = 6 = n(B’)
(v) A ∪ A’ = ξ
A ∩ A’ = ϕ
B ∪ B’ = ξ, B ∩ B’ = ϕ
(vi) (A ∪ B)’ = {-1, 1} and A’ ∩ B’= {-1, 1}
Yes both are equal.
(vii) (A ∩ B)’ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5}
A’ ∪ B’ = {-5, -4, -3, -2, -1, 1, 2, 3, 4, 5}
Yes, (A ∩ B)’ = A’ ∪ B’
Question 3.
If ξ = {x | x ϵ N, x ≤ 12}, A = {prime numbers} and B = {odd numbers}, then
(i) Find A ∪ B and A ∩ B.
(ii) Verify that n(A ∪ B) + n(A ∩ B) = n(A) + n(B).
(iii) Find A’ and B’.
(iv) Find (A ∪ B)’ and verify that n(A ∪ B) + n(A ∪ B)’ = n(ξ).
(v) Find (A ∩ B)’ and A’ ∪ B’. Are they equal ?
Solution:
Here,
ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
A = {2, 3, 5, 7, 11}
B = {1, 3, 5, 7, 9, 11}
(i) A ∪ B = {1, 2, 3, 5, 7, 9,11}
and A ∩ B = {3, 5, 7, 11}
(ii) n(A ∪ B) + n(A ∩ B) = n(A) + n(B)
7 + 4 = 5 + 6
11 = 11
Hence, verified.
(iii) A’ = ξ – A = {1, 4, 6, 8, 9, 10, 12}
and B’ = ξ – B = {2, 4, 6, 8, 10, 12}
(iv) (A ∪ B)’ = ξ – (A ∪ B)= {4, 6, 8, 10, 12}
To show,
n(A ∪ B) + n(A ∪ B)’ = n(ξ)
7 + 5 = 12
12 = 12
Hence verified
(v) (A ∩ B)’ = ξ – (A ∪ B)
= {1, 2, 4, 6, 8, 9, 10, 12}
A’ ∪ B’ = {1, 2, 4, 6, 8, 9, 10, 12}
Yes, the given sets are equal
because they have the same elements.
Question 4.
If ξ = {x : x ϵ N, x ≤ 12}, A= {x : x ≥ 7} and B = {x : 4 < x < 10}. 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:
(i) (A ∪ B)’ = A’ ∩ B’
(ii) A – B = A ∩ B’
(iii) n(A ∪ B) + n((A ∪ B)’) = n(ξ)
(iv) n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B).
Solution:
The given sets in the roster form are :
ξ = {1,2, 3, …, 12}
A = {7, 8, 9, 10, 11, 12}
B = {5, 6, 7, 8, 9}
(i) A’ = ξ – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} – {7,8,9,10,11,12}
= {1,2, 3, 4, 5, 6}
(ii) B’ = ξ – B
= {1, 2, 3, …, 12} – {5, 6, 7, 8, 9}
= {1, 2, 3, 4, 10, 11, 12}
(iii) A ∪ B = {7, 8, 9, 10, 11, 12} ∪ {5, 6, 7, 8, 9}
= {5, 6, 7, 8, 9, 10, 11, 12}
(iv) A ∩ B = {7, 8, 9, 10, 11, 12} ∩ {5, 6, 7, 8, 9}
= {7, 8, 9}
(v) A – B = {7, 8, 9, 10, 11, 12} – {5, 6, 7, 8, 9}
= {10, 11, 12}
(vi) B – A = {5, 6, 7, 8, 9} – {7, 8, 9, 10, 11, 12}
= {5, 6}
(vii) (A ∪ B)’ = ξ – A ∪ B
= {1,2,3,…, 12} – {5, 6, 7, 8, 9, 10, 11, 12}
= {1,2,3, 4}
(viii) A’ ∩ B’= {1,2, 3, 4, 5, 6} ∩ {1,2, 3, 4, 10, 11, 12}
= {1, 2, 3, 4}
Verification :
(i) (A ∪ B)’ = {1, 2, 3, 4} (By VII part)
and A’ ∩ B’ = {1, 2, 3, 4} (By VIII part)
Hence (A ∪ B)’ = A’ ∩ B’ (Verified)
(ii) A – B = {10, 11, 12} (By (v) part)
and A ∩ B’ = {7, 8, 9, 10, 11, 12} ∩ {1, 2, 3, 4, 10, 11, 12} (By (ii) part)
= {10, 11, 12}
Hence A – B = A ∩ B’ (Verified)
(iii) n(A ∪ B) + n((A ∪ B)’) = 8 + 4=12
n(ξ) = 12
Hence n(A ∪ B) + n((A ∪ B)’) = n(ξ) (Verified)
(iv) n(A ∪ B) = 8
and n(A – B) + n(B – A) + n(A ∩ B)
= 3 + 2 + 3 = 8
Hence n(A ∪ B) = n(A – B) + n(B – A) + n(A ∩ B) (Verified)
Question 5.
Given A = {students who like cricket} and B = {students who like tennis}; n(A) = 20, n(B) = 15 and n(A ∩ B)= 5.
Illustrate this through a Venn diagram. Hence find n(A ∪ B).
Solution:
Here, n(A) = 20
n(B) = 15
n(A ∩ B) = 5
Now, only A i.e. students who like cricket only
= n(A) – n(A ∩ B) = 20 – 5 = 15
Only B i.e. students who like tennis only
= n(B) – n(A ∩ B) = 15 – 5 = 10
n(A ∪ B)= 15 + 5 + 10 (from Venn diagram)
Using formula
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 20 + 15 – 5 = 35 – 5 = 30.
Question 6.
If n(ξ) = 50, n(A) = 15, n(B) = 13 and n(A ∩ B) = 10. Find n(A’), n(B’) and n(A ∪ B).
Solution:
Here, n(ξ) = 50
n(A) = 15
n(B) = 13
n(A ∩ B) = 10
Using formula,
n(A’) = n(ξ) – n(A)
= 50 – 15 = 35
n(B’) = n(ξ) – n(B)
= 50 – 13 = 37
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 15 + 13 – 10 = 18.
Question 7.
If n(ξ) = 60, n(A) = 35, n(B’) = 36 and n((A ∩ B)’) = 51, find :
(i) n(B)
(ii) n(A ∩ B)
(iii) n(A ∪ B)
(iv) n(A – B)
Solution:
Given n(ξ) = 60
n(A) = 35
n(B’) = 36
and n((A ∩ B)’) = 51
(i) n(B) = n(ξ) – n(B’) = 60 – 36 = 24
(ii) n(A ∩ B) = n(ξ) – n((A ∩ B)’) = 60 – 51 = 9
(iii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 35 + 24 – 9 = 59 – 9 = 50
(iv) n(A – B) = n(A) – n(A ∩ B) = 35 – 9 = 26
Question 8.
In a city, 50% people read newspaper A, 45% read newspaper B, and 25% read neither A nor B. What percentage of people read both the newspapers A as well as B ?
Solution:
Let A = Those people who read newspaper A
B = Those people who read newspaper B
Here. n(ξ) = 100%
n(A) = 50%
n(B) = 45%
n(A ∪ B)’ = 25%
∴ n(A ∪ B) = n(ξ) – n((A ∪ B)’)
= 100% – 25% = 75%
We know that,
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ 75% = 50% + 45% – n(A ∩ B)
⇒ n(A ∩ B) = 95% – 75%
⇒ n(A ∩ B) = 20%