Selina Concise Class 8 Math Chapter 6 Sets Exercise 6D Solutions
EXERCISE 6D
(1) (i) A = {4, 5, 6}
B = {0, 1, 2, 3}
(ii) A ∪ B = {0, 1, 2, 3, 4, 5, 6}
(iii) A ∩ B = {Ф}
(iv) A – B = {4, 5, 6}
(v) B – A = {0, 1, 2, 3}
(2) (i) P = {4, 5, 6, 7, 8}
Q = {1, 2, 3, 4, 5}
P ∪ Q = {1, 2, 3, 4, 5, 6, 7, 8}
P ∩ Q = {4, 5}
(ii) Yes, all the element of set P ∪ Q are contained in P ∩ Q. Therefore, P ∪ Q is a proper set of P ∩ Q.
(3) A = {5, 6, 7, 8, 9}
B = {4, 5, 6, 7}
C = {1, 2, 3, 4, 5}
(i) A ∪ B = {4, 5, 6, 7, 8, 9}
(A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(ii) B ∪ C = {1, 2, 3, 4, 5, 6, 7}
A ∪ (B ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(iii) A ∩ B = {5, 6, 7}
(A ∩ B) ∩ C = {5}
(iv) B ∩ C = {4 , 5}
A ∩ (B ∩ C) = {5}
(v) (A ∪ B) ∪ C = A ∪ (B ∪ C)
⇒ {1, 2, 3, 4, 5, 6, 7, 8, 9} = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Hence, these are equal.
(vi) (A ∩ B) ∩ C = A ∩ (B ∩ C)
⇒ {5 } = {5}
Hence, these are equal.
(4) A = {0, 1, 2, 4, 5}
B = {0, 2, 4, 6, 8}
C = {0, 3, 6, 9}
(i) L.H.S = A ∪ (B ∪ C) = {0, 1, 2, 3, 4, 5, 6, 8, 9}
R.H.S. = (A ∪ B) ∪ C = {0, 1, 2, 3, 4, 5, 6, 8, 9}
∴ L.H.S. = R.H.S. (Proved)
(ii) L.H.S. = A ∩ (B ∩ C) = {0}
R.H.S = (A ∩ B) ∩ C = {0}
∴ L.H.S. = R.H.S. (Proved)
(5) A = {6, 7, 8, 9}
B = {3, 4, 5, 6, 7}
C = {x = 2n; n ∈ N and n ≤ 4}
x = 2n
When, n = 1, x = 2 × 1 = 2
When, n = 2, x = 2 × 2 = 4
When, n = 3, x = 2 × 3 = 6
When, n = 4, x = 2 × 4 = 8
C = {2, 4, 6, 8}
(i) A ∩ (B ∪ C) = {6, 7, 8, 9} ∩ {2, 3, 4, 5, 6, 7, 8}
= {6, 7, 8}
(ii) (B ∪ A) ∩ (B ∪ C)
= {3, 4, 5, 6, 7, 8, 9} ∩ {2, 3, 4, 5, 6, 7, 8}
= {3, 4, 5, 6, 7, 8}
(iii) B ∪ (A ∩ C)
= {3, 4, 5, 6, 7} ∪ {6, 8}
= {3, 4, 5, 6, 7, 8}
(iv) (A ∩ B) ∪ (A ∩ C)
= {7} ∪ {6, 8}
= {6, 7, 8}
(6) P = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Q = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
(i) P ∪ Q = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 36, 48}
(ii) P ∩ Q = {1, 2, 3, 4, 6, 12}
(iii) Q – P= {1, 2, 3, 4, 6, 9, 12, 18, 36} – {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
= {8, 16, 24, 48}
(iv) P’ ∩ Q = Q – P
= {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} – {1, 2, 3, 4, 6, 9, 12, 18, 36}
= {8, 16, 24, 48}
(7) A = {6, 7, 8, 9}
B = {4, 6, 8, 10}
C = {3, 4, 5, 6, 7}
(i) A – B
= {6, 7, 8, 9} – {4, 6, 8, 10}
= {7, 9}
(ii) B – C
= {4, 6, 8, 10} – {3, 4, 5, 6, 7}
= {8, 10}
(iii) B – (A – C)
= {4, 6, 8, 10} – [{6, 7, 8, 9} – {3, 4, 5, 6, 7}]
= {4, 6, 8, 10} – {8, 9}
= {4, 6, 10}
(iv) A – (B ∪ C)
= {6, 7, 8, 9} – {3, 4, 5, 6, 7, 8, 10}
= {9}
(v) B – (A ∩ C)
= {4, 6, 8, 10} – {6, 7}
= {4, 8, 10}
(vi) B – B = {4, 6, 8, 10} – {4, 6, 8, 10} = Ф
(8) A = {1, 2, 3, 4, 5}
B = {2, 4, 6, 8}
C = {3, 4, 5, 6}
(i) L.H.S. = A – (B ∪ C)
= {1, 2, 3, 4, 5} – {2, 3, 4, 5, 6, 8}
= {1}
R.H.S. = (A – B) ∩ (A – C)
= {1, 3, 5} ∩ {1, 2}
= {1}
∴ L.H.S. = R.H.S (Proved)
(ii) L.H.S = A – (B ∩ C)
= {1, 2, 3, 4, 5} – {4, 6}
= {1, 2, 3, 5}
R.H.S = (A – B) ∪ (A – C)
= {1, 3, 5} ∪ {1, 2}
= {1, 2, 3, 5}
∴ L.H.S. = R.H.S (Proved)
(9) A = {1, 2, 3, 4, 5}
B = {3, 6, 9}
C = {x ∈ N : 2x – 5 ≤ 8}
2x – 5 ≤ 8
⇒ 2x – 5 + 5 ≤ 8 + 5
⇒ 2x ≤ 13
⇒ x ≤ 13/2
⇒ x ≤ 6.5
∴ C = {1, 2, 3, 4, 5, 6}
(i) L.H.S. = A ∪ (B ∩ C)
= {1, 2, 3, 4, 5} ∪ [{3, 6, 9} ∩ {1, 2, 3, 4, 5, 6}]
= {1, 2, 3, 4, 5} ∪ {3, 6}
= {1, 2, 3, 4, 5, 6}
R.H.S. = (A ∪ B) ∩ (A ∪ C)
= {1, 2, 3, 4, 5, 6} ∩ {1, 2, 3, 4, 5, 6}
= {1 , 2, 3, 4, 5, 6}
∴ L.H.S. = R.H.S (Proved)
(ii) L.H.S = A ∩ (B ∪ C)
= {1, 2, 3, 4, 5} ∩ {1, 2, 3, 4, 5, 6, 9}
= {1, 2, 3, 4, 5}
R.H. S. = (A ∩ B) ∪ (A ∩ C)
= {3} ∪ {1, 2, 3, 4, 5}
= {1, 2, 3, 4, 5}
∴ L.H.S. = R.H.S (Proved)
