**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)