Selina Concise Class 7 Math Chapter 13 Set Concepts Exercise 13D Solution
EXERCISE 13D
(1) If A = {4, 5, 6, 7, 8} and B = {6, 8, 10, 12}, find:
(i) {4, 5, 6, 8, 10, 12}
(ii) {6, 8}
(iii) {4, 5, 7}
(iv) {10, 12}
(2) If A = {3, 5, 7, 9, 11} and B = {4, 7, 10}, find:
(i) n(A) = 5
(ii) n(B) = 3
(iii) A ∪ B = 3, 4, 5, 7, 9, 10, 11 And n(A ∪ B) = 7
(iv) A ∩ B = 7 And n(A ∩ B) = 1
(3) If A = {2, 4, 6, 8} and B = {3, 6, 9, 12}, find:
(i) A ∩ B = 6 And n(A ∩ B) = 1
(ii) (A – B) = {2, 4, 8} And n(A – B) = 3
(iii) n(B)= 4
(4) If P = {x : x is a factor of 12} and Q = {x : x is factor of 16}, find:
(i) P = {1, 2, 3, 4, 6, 12}
Then, n(P) = 6
(ii) Q = {1, 2, 4, 8, 16}
Then, n(Q) = 5
(iii) Q – P = {8 , 16} and n(Q – P) = 2
(5) M = {x : x is a natural number between 0 and 8} and B = {x : x is a natural number from 5 to 10}, find:
M = {0, 1, 2, 3, 4, 5, 6, 7} And N = {5, 6, 7, 8, 9, 10}
(i) M – N = {1, 2, 3, 4} And n(M – N) = 4
(ii) N – M = {8, 9, 10} And n(N – M) = 3
(6) If A = {x : x is natural number divisible by 2 and x < 16} and B = {x : x is a whole number divisible by 3 and x < 18}, find:
(i) A = {2, 4, 6, 8, 10, 12, 14} and n(A) = 7
(ii) B = {3, 6, 9, 12, 15, 18} and n(B) = 6
(iii) A ∩ B = {6, 12} n(A ∩ B) = 2
(iv) A – B = {2, 4, 8, 10, 14} then, n(A – B)= 5
(7) Let A and B be two sets such that n(A) = 75, n(B) = 65 and n(A ∩ B) = 45, find:
(i) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 75 + 65 – 45
= 140 – 45 = 95
(ii) n(A – B) = n(A) – n (A ∩ B)
= 75 – 45 = 30
(iii) n(B – A) = n(B) – n(A ∩ B)
= 65 – 45 = 20
(8) Let n(A) = 45, n(B) = 38 and n(A ∪ B) = 70 find:
(i) n(A ∩ B) = n(A) + n(B) – n(A ∩ B)
= 45 + 38 – 70
= 83 – 45 = 38
(ii) n(A – B) = n(A ∪ B) – n(B)
= 70 – 38 = 32
(iii) n(B – A) = n(A ∪ B) – n(A)
= 70 – 45 = 25
(9) Let n(A) = 30, n(B) = 27 and n(A ∪ B) = 45, find:
(i) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 30 + 27 – 45
= 57 – 45 = 12
(ii) n(A – B) = n(A ∪ B) – n(B)
= 45 – 27 = 18
(10) Let n(A) = 31, n(B) = 20 and n(A ∩ B) = 6, find:
(i) n(A – B) = n(A) – n(A ∩ B)
= 31 – 6 = 25
(ii) n(B – A) = n(B) – n(A ∩ B)
= 20 – 6 = 14
(iii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 31 + 20 – 6
= 51 – 6 = 45