Selina Concise Class 7 Math Chapter 13 Set Concepts Exercise 13D Solution

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

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