PRACTICE QUESTIONS ON CARDINALITY OF SETS

Question 1 :

(i) If n(A) = 25, n(B) = 40, n(A∪B) = 50 and n(B′) = 25 , find n(A∩B) and n(U).

Solution :

n(AUB)  =  n(A) + n (B) -  n(AnB)

50  =  25 + 40 -  n(AnB)

n(A n B)  =  65 - 50

n(A n B)  =  15

n (U)  =  n(B) + n(B')

n (U)  =  40 + 25

=  65

(ii) If n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B′) = 350, find n(B) and n(U).

Solution :

n(AUB)  =  n(A) + n (B) -  n(AnB)

500  =  300 + n(B) - 50

500  =  250 + n(B)

n(B)  =  500 - 250

n(B)  =  250

n(U)  =  n(B) + n(B')

=  250 + 350

n(U)  =  600

Question 2 :

If U = {x : x ∈ N, x ≤ 10}, A = {2, 3, 4, 8, 10} and B = {1, 2, 5, 8, 10}, then verify that n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Solution :

U = {x : x ∈ N, x ≤ 10}

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {2, 3, 4, 8, 10} and B = {1, 2, 5, 8, 10},

A U B  =  {1, 2, 3, 4, 5, 8, 10}

A n B  =  {2, 8, 10}

n(AUB)  =  7

n(A n B)  =  3

n(A)  =  5

n(B)  =  5

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

7  =  5 + 5 - 3

7  =  10 - 3

7  =  7

Hence verified. 

Question 3 :

Verify n(A U B U C) = n(A) + n(B) + n(C) − n(A n B) − n(B n C) − n(A n C) + n(A n B n C) for the following sets.

(i) A = {a, c, e, f, h}, B = {c, d, e, f } and C = {a, b, c, f }

Solution :

n(A)  =  5, n (B)  =  4, n (C)  =  4

A n B  =  {c, e , f}  ==>  n(A n B)  =  3

B n C  =  {c, f}  ==>  n(B n C)  =  2

A n C  =  {a, c, f}  ==>  n(An C)  =  3

A n B n C  =  {c, f}  ==>  n(A n B n C)  =  2

A U B U C  =  {a, b, c, d, e, f, h}  ==>  n(A U B U C)  =  7

n(A U B U C) = n(A) + n(B) + n(C) − n(A n B) − n(B n C) − n(A n C) + n(A n B n C) 

7  =  5 + 4 + 4 - 3 - 2 - 3 + 2

7  =  13 - 6

7  =  7

Hence proved.

(ii) A = {1, 3, 5}, B = {2, 3, 5, 6} and C = {1, 5, 6, 7}

Solution :

n(A)  =  3, n (B)  =  4, n (C)  =  4

A n B  =  {3, 5}  ==>  n(A n B)  =  2

B n C  =  {5, 6}  ==>  n(B n C)  =  2

A n C  =  {1, 5}  ==>  n(An C)  =  2

A n B n C  =  {5}  ==>  n(A n B n C)  =  1

A U B U C  =  {1, 2, 3, 5, 6, 7}  ==>  n(A U B U C)  =  6

n(A U B U C) = n(A) + n(B) + n(C) − n(A n B) − n(B n C) − n(A n C) + n(A n B n C) 

6  =  3 + 4 + 4 - 2 - 2 - 2 + 1

6  =  11 - 6 + 1

6  =  6

Hence proved.

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