Question

29. If A,B,C are sets such that n(A) = 12, n (B) = 16, n (C) = 18, n(AnB)=6, n(BnC)=8. n( Cna)=10 and n( AnBnC) = 4,then the number of elements belonging to exactly one of A,B and C
A) 8 B) 10 C) 15 D) 6​

Answer 1

Answer:

Set A has 7 elements & set B has 8 elements.

Minimum A∪B elements condition: Out of 8 elements of set B, 7 elements are identical to that of set A. In this case A∪B=B & it will have 8 elements.

Maximum A∪B elements condition: A & B are disjoint sets, that no element will be common to them. Then:

n(A∪B)=n(A)+n(B).

A∪B has 15 elements

Answer 2

Concept

An expression n(x) represents the number of elements in set X. n(X∩Y)  represents the number of elements that are common in set X and Y.

Given

n(A) = 12, n (B) = 16, n (C) = 18, n(A∩B)=6, n(B∩C)=8. n(C∩A)=10 and n(A∩B∩C) = 4

Find

we need to find the number of elements belonging to exactly one of A,B and C.

Solution

We have

n(A∩B)=6, but this also includes n(A∩B∩C)

⇒ n(only A∩B) = 6-4 =2

Similarly,

n(only B∩C)= 8-4 = 4

n(only C∩A)= 10-4 = 6

Now,

n(A) = 12 = n(only A) + n(only A∩B) + n(only C∩A) +  n(A∩B∩C)

12 = n(only A) + 2 + 6 + 4

n(only A) = 12 - 12 = 0

Similarly

n(B) = 16 = n(only B) + n(only A∩B) + n(only C∩B) +  n(A∩B∩C)

16 = n(only B) + 2 + 4 + 4

n(only B) = 6

n(C) = 18 = n(only C) + n(only A∩C) + n(only C∩B) +  n(A∩B∩C)

18 = n(only C) + 6 + 4 + 4

n(only C) = 4

Now,

n(only A) + n(only B) + n(only C) = 0 + 6 + 4 = 10

Thus, number of elements belonging to exactly one of A,B and C is 10, that is, option 2.

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