If A and B are two sets such that n(A) = 28, n(B) = 32 and n(A ∪ B) = 50, find n(A ∩ B).
Answer
Given,
n(A) = 28, n(B) = 32 and n(A ∪ B) = 50
We know that,
⇒ n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Substituting values we get :
⇒ 50 = 28 + 32 − n(A ∩ B)
⇒ 50 = 60 − n(A ∩ B)
⇒ n(A ∩ B) = 60 − 50 = 10
Hence, n(A ∩ B) = 10.
If X and Y are two sets such that X has 21 elements, Y has 32 elements and X ∩ Y has 11 elements, how many elements does X ∪ Y have?
Answer
Given,
n(X) = 21, n(Y) = 32 and n(X ∩ Y) = 11
We know that,
n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y)
⇒ n(X ∪ Y) = 21 + 32 − 11
⇒ n(X ∪ Y) = 53 − 11 = 42
Hence, X ∪ Y has 42 elements.
If n(A) = 20, n(B) = 18 and n(A ∩ B) = 5, calculate
(i) n(A ∪ B)
(ii) n(A − B)
(iii) n(B − A)
Answer
Given,
n(A) = 20, n(B) = 18 and n(A ∩ B) = 5
(i) n(A ∪ B)
We know that,
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
⇒ n(A ∪ B) = 20 + 18 − 5 = 33
Hence, n(A ∪ B) = 33.
(ii) n(A − B)
The set A − B is obtained from A by removing those elements of A which also belong to B, i.e. the elements of A ∩ B.
∴ n(A − B) = n(A) − n(A ∩ B) = 20 − 5 = 15
Hence, n(A − B) = 15.
(iii) n(B − A)
Similarly, the set B − A is obtained from B by removing the elements of A ∩ B.
∴ n(B − A) = n(B) − n(A ∩ B) = 18 − 5 = 13
Hence, n(B − A) = 13.
If n(A ∪ B) = 18, n(A − B) = 5, n(B − A) = 3, find n(A ∩ B).
Answer
Given,
n(A ∪ B) = 18, n(A − B) = 5 and n(B − A) = 3
The sets A − B, A ∩ B and B − A are mutually disjoint and together they make up A ∪ B.
∴ n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B)
⇒ 18 = 5 + 3 + n(A ∩ B)
⇒ 18 = 8 + n(A ∩ B)
⇒ n(A ∩ B) = 18 − 8 = 10
Hence, n(A ∩ B) = 10.
If A and B are two sets such that n(A − B) = 5, n(B − A) = 3 and n(A ∩ B) = 10, then find the following:
(i) n(A ∪ B)
(ii) n(A)
(iii) n(B).
Answer
Given,
n(A − B) = 5, n(B − A) = 3 and n(A ∩ B) = 10
(i) n(A ∪ B)
The sets A − B, A ∩ B and B − A are mutually disjoint and together they make up A ∪ B.
∴ n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B)
⇒ n(A ∪ B) = 5 + 3 + 10 = 18
Hence, n(A ∪ B) = 18.
(ii) n(A)
The set A is made up of the two disjoint parts A − B and A ∩ B.
∴ n(A) = n(A − B) + n(A ∩ B) = 5 + 10 = 15
Hence, n(A) = 15.
(iii) n(B)
The set B is made up of the two disjoint parts B − A and A ∩ B.
∴ n(B) = n(B − A) + n(A ∩ B) = 3 + 10 = 13
Hence, n(B) = 13.
If n(ξ) = 50, n(A) = 28 and n(B) = 30, then what is the greatest value of n(A ∪ B)?
Answer
Given,
n(ξ) = 50, n(A) = 28 and n(B) = 30
It is understood that every set under consideration is a subset of the universal set ξ.
∴ A ∪ B ⊂ ξ
⇒ n(A ∪ B) ≤ n(ξ)
⇒ n(A ∪ B) ≤ 50
So n(A ∪ B) can never exceed 50, and the value 50 is actually possible.
[Taking n(A ∪ B) = 50 gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 28 + 30 − 50 = 8, which is a permissible value since 8 ≤ 28 and 8 ≤ 30.]
Hence, the greatest value of n(A ∪ B) is 50.
If n(ξ) = 48 and n(A' ∪ B') = 36, then find n(A ∩ B).
Answer
Given,
n(ξ) = 48 and n(A' ∪ B') = 36
By De Morgan's law,
A' ∪ B' = (A ∩ B)'
∴ n((A ∩ B)') = 36
We know that,
n(X') = n(ξ) − n(X)
∴ n((A ∩ B)') = n(ξ) − n(A ∩ B)
⇒ 36 = 48 − n(A ∩ B)
⇒ n(A ∩ B) = 48 − 36 = 12
Hence, n(A ∩ B) = 12.
If n(ξ) = 40, n(A) = 22, n(A ∩ B) = 8 and n((A ∪ B)') = 6, determine n(B).
Answer
Given,
n(ξ) = 40, n(A) = 22, n(A ∩ B) = 8 and n((A ∪ B)') = 6
We know that,
n((A ∪ B)') = n(ξ) − n(A ∪ B)
⇒ 6 = 40 − n(A ∪ B)
⇒ n(A ∪ B) = 40 − 6 = 34
Also we know that
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
⇒ 34 = 22 + n(B) − 8
⇒ 34 = 14 + n(B)
⇒ n(B) = 34 − 14 = 20
Hence, n(B) = 20.
If n(ξ) = 50, n(A) = 20, n(B) = 26 and n((A ∪ B)') = 6, find
(i) n(A ∩ B)
(ii) n(A − B).
Answer
Given,
n(ξ) = 50, n(A) = 20, n(B) = 26 and n((A ∪ B)') = 6
We know that,
n((A ∪ B)') = n(ξ) − n(A ∪ B)
⇒ 6 = 50 − n(A ∪ B)
⇒ n(A ∪ B) = 50 − 6 = 44
(i) n(A ∩ B)
We know that,
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
⇒ 44 = 20 + 26 − n(A ∩ B)
⇒ 44 = 46 − n(A ∩ B)
⇒ n(A ∩ B) = 46 − 44 = 2
Hence, n(A ∩ B) = 2.
(ii) n(A − B)
The set A − B is obtained from A by removing the elements of A ∩ B.
∴ n(A − B) = n(A) − n(A ∩ B) = 20 − 2 = 18
Hence, n(A − B) = 18.