A, B, C are three sets such that n(A)=25, n(B)=20, n(c)=27, n(A ∩ B)=5, n(B ∩ C)=7 and A ∩ C= φ then n(A ∪ B ∪ C) is equal to

Question

A, B, C are three sets such that n(A)=25, n(B)=20, n(c)=27, n(A ∩ B)=5, n(B ∩ C)=7 and A ∩ C= φ then n(A ∪ B ∪ C) is equal to (A) 60 (B) 65 (C) 67 (D) 72 .

Tardigrade Admin · Accepted Answer

A ∩ C=φ ⇒ A ∩ B ∩ C=φ. ∴ n(A ∪ B ∪ C)=n(A)+n(B)+n(C)-n(A ∩ B)-n(B ∩ C) -n(A ∩ C)+n(A ∩ B ∩ C) = 25+20+27-5-7-0+0=60 .

Q. $A, B, C$ are three sets such that $n (A) = 25$ , $n (B) = 20, n (c) = 27, n (A \cap B) = 5, n (B \cap C) = 7$ and $A \cap C =$ $ϕ$ then $n (A \cup B \cup C)$ is equal to

60

65

67

72 .

$A \cap C = ϕ$
$\Rightarrow A \cap B \cap C = ϕ$ .
$∴ n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)$
$= 25 + 20 + 27 - 5 - 7 - 0 + 0 = 60.$

Q. A,B,C are three sets such that n(A)=25, n(B)=20,n(c)=27,n(A∩B)=5,n(B∩C)=7 and A∩C= ϕ then n(A∪B∪C) is equal to

60

65

67

72 .

A∩C=ϕ ⇒A∩B∩C=ϕ. ∴n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C) =25+20+27−5−7−0+0=60.

Q. $A, B, C$ are three sets such that $n (A) = 25$ , $n (B) = 20, n (c) = 27, n (A \cap B) = 5, n (B \cap C) = 7$ and $A \cap C =$ $ϕ$ then $n (A \cup B \cup C)$ is equal to

$A \cap C = ϕ$
$\Rightarrow A \cap B \cap C = ϕ$ .
$∴ n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)$
$= 25 + 20 + 27 - 5 - 7 - 0 + 0 = 60.$