If nA - 50- nB - 20- nA - B - 10- then nA - B -

Given: n(A) = 50, n(B) = 20, n(A ∩ B) = 10 Using the relation: n(A Δ B) = n(A ∪ B) – n(A ∩ B) ⇒ n(A Δ B)=(n(A)+n(B) n(A∩ B))– n(A∩ B) ⇒ n(A Δ B)=50+20 10 ...

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Standard XII

Mathematics

Properties of Symmetric Difference of Sets

If nA = 50, n...

Question

If $n (A) = 50, n (B) = 20, n (A \cap B) = 10,$ then $n (A Δ B) =$

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Solution

The correct option is A $50$
Given: $n (A) = 50, n (B) = 20, n (A \cap B) = 10$
Using the relation:
$n (A Δ B) = n (A \cup B) - n (A \cap B) \Rightarrow n (A Δ B) = (n (A) + n (B) - n (A \cap B)) - n (A \cap B) \Rightarrow n (A Δ B) = 50 + 20 - 10 - 10 = 50$

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If n(A)=50,n(B)=20,n(A∩B)=10, then n(A Δ B)=

The correct option is A 50Given: n(A)=50,n(B)=20,n(A∩B)=10 Using the relation: n(A Δ B)=n(A∪B)–n(A∩B)⇒n(A Δ B)=(n(A)+n(B)−n(A∩B))–n(A∩B)⇒n(A Δ B)=50+20−10−10=50

If $n (A) = 50, n (B) = 20, n (A \cap B) = 10,$ then $n (A Δ B) =$

The correct option is A $50$
Given: $n (A) = 50, n (B) = 20, n (A \cap B) = 10$
Using the relation:
$n (A Δ B) = n (A \cup B) - n (A \cap B) \Rightarrow n (A Δ B) = (n (A) + n (B) - n (A \cap B)) - n (A \cap B) \Rightarrow n (A Δ B) = 50 + 20 - 10 - 10 = 50$