Question

if n (S) =100n(A-B)=10 n (B-A) =26 and n(AUB)'=52 then n(A intersection B) is equal to

Answer 1

We have,

n(S) = 100
n(A-B) = 10
⇒n(A) - n(A∩B) = 10 ----------(i)

n(B-A) = 26
⇒n(B) - n(A∩B) = 26 ----------(ii)

(i)+(ii), we get

n(A) + n(B) - 2n(A∩B) = 36.............(iii)

n(A∪B)' = 52
⇒n(S) - n(A∪B) = 52
⇒100 - n(A∪B) = 52
⇒100 - 52 = n(A∪B)
⇒48 = n(A) + n(B) - n(A∩B)...............(iv)

(iii) - (iv), we get

-2n(A∩B) + n(A∩B) = 36 - 48

⇒-n(A∩B) = -12
⇒n(A∩B) = 12

Thus, A intersection B is 12.