Question

if n(aub) = 18 ,n(a-b) = 5, n(b-a) =3, find n(anb).

Answer 1

Answer:

Step-by-step explanation:

Given: n(A - B) =18, n(A∪B) =70, n(A ∩ B) =25

n(A∪B) = n(A - B) + n(A ∩ B) + n(B - A)  

  70 = 18 + 25 + n(B - A)  

    70 = 43 + n(B - A)  

    n(B - A) = 70 - 43  

     n(B - A) = 27  

Now n(B) = n(A ∩ B) + n(B - A)  

       n(B)  = 25 + 27  

         n(B)   = 52  

Answer 2

Given:

$n(A\cup B)=18; n(B-A)3$

$n(A-B)=5;$

$n(A\cup B)=n(A-B)+n(B-A)+n(A\cap B)$

$18=5+3+n(A \cap B)$

$n(A \cap B)=18-8$

$n(A\cap B)=10$