Question

If n(A∆B)=12 and n of A intersection B =3 then find the greatest possible value of n of A×B

Answer 1

Answer:

Ifn(A)=7,n(A∪B)=11,andn(B)=5,thenwhatisn(A∩B)?

Solution

n(A)=7

n(B)=5

n(A∪B)=11

(A∩B)=?

Let the intersection of A and B = n,n(Aonly)=7−n,n(Bonly)=5−n

n(AUB)=n(Aonly)+n(Bonly)+n(AnB)

After you replace the equations in the formula above,

n(AUB)=n(Aonly)+n(Bonly)+n(AnB)

n(AUB)=7−n+5−n+n

Remember our n(AUB)=11,,

11=7−n+5−n+n

Revise the formula for easy understanding starting from n,so

n+5−n+7−n=11

n−n−n+5+7=11

The first two n (s) cancel out and we remain with one, as shown below

5+7−n=11

But 5+7=12

Therefore,

12−n=11

Here n crosses to become a positive (+n) and 11 crosses to the opposite side to become a negative (-11) as shown below,

12−n=11

12−11=n

but 12−11=1

1=n

n=1

But remember we said let the intersection of A and B = n

so n(AnB)=n

Theren(AnB)=1.

So Ifn(A)=7,n(A∪B)=11,andn(B)=5,thenn(A∩B)=1.