Question

If n(u)=40,n(a)=25 and n(b)=20 what is the least value of n(a intersection b)

Answer 1

Answer:

Hope it will help u

n(u)=40

n(a) =40-25

=15

n(b)=40-20

=20

.

. .(a n b)=n(u)-n(b)+n(a)

=40- 20+15

=5

Answer 2

The least value of n(anb) is 5.

Given, n(u)=40, n()=25 and n(b)=20

We have to find the least value of n(anb)

We know that,

n(aub) = n(a) + n(b) - n(anb)

n(aub) =  n(a) + n(b)

           = 25 + 20

           = 45

Given that, n(u)=40

∴  n(aub) ≤ 40

⇒ n(a) + n(b) - n(anb) ≤ 40

⇒    25 + 20 - n(anb) ≤ 40

⇒             45 - n(anb) ≤ 40

⇒                    45- 40 ≤ n(anb)

⇒                             5  ≤ n(anb)

Thus, n(anb) has the least value as 5.

#SPJ3