If n(U)=360, n(A)=240, n(B)=160, find the maximum value of n(A union B ) and minimum value of - Maths - Sets

Question

If n(U)=360, n(A)=240, n(B)=160, find the maximum value of n(A
union B ) and minimum value of n(A intersection B) When will be the
value of n(A intersection B) be maximum and find its value

Abhinav Uniyal · Answer

We know n(AUB) = n(A)+n(B)-n(A∩B) 1) From the formula it is
clear that AUB will be maximum when A∩B is least Now in this
question AUB max will be when AUB=U This implies max n(AUB) = 360
Now when AUB is maximum A∩B is least Substituting value of AUB
in 1) 360=240+160-n(A∩B) Hence solving we get Min value of
A∩B =40 Similarly max value of A∩B is when B is a subset of
A Hence A∩B maximum=n(B)=160 Hope it helps!!!

If n(U)=360, n(A)=240, n(B)=160, find the maximum value of n(A union B ) and minimum value of n(A intersection B) . When will be the value of n(A intersection B) be maximum and find its value