Let us recall the formula
\(\displaystyle{n}{\left({A}{U}{B}\right)}={n}{\left({A}\right)}+{n}{\left({B}\right)}-{n}{\left({A}⋂{B}\right)}\)

It is given that n(U)=100, n(B')=30

Thus n(B)=n(U)-n(B')=100-30=70

Hence we obtain \(\displaystyle{n}{\left({A}⋂{B}\right)}={n}{\left({A}\right)}+{n}{\left({B}\right)}-{n}{\left({A}{U}{B}\right)}\) =60+70-75 =55

It is given that n(U)=100, n(B')=30

Thus n(B)=n(U)-n(B')=100-30=70

Hence we obtain \(\displaystyle{n}{\left({A}⋂{B}\right)}={n}{\left({A}\right)}+{n}{\left({B}\right)}-{n}{\left({A}{U}{B}\right)}\) =60+70-75 =55