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We have,

No of elements in Z = 90

and, No of elements in Y = 16 + 18 + 17 + b = 51 + b

So,

$\frac{n (Z)}{n (Y)} = \frac{4}{5} \Rightarrow \frac{51 + b}{90} = \frac{4}{5} \Rightarrow 51 + b = 72 \Rightarrow b = 21$

22.

Consider the proper subsets of {1, 2, 3, 4). How many of these subsets are superset of the set {3} ?