Number’s divisible by $2$ in $X = 61$ [ = integer(123/2) ]
Number’s divisible by $3$ in $X = 41$
Number’s divisible by $5$ in $X = 24$
Number’s divisible by $2$ and $3$ i.e. by $6 = 20$
Number’s divisible by $2$ and $5$ i.e by $10 = 12$
Number’s divisible by $3$ and $5$ i.e by $15 = 8$
Number’s divisible by $2$ and $3$ and $5$ i.e by $30 = 4$
Number’s divisible by either $2$ or $3$ or $5$ = $N(AUBUC)$ = $N(A) +N(B)+N(C) -N(A∩B)-N(B∩C)-N(A∩C)+ N(A∩B∩C) $
$= 61 +41+24 -20-12-8 +4 = 90$
$X$={ $n ,1 ≤ n ≤ 123, n$ is not divisible by either $2, 3$ or $5$ }
Cardinality = $123-90$ =$33$
Correct Answer: $B$