The mode of the following data is $33 \frac{1}{3}$ and the total frequency is $100$. Find the missing frequencies $x$ and $y$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Frequency	$7$	$12$	$x$	$28$	$y$	$9$

(A) Here,the mode $33 \frac{1}{3}$ lies in the class $30-40$.
Therefore,$30-40$ is the modal class.
Now,$c$ (class length) $= 10$,$l$ (lower limit of the modal class) $= 30$,$f_1$ (frequency of the modal class) $= 28$,$f_0$ (frequency of the class preceding the modal class) $= x$,and $f_2$ (frequency of the class succeeding the modal class) $= y$.
Substituting the values in the mode formula $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times c$:
$33 \frac{1}{3} = 30 + \left( \frac{28 - x}{2(28) - x - y} \right) \times 10$
$3 \frac{1}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{10}{3} = \frac{28 - x}{56 - x - y} \times 10$
$\frac{1}{3} = \frac{28 - x}{56 - x - y}$
$56 - x - y = 84 - 3x$
$2x - y = 28$ ... $(1)$
Moreover,the total frequency is $100$.
$7 + 12 + x + 28 + y + 9 = 100$
$56 + x + y = 100$
$x + y = 44$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$3x = 72 \implies x = 24$
Substituting $x = 24$ in equation $(2)$:
$24 + y = 44 \implies y = 20$
Thus,the missing frequencies are $x = 24$ and $y = 20$.