The mode of the following frequency distribution is $46$ and the total frequency is $150$. Find the missing frequencies $x$ and $y$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$6$	$8$	$17$	$x$	$42$	$30$	$y$	$8$

(A) Given total frequency $N = 150$.
Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$
$111 + x + y = 150 \implies x + y = 39$ --- $(1)$
Since the mode is $46$,the modal class is $40-50$.
Here,$l = 40, f_1 = 42, f_0 = x, f_2 = 30, h = 10$.
Mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$46 = 40 + \left( \frac{42 - x}{2(42) - x - 30} \right) \times 10$
$6 = \left( \frac{42 - x}{84 - 30 - x} \right) \times 10$
$6 = \frac{10(42 - x)}{54 - x} \implies 3(54 - x) = 5(42 - x)$
$162 - 3x = 210 - 5x$
$2x = 48 \implies x = 24$
Substituting $x = 24$ in $(1)$: $24 + y = 39 \implies y = 15$.
Thus,$x = 24$ and $y = 15$.