The median of the following frequency distribution is $46$ and the total frequency is $230$. Find the missing frequencies $x$ and $y$.

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$12$	$30$	$x$	$65$	$y$	$25$	$18$

(N/A) Given total frequency $N = 230$.
Sum of frequencies: $12 + 30 + x + 65 + y + 25 + 18 = 230$
$150 + x + y = 230 \implies x + y = 80$ ---(Equation $1$)
Since the median is $46$,the median class is $40-50$.
Here,$l = 40$,$h = 10$,$f = 65$,$N = 230$,and cumulative frequency of the class preceding the median class $cf = 12 + 30 + x = 42 + x$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$46 = 40 + \left( \frac{115 - (42 + x)}{65} \right) \times 10$
$6 = \left( \frac{73 - x}{65} \right) \times 10$
$6 \times 6.5 = 73 - x$
$39 = 73 - x \implies x = 34$.
Substituting $x = 34$ in Equation $1$: $34 + y = 80 \implies y = 46$.
Thus,$x = 34$ and $y = 46$.