The median of the following frequency distribution is $525$ and the total frequency is $100$. Find the missing frequencies $x$ and $y$.

Class	$0-100$	$100-200$	$200-300$	$300-400$	$400-500$	$500-600$	$600-700$	$700-800$	$800-900$	$900-1000$
Frequency	$2$	$5$	$x$	$12$	$17$	$20$	$y$	$9$	$7$	$4$

(A) Given,total frequency $N = 100$.
Sum of frequencies: $2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 100$
$76 + x + y = 100 \implies x + y = 24$ --- $(1)$
Since the median is $525$,the median class is $500-600$.
Here,$l = 500$,$f = 20$,$cf = (2 + 5 + x + 12 + 17) = 36 + x$,$h = 100$,and $N/2 = 50$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x \implies x = 9$
Substituting $x = 9$ in $(1)$: $9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.