The median of the following data is $525$. Find the values of $x$ and $y$,if the total frequency is $100$.

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

(X=9, Y=15) First,we construct the cumulative frequency table:

Class intervals	Frequency	Cumulative frequency
$0-100$	$2$	$2$
$100-200$	$5$	$7$
$200-300$	$x$	$7+x$
$300-400$	$12$	$19+x$
$400-500$	$17$	$36+x$
$500-600$	$20$	$56+x$
$600-700$	$y$	$56+x+y$
$700-800$	$9$	$65+x+y$
$800-900$	$7$	$72+x+y$
$900-1000$	$4$	$76+x+y$

Given that the total frequency $n = 100$,we have:
$76 + x + y = 100 \implies x + y = 24$ ........... $(1)$
The median is $525$,which lies in the class $500-600$. Thus,$l = 500$,$f = 20$,$cf = 36 + x$,and $h = 100$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$525 = 500 + \left( \frac{50 - (36 + x)}{20} \right) \times 100$
$25 = (14 - x) \times 5$
$5 = 14 - x$
$x = 9$
Substituting $x = 9$ into equation $(1)$:
$9 + y = 24 \implies y = 15$.
Thus,$x = 9$ and $y = 15$.