The mean of the following data is $26.5$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$6$	$x$	$17$	$y$	$8$

(A) Given total frequency $N = 60$.
Sum of frequencies: $6 + x + 17 + y + 8 = 60 \implies x + y + 31 = 60 \implies x + y = 29$ (Equation $1$).
The class marks $(f_i)$ are $5, 15, 25, 35, 45$.
Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 26.5$.
$\sum f_i x_i = (6 \times 5) + (x \times 15) + (17 \times 25) + (y \times 35) + (8 \times 45) = 30 + 15x + 425 + 35y + 360 = 815 + 15x + 35y$.
$\frac{815 + 15x + 35y}{60} = 26.5 \implies 815 + 15x + 35y = 1590 \implies 15x + 35y = 775$.
Dividing by $5$: $3x + 7y = 155$ (Equation $2$).
From Equation $1$,$x = 29 - y$. Substituting into Equation $2$: $3(29 - y) + 7y = 155 \implies 87 - 3y + 7y = 155 \implies 4y = 68 \implies y = 17$.
Then $x = 29 - 17 = 12$.
Thus,$x = 12$ and $y = 17$.