The mode of the following frequency distribut | vedclass

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(X=43, Y=18) Given total frequency $N = 200.$Sum of frequencies: $8 + 12 + 27 + x + 55 + 37 + y = 200 \implies 139 + x + y = 200 \implies x + y = 61$

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(X=43, Y=18) Given total frequency $N = 200.$
Sum of frequencies: $8 + 12 + 27 + x + 55 + 37 + y = 200 \implies 139 + x + y = 200 \implies x + y = 61$ (Equation $1$).
Since the mode is $64,$ the modal class is $60-70.$
Formula for mode: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h.$
Here,$l = 60, f_1 = 55, f_0 = x, f_2 = 37, h = 10.$
$64 = 60 + \left( \frac{55 - x}{2(55) - x - 37} \right) \times 10.$
$4 = \left( \frac{55 - x}{110 - x - 37} \right) \times 10 \implies 4 = \frac{550 - 10x}{73 - x}.$
$292 - 4x = 550 - 10x \implies 6x = 258 \implies x = 43.$
Substituting $x = 43$ in Equation $1$: $43 + y = 61 \implies y = 18.$
Thus,$x = 43$ and $y = 18.$

Class	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$
Frequency	$8$	$12$	$27$	$x$	$55$	$37$	$y$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$7$	$16$	$25$	$32$	$20$

Class	$0-100$	$100-200$	$200-300$	$300-400$	$400-500$	$500-600$
Frequency	$12$	$18$	$27$	$20$	$17$	$6$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Frequency	$4$	$7$	$x$	$18$	$15$	$8$	$7$

The mode of the following frequency distribution is $64$ and the total frequency is $200.$ Find the missing frequencies $x$ and $y.$

Class $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$

Frequency $8$ $12$ $27$ $x$ $55$ $37$ $y$

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The mode of the following frequency distribution is $64$ and the total frequency is $200.$ Find the missing frequencies $x$ and $y.$ Class $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ Frequency $8$ $12$ $27$ $x$ $55$ $37$ $y$