If the mean of the data:Class5 - 1010 - 1515 | vedclass

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(C) The mean of the data is given by $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.Mid-values $(x_i)$: $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$.Total frequency

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$2x^2 - 19x - 10 = 0$

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(C) The mean of the data is given by $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.Mid-values $(x_i)$: $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$.Total frequency $\sum f_i = 2 + k + 28 + 54 + (k + 1) + 5 = 90 + 2k$.Sum of $f_i x_i = (2 	imes 7.5) + (k 	imes 12.5) + (28 	imes 17.5) + (54 	imes 22.5) + ((k + 1) 	imes 27.5) + (5 	imes 32.5) = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5 = 1910 + 40k$.Given $\bar{x} = 21$, so $\frac{1910 + 40k}{90 + 2k} = 21$.$1910 + 40k = 21(90 + 2k) \Rightarrow 1910 + 40k = 1890 + 42k$.$2k = 20 \Rightarrow k = 10$.Testing $k = 10$ in the given equations:$A) 2(10)^2 - 23(10) - 10 = 200 - 230 - 10 = -40 
eq 0$.$B) 4(10)^2 - 35(10) + 24 = 400 - 350 + 24 = 74 
eq 0$.$C) 2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$.$D) 2(10)^2 - 35(10) + 98 = 200 - 350 + 98 = -52 
eq 0$.Therefore, $k = 10$ is a root of the equation $2x^2 - 19x - 10 = 0$.

Class	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Frequency	$2$	$k$	$28$	$54$	$k + 1$	$5$

$x_i$	$1$	$2$	$3$	$4$	$5$
$f_i$	$5$	$4$	$f$	$2$	$3$

If the mean of the data:
Class $5 - 10$ $10 - 15$ $15 - 20$ $20 - 25$ $25 - 30$ $30 - 35$
Frequency $2$ $k$ $28$ $54$ $k + 1$ $5$

is $21$, then $k$ is one of the roots of the equation:

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If the mean of the data:Class$5 - 10$$10 - 15$$15 - 20$$20 - 25$$25 - 30$$30 - 35$Frequency$2$$k$$28$$54$$k + 1$$5$is $21$, then $k$ is one of the roots of the equation: