Mean and Median Questions in English

(B) The original set has $19$ observations,so the median is the $10^{th}$ observation when arranged in ascending order.
When two new observations $8$ and $32$ are added,the total number of observations becomes $21$.
The new median will be the $\frac{21+1}{2} = 11^{th}$ observation.
Since $8 < 30$ and $32 > 30$,the original $10^{th}$ observation (which is $30$) remains the median of the new set because one smaller value is added to the left and one larger value is added to the right,shifting the position of the median by one index but keeping the value the same.

(A) Let the weight of each natural number be $W$.
The weighted mean is given by the formula:
$\text{Weighted Mean} = \frac{\sum_{i=1}^{n} (i \times W)}{\sum_{i=1}^{n} W}$
$= \frac{W(1 + 2 + 3 + ... + n)}{n \times W}$
$= \frac{\frac{n(n + 1)}{2}}{n}$
$= \frac{n + 1}{2}$

(D) The weighted mean $(W.M.)$ of a set of values $x_1, x_2, ..., x_n$ with corresponding weights $w_1, w_2, ..., w_n$ is calculated by the formula:
$W.M. = \frac{\sum_{i=1}^{n} w_i x_i}{\sum_{i=1}^{n} w_i}$
Therefore,the correct option is $D$.

(D) Here,$N = \Sigma f_i = 4 + 5 + 6 + 10 + 20 = 45$.
$\Sigma f_i x_i = (5 \times 4) + (8 \times 5) + (11 \times 6) + (14 \times 10) + (17 \times 20)$
$= 20 + 40 + 66 + 140 + 340 = 606$.
$\therefore \bar{x} = \frac{\Sigma f_i x_i}{N} = \frac{606}{45} = 13.466... \approx 13.47$.

(B) To find the median,we first calculate the cumulative frequency $(c.f.)$:

Class	$f_i$	$c.f.$
$0 - 10$	$8$	$8$
$10 - 20$	$30$	$38$
$20 - 30$	$40$	$78$
$30 - 40$	$12$	$90$
$40 - 50$	$10$	$100$

Here,$N = 100$,so $\frac{N}{2} = \frac{100}{2} = 50$.
The cumulative frequency just greater than $50$ is $78$,which corresponds to the class interval $20 - 30$.
Thus,the median class is $20 - 30$.
Here,$l = 20$,$f = 40$,$F = 38$,and $h = 10$.
Median $= l + \left( \frac{\frac{N}{2} - F}{f} \right) \times h$
$= 20 + \left( \frac{50 - 38}{40} \right) \times 10$
$= 20 + \left( \frac{12}{40} \right) \times 10$
$= 20 + 3 = 23$.

(B) The formula for the weighted mean is $\bar{x}_w = \frac{\sum w_i x_i}{\sum w_i}$.
Given weights $(w_i)$ are $2, 1, 1, 2$ and scores $(x_i)$ are $60, 70, 70, 80$.
Weighted Mean $= \frac{(2 \times 60) + (1 \times 70) + (1 \times 70) + (2 \times 80)}{2 + 1 + 1 + 2}$.
$= \frac{120 + 70 + 70 + 160}{6}$.
$= \frac{420}{6} = 70$.

(B) Given that the mean of $x_1, x_2, \dots, x_{10}$ is $20$,we have:
$\frac{x_1 + x_2 + \dots + x_{10}}{10} = 20 \implies \sum_{i=1}^{10} x_i = 200$
The new set of observations is $(x_1 + 4), (x_2 + 8), \dots, (x_{10} + 40)$.
The mean of the new set is:
$\text{Mean} = \frac{(x_1 + 4) + (x_2 + 8) + \dots + (x_{10} + 40)}{10}$
$\text{Mean} = \frac{(x_1 + x_2 + \dots + x_{10}) + (4 + 8 + \dots + 40)}{10}$
$\text{Mean} = \frac{\sum x_i}{10} + \frac{4(1 + 2 + \dots + 10)}{10}$
Using the sum of first $n$ natural numbers formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we get $\sum_{k=1}^{10} k = \frac{10 \times 11}{2} = 55$.
$\text{Mean} = 20 + \frac{4 \times 55}{10} = 20 + \frac{220}{10} = 20 + 22 = 42$.

(C) The combined mean $\bar{X}$ of two series with means $\bar{X}_1$ and $\bar{X}_2$ and sizes $n_1$ and $n_2$ is given by the formula:
$\bar{X} = \frac{n_1 \bar{X}_1 + n_2 \bar{X}_2}{n_1 + n_2}$
Since $\bar{X}$ is a weighted average of $\bar{X}_1$ and $\bar{X}_2$,it must lie between the two values.
Given $\bar{X}_1 < \bar{X}_2$,it follows that $\bar{X}_1 < \bar{X} < \bar{X}_2$.

(C) Let the age of each student be $x$ years. Then the teacher's age is $(x + 20)$ years.
The average age is given by:
$\frac{(x + 20) + 3x}{4} = 20$
Solving for $x$:
$4x + 20 = 80$
$4x = 60$
$x = 15$
Therefore,the teacher's age is $x + 20 = 15 + 20 = 35$ years.

(B) The weighted mean is given by $\bar{x}_w = \frac{\sum_{i=1}^{n} i \cdot i^2}{\sum_{i=1}^{n} i^2}$.
This simplifies to $\bar{x}_w = \frac{\sum_{i=1}^{n} i^3}{\sum_{i=1}^{n} i^2}$.
Using the standard summation formulas $\sum i^3 = [\frac{n(n+1)}{2}]^2$ and $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$\bar{x}_w = \frac{[n(n+1)/2]^2}{n(n+1)(2n+1)/6} = \frac{n^2(n+1)^2}{4} \cdot \frac{6}{n(n+1)(2n+1)}$.
Simplifying the expression,we obtain $\bar{x}_w = \frac{3n(n+1)}{2(2n+1)}$.

(C) Let the man spend $x$ Rs on each type of pen.
The number of pens bought at $5$ Rs/pen is $\frac{x}{5}$.
The number of pens bought at $10$ Rs/pen is $\frac{x}{10}$.
The number of pens bought at $20$ Rs/pen is $\frac{x}{20}$.
Total cost = $x + x + x = 3x$.
Total number of pens = $\frac{x}{5} + \frac{x}{10} + \frac{x}{20} = \frac{4x + 2x + x}{20} = \frac{7x}{20}$.
Average cost per pen = $\frac{\text{Total cost}}{\text{Total number of pens}} = \frac{3x}{\frac{7x}{20}} = 3x \times \frac{20}{7x} = \frac{60}{7}$ Rs/pen.

(D) Let the set of $n$ numbers be $x_1, x_2, ..., x_n$.
The mean is given by $\bar x = \frac{1}{n} \sum_{i=1}^{n} x_i$.
If $\lambda$ is subtracted from each number,the new set becomes $(x_1 - \lambda), (x_2 - \lambda), ..., (x_n - \lambda)$.
The new mean $\bar x_{new}$ is $\frac{1}{n} \sum_{i=1}^{n} (x_i - \lambda)$.
This simplifies to $\frac{1}{n} (\sum_{i=1}^{n} x_i - \sum_{i=1}^{n} \lambda) = \frac{1}{n} (\sum_{i=1}^{n} x_i - n\lambda)$.
Thus,$\bar x_{new} = \frac{1}{n} \sum_{i=1}^{n} x_i - \frac{n\lambda}{n} = \bar x - \lambda$.

(D) For a series with an even number of observations $n$,the data is first arranged in ascending or descending order.
Since $n$ is even,there are two middle terms located at the $(\frac{n}{2})$-th position and the $(\frac{n}{2}+1)$-th position.
The median is calculated as the arithmetic mean (average) of these two middle terms.
Therefore,$\text{Median} = \frac{(\frac{n}{2}\text{-th term}) + ((\frac{n}{2}+1)\text{-th term})}{2}$.

(C) The mean of a set of numbers is calculated as the sum of the numbers divided by the count of the numbers.
For the second set of numbers,the sum is $130 + 126 + 68 + 50 + 1 = 375$.
The count of the numbers is $5$.
Therefore,the mean is $\frac{375}{5} = 75$.

(B) The mean of the five observations is given by:
$\frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} = 11$
Simplifying the numerator:
$\frac{5x + 20}{5} = 11$
$x + 4 = 11$
$x = 7$
The last three observations are $(x + 4), (x + 6)$,and $(x + 8)$.
Substituting $x = 7$,these are $11, 13$,and $15$.
The mean of the last three observations is:
$\frac{11 + 13 + 15}{3} = \frac{39}{3} = 13$

(B) The mean deviation of a distribution is minimum when it is taken about the median of the distribution.
Given the observations are $x_1 < x_2 < \dots < x_{101}$,the total number of observations $n = 101$.
The median of $n$ observations is the $\left( \frac{n+1}{2} \right)^{th}$ observation.
For $n = 101$,the median is the $\left( \frac{101+1}{2} \right)^{th} = 51^{st}$ observation.
Therefore,$k = x_{51}$.

(B) Given that the mean of $x_1, x_2, ......., x_n$ is $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$.
Let the new series be $y_i = x_i + 2i$ for $i = 1, 2, ......., n$.
The mean of the new series is $\bar{y} = \frac{1}{n} \sum_{i=1}^{n} (x_i + 2i)$.
$\bar{y} = \frac{1}{n} (\sum_{i=1}^{n} x_i + 2 \sum_{i=1}^{n} i)$.
Using the sum of first $n$ natural numbers formula $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$:
$\bar{y} = \frac{\sum x_i}{n} + \frac{2}{n} \times \frac{n(n+1)}{2}$.
$\bar{y} = \bar{x} + (n+1)$.

(A) To find the median,we first construct the cumulative frequency table:

Marks obtained	Frequency $(f)$	Cumulative Frequency $(cf)$
$0-10$	$2$	$2$
$10-20$	$18$	$20$
$20-30$	$30$	$50$
$30-40$	$45$	$95$
$40-50$	$35$	$130$
$50-60$	$20$	$150$
$60-70$	$6$	$156$
$70-80$	$3$	$159$

Here,$N = \sum f = 159$.
We calculate $\frac{N}{2} = \frac{159}{2} = 79.5$.
The cumulative frequency just greater than $79.5$ is $95$,which corresponds to the class interval $30-40$.
Thus,the median class is $30-40$.
Here,$l = 30$,$f = 45$,$cf = 50$,and $h = 10$.
Using the median formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
$\text{Median} = 30 + \left( \frac{79.5 - 50}{45} \right) \times 10$
$\text{Median} = 30 + \left( \frac{29.5}{45} \right) \times 10 = 30 + \frac{295}{45} = 30 + 6.555... \approx 36.56$.
Rounding to two decimal places,we get $36.56$. Since $36.55$ is the closest option provided,the correct answer is $A$.

(C) We know that the mean $\mu$ is defined as $\mu = \frac{\sum f_i y_i}{\sum f_i}$,which implies $\sum f_i y_i = \mu \sum f_i$.
Now,consider the expression $\sum f_i(y_i - \mu)$.
Expanding the summation,we get $\sum f_i y_i - \sum f_i \mu$.
Since $\mu$ is a constant,this becomes $\sum f_i y_i - \mu \sum f_i$.
Substituting $\sum f_i y_i = \mu \sum f_i$,we get $\mu \sum f_i - \mu \sum f_i = 0$.

(C) The sequence has $34$ terms. The median is the average of the $17^{\text{th}}$ and $18^{\text{th}}$ terms.
Given $x_1 = x_2 = \dots = x_{10} = 150$.
For $i \ge 10$,the sequence follows $x_{i+1} = x_i - 2$,which is an arithmetic progression with first term $a = x_{10} = 150$ and common difference $d = -2$.
The $n^{\text{th}}$ term of this progression (starting from $i=10$) is $x_{10+k} = 150 + k(-2)$.
For the $17^{\text{th}}$ term,$10+k = 17 \implies k = 7$,so $x_{17} = 150 + 7(-2) = 150 - 14 = 136$.
For the $18^{\text{th}}$ term,$10+k = 18 \implies k = 8$,so $x_{18} = 150 + 8(-2) = 150 - 16 = 134$.
The median is $\frac{x_{17} + x_{18}}{2} = \frac{136 + 134}{2} = \frac{270}{2} = 135$.

(B) The formula for weighted $A.M.$ is given by $\frac{\sum w_i x_i}{\sum w_i}$.
Here,the weights $(w_i)$ are $2, 1, 1, 2$ and the scores $(x_i)$ are $60, 70, 70, 80$.
Weighted $A.M. = \frac{(2 \times 60) + (1 \times 70) + (1 \times 70) + (2 \times 80)}{2 + 1 + 1 + 2}$
$= \frac{120 + 70 + 70 + 160}{6}$
$= \frac{420}{6} = 70$.

(B) First,arrange the data in ascending order of magnitude and calculate the cumulative frequency $(cf)$:

Height (in $cm$)	$150$	$152$	$154$	$155$	$156$	$160$	$161$
No of students $(f)$	$8$	$4$	$3$	$7$	$3$	$12$	$4$
Cumulative Frequency $(cf)$	$8$	$12$	$15$	$22$	$25$	$37$	$41$

Here,the total number of observations $N = 41$,which is an odd number.
The median is the value of the $\left(\frac{N+1}{2}\right)^{\text{th}}$ observation.
Median $= \left(\frac{41+1}{2}\right)^{\text{th}} = 21^{\text{st}}$ observation.
From the cumulative frequency table,we observe that the $21^{\text{st}}$ observation falls in the range where the height is $155$ (since $cf$ for $154$ is $15$ and $cf$ for $155$ is $22$).
Thus,the median is $155$.

(D) For a set of $13$ observations arranged in decreasing order,the median is the $7^{th}$ observation.
Let the observations be $x_1 \ge x_2 \ge \dots \ge x_7 \ge \dots \ge x_{13}$.
The median is $x_7 = 18.6$.
When the first two observations $(x_1, x_2)$ are increased and the last three observations $(x_{11}, x_{12}, x_{13})$ are decreased,the relative order of the $7^{th}$ observation remains unaffected because the changes occur at the extreme ends of the sorted list.
Since the $7^{th}$ observation remains the same,the median remains unchanged.

(C) Let $x_{1}, x_{2}, \dots, x_{n}$ be $n$ observations,then the original mean is $\bar{X} = \frac{1}{n} \sum_{i=1}^{n} x_{i}$.
Let the new observations be $y_{i} = \frac{x_{i}}{\alpha} + 10$.
The new mean $\bar{Y}$ is given by:
$\bar{Y} = \frac{1}{n} \sum_{i=1}^{n} y_{i} = \frac{1}{n} \sum_{i=1}^{n} \left( \frac{x_{i}}{\alpha} + 10 \right)$
$\bar{Y} = \frac{1}{\alpha} \left( \frac{1}{n} \sum_{i=1}^{n} x_{i} \right) + \frac{1}{n} \sum_{i=1}^{n} 10$
$\bar{Y} = \frac{1}{\alpha} \bar{X} + \frac{10n}{n} = \frac{\bar{X}}{\alpha} + 10$
Combining the terms,we get $\bar{Y} = \frac{\bar{X} + 10\alpha}{\alpha}$.

(D) First,we calculate the cumulative frequency $(c.f.)$:

Interval	$f$	$c.f.$
$10-20$	$180$	$180$
$20-30$	$82$	$262$
$30-40$	$34$	$296$
$40-50$	$180$	$476$
$50-60$	$136$	$612$
$60-70$	$23$	$635$
$70-80$	$50$	$685$

Total frequency $N = 685$.
$\frac{N}{2} = \frac{685}{2} = 342.5$.
The cumulative frequency just greater than $342.5$ is $476$,which corresponds to the class interval $40-50$.
Here,$l = 40$,$f = 180$,$cf = 296$,and $h = 10$.
Median $= l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
Median $= 40 + \left( \frac{342.5 - 296}{180} \right) \times 10$
Median $= 40 + \left( \frac{46.5}{180} \right) \times 10 = 40 + \frac{46.5}{18} = 40 + 2.5833... \approx 42.58$.

(A) Given the set of $10$ numbers: $10, 22, 26, 29, 34, x, 42, 67, 70, y$ in increasing order.
$1$. Mean calculation:
$\text{Mean} = \frac{10 + 22 + 26 + 29 + 34 + x + 42 + 67 + 70 + y}{10} = 42$
$300 + x + y = 420$
$x + y = 120 \quad \dots (i)$
$2$. Median calculation:
For $10$ observations,the median is the average of the $5^{th}$ and $6^{th}$ terms.
$\text{Median} = \frac{34 + x}{2} = 35$
$34 + x = 70$
$x = 36$
$3$. Finding $y$:
Substitute $x = 36$ into equation $(i)$:
$36 + y = 120$
$y = 84$
$4$. Final ratio:
$\frac{y}{x} = \frac{84}{36} = \frac{7}{3}$

(A) The sum of frequencies is given by $\sum f_i = (x+1)^2 + (2x-5) + (x^2-3x) + x = 20$.
Expanding this,we get $x^2 + 2x + 1 + 2x - 5 + x^2 - 3x + x = 20$.
$2x^2 + 2x - 4 = 20$ $\Rightarrow 2x^2 + 2x - 24 = 0$ $\Rightarrow x^2 + x - 12 = 0$.
Factoring the quadratic equation: $(x+4)(x-3) = 0$.
Since frequency cannot be negative,$2x-5 > 0$ implies $x > 2.5$,so we take $x = 3$.
Substituting $x = 3$ into the frequencies:
Marks $2$: $f_1 = (3+1)^2 = 16$
Marks $3$: $f_2 = 2(3)-5 = 1$
Marks $5$: $f_3 = 3^2 - 3(3) = 0$
Marks $7$: $f_4 = 3$
Total students: $16 + 1 + 0 + 3 = 20$.
The mean $\bar{x} = \frac{\sum x_i f_i}{\sum f_i} = \frac{(2 \times 16) + (3 \times 1) + (5 \times 0) + (7 \times 3)}{20} = \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$.

(C) Let the $11$ distinct positive integers be arranged in increasing order: $x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10} < x_{11}$.
The median of these $11$ integers is $x_6$.
The other $10$ integers are $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}$.
The median of these $10$ integers is $m = \frac{x_5 + x_6}{2}$.
Since $x_5 < x_6$,it follows that $x_5 < \frac{x_5 + x_6}{2} < x_6$,so $x_5 < m < x_6$.
Replacing the largest integer $x_{11}$ with $m$,the new set of $11$ integers in increasing order is $x_1, x_2, x_3, x_4, x_5, m, x_6, x_7, x_8, x_9, x_{10}$.
The new median is the $6^{th}$ term,which is $m$.
Since $m < x_6$,the median decreases.

(D) The total frequency $N = 3 + 9 + 10 + 8 + 6 = 36$.

Class	Frequency	Cumulative Frequency
$0-4$	$3$	$3$
$4-8$	$9$	$12$
$8-12$	$10$	$22$
$12-16$	$8$	$30$
$16-20$	$6$	$36$

Since $\frac{N}{2} = \frac{36}{2} = 18$,the median class is $8-12$.
The median formula is $M = l + \left( \frac{\frac{N}{2} - C}{f} \right) \times h$,where $l = 8$,$C = 12$,$f = 10$,and $h = 4$.
$M = 8 + \left( \frac{18 - 12}{10} \right) \times 4 = 8 + \left( \frac{6}{10} \right) \times 4 = 8 + 2.4 = 10.4$.
Therefore,$20M = 20 \times 10.4 = 208$.

(C) Given the total number of students is $20$,the sum of frequencies must be $20$:
$(x+1)^2 + (2x-5) + (x^2-3x) + x = 20$
$(x^2+2x+1) + 2x - 5 + x^2 - 3x + x = 20$
$2x^2 + 2x - 4 = 20$
$2x^2 + 2x - 24 = 0$
$x^2 + x - 12 = 0$
$(x+4)(x-3) = 0$
Since $x > 0$,we have $x = 3$.
Now,substitute $x=3$ into the frequency distribution:
Marks $(x_i)$: $2, 3, 5, 7$
Frequencies $(f_i)$: $(3+1)^2=16, (2(3)-5)=1, (3^2-3(3))=0, 3$
Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2(16) + 3(1) + 5(0) + 7(3)}{20}$
$= \frac{32 + 3 + 0 + 21}{20} = \frac{56}{20} = 2.8$

(A) The given values are $\sin^2 10^{\circ}, \sin^2 20^{\circ}, \sin^2 30^{\circ}, \dots, \sin^2 90^{\circ}$.
There are $9$ terms in total.
Sum $= \sin^2 10^{\circ} + \sin^2 20^{\circ} + \sin^2 30^{\circ} + \sin^2 40^{\circ} + \sin^2 50^{\circ} + \sin^2 60^{\circ} + \sin^2 70^{\circ} + \sin^2 80^{\circ} + \sin^2 90^{\circ}$.
Using $\sin(90^{\circ} - \theta) = \cos \theta$,we have $\sin^2 80^{\circ} = \cos^2 10^{\circ}$,$\sin^2 70^{\circ} = \cos^2 20^{\circ}$,$\sin^2 60^{\circ} = \cos^2 30^{\circ}$,and $\sin^2 50^{\circ} = \cos^2 40^{\circ}$.
Sum $= (\sin^2 10^{\circ} + \cos^2 10^{\circ}) + (\sin^2 20^{\circ} + \cos^2 20^{\circ}) + (\sin^2 30^{\circ} + \cos^2 30^{\circ}) + (\sin^2 40^{\circ} + \cos^2 40^{\circ}) + \sin^2 90^{\circ}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin 90^{\circ} = 1$,
Sum $= 1 + 1 + 1 + 1 + (1)^2 = 5$.
Mean $= \frac{\text{Total Sum}}{\text{Number of terms}} = \frac{5}{9}$.

(D) Let the observations be $x_1, x_2, x_3, \ldots, x_n$.
The mean is given by $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = 10$.
When each observation is multiplied by $2$,the new observations become $2x_1, 2x_2, 2x_3, \ldots, 2x_n$.
The new mean $\bar{x}_{new} = \frac{\sum_{i=1}^{n} (2x_i)}{n} = 2 \times \left( \frac{\sum_{i=1}^{n} x_i}{n} \right)$.
Substituting the value of the original mean,we get $\bar{x}_{new} = 2 \times 10 = 20$.

(B) Let the $n$ numbers be $x_1, x_2, x_3, \ldots, x_n$.
The mean of these numbers is $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$.
When each number is divided by $5$,the new set of numbers is $\frac{x_1}{5}, \frac{x_2}{5}, \ldots, \frac{x_n}{5}$.
The mean of this new set is given as $\frac{X}{5}$.
Therefore,$\frac{\frac{x_1}{5} + \frac{x_2}{5} + \ldots + \frac{x_n}{5}}{n} = \frac{X}{5}$.
$\frac{1}{5} \left( \frac{\sum_{i=1}^{n} x_i}{n} \right) = \frac{X}{5}$.
Multiplying both sides by $5$,we get $\frac{\sum_{i=1}^{n} x_i}{n} = X$.
Thus,the mean of the original $n$ numbers is $X$.

(C) Given data: $12, 14, 20, 23, 25, 32$
The arithmetic mean $\bar{x}$ is calculated as:
$\bar{x} = \frac{\sum x_i}{n} = \frac{12 + 14 + 20 + 23 + 25 + 32}{6}$
$\bar{x} = \frac{126}{6} = 21$
Therefore,the arithmetic mean is $21$.

(B) The formula for the mean of $n$ observations is given by $\text{Mean} = \frac{\sum x_i}{n}$.
Given the observations are $8, 6, 7, 5, x, 4$ and the mean is $7$.
Number of observations $n = 6$.
$\frac{8 + 6 + 7 + 5 + x + 4}{6} = 7$
$\frac{30 + x}{6} = 7$
$30 + x = 42$
$x = 42 - 30$
$x = 12$
Hence,option $B$ is correct.

(C) Given that the number $i$ is repeated $i$ times for $i = 1, 2, \ldots, n$.
The sum of the observations is $\sum_{i=1}^{n} i \times i = \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$.
The total number of observations is $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.
The mean $\bar{X}$ is given by the ratio of the sum of observations to the total number of observations:
$\bar{X} = \frac{\sum i^2}{\sum i} = \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{2n+1}{3}$.