MCQ 11 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : Consider the following frequency distribution :
| Class interval |
$3-6$ |
$6-9$ |
$9-12$ |
$12-15$ |
$15-18$ |
$18-21$ |
| Frequency |
$2$ |
$5$ |
$21$ |
$23$ |
$10$ |
$12$ |
he mode of the above data is $12.4.$
Reason : The value of the variable which occurs most often is the mode. - A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- ✓
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: B. Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
The maximum frequency is $23$ and the modal class is $12 - 15.$
So $,\text{l} = 12,\text{ f}_\text{i} = 23,\text{ f}_0 = 21, \text{ f}_2 = 10 $ and $\text{h}=3$
$\therefore\text{ Mode}=12+\bigg(\frac{23-21}{2\times23-21-10}\bigg)\times3$
$=\bigg(12+3\times\frac{2}{15}\bigg)$
$=12.4$
View full question & answer→MCQ 21 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as:
Assertion : If the value of mode and mean is $60$ and $66$ respectively, then the value of median is $64.$
Reason : Median $=\ ($mode $+\ 2$ mean$)$
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- ✓
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: C. Assertion $(A)$ is true but reason $(R)$ is false.
Median $=\frac{1}{3}(\text{mode + 2 mean})$
$=\frac{1}{3}(60+2\times66)$
$=64$
View full question & answer→MCQ 31 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as:
Assertion : If the value of mode and mean is $60$ and $66$ respectively, then the value of median is $64.$
Reason : Median $= ($mode $+\ 2 $ mean$)$
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- ✓
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: C. Assertion $(A)$ is true but reason $(R)$ is false.
Median $=\frac{1}{3}(\text{mode + 2 mean)}$
$=\frac{1}{3}(60+2\times66)=64$
View full question & answer→MCQ 41 Mark
Directions: In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : Consider the following frequency distribution:
| Class interval |
$10-15$ |
$15-20$ |
$20-25$ |
$25-30$ |
$30-35$ |
| Frequency |
$5$ |
$9$ |
$12$ |
$6$ |
$8$ |
Reason : The class having maximum frequency is called the modal class. - A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true.
The maximum frequency is $12,$ which lies in the interval $20 - 25.$
So, the modal class is $20 - 25.$
View full question & answer→MCQ 51 Mark
Directions: In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R)$. Mark the correct choice as :
Assertion: Consider the following frequency distribution:
| Class interval |
$0-4$ |
$4-8$ |
$8-12$ |
$12-16$ |
$16-20$ |
| Frequency |
$6$ |
$3$ |
$5$ |
$20$ |
$10$ |
The median class is $12-16.$
Reason : Let $\text{n}=\sum\text{f}_\text{i}$ Then, the class whose cumulative frequency is just lesser than$\big(\frac{\text{n}}{2}\big)$is the median class. - A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- ✓
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: C. Assertion $(A)$ is true but reason $(R)$ is false.
We know that, the class whose cumulative frequency is just greater than$\frac{\text{n}}{2}$ is the median class.
So, Reason is wrong.
The cumulative frequency distribution table from the given data can be drawn as :
| Class Interval |
Frequency |
Cumulative Frequency |
| $0-4$ |
$6$ |
$6$ |
| $4-8$ |
$3$ |
$9$ |
| $8-12$ |
$5$ |
$14$ |
| $12-16$ |
$20$ |
$34$ |
| $16-20$ |
$10$ |
$44$ |
Here $\text{n}=44$
$\Rightarrow\frac{\text{n}}{2}=22,$ which lies in the interval $12 - 16.$
So, it is the median class. View full question & answer→MCQ 61 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R)$.Mark the correct choice as:
Assertion : If the arithmetic mean of $5,7,\text{x},10,15$ is $\text{x}$ then $\text{x}=9.25$
Reason : If $\text{x}_1,\text{x}_2,\text{x}_3...\text{x}_\text{n}$ are nn values of a variable $\text{X},$ then the arithmetic mean of these values is given by
$\frac{\text{x}_1+\text{x}_2+\text{x}_3+...+\text{x}_\text{n}}{2\text{n}}$
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- ✓
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: C. Assertion $(A)$ is true but reason $(R)$ is false.
If $\text{x}_1,\text{x}_2,\text{x}_3...\text{x}_\text{n}$ are $\text{n}$ values of a variable $\text{X},$
then the arithmetic mean of these values is given by
$\frac{\text{x}_1+\text{x}_2+\text{x}_3+...+\text{x}_\text{n}}{2\text{n}}$
$\therefore$ Reason is wrong.
Now, mean of given values $=\text{x}\ [$Given$]$
$\Rightarrow\frac{5+7+\text{x}+10+15}{5}=\text{x}$
$\Rightarrow\frac{37+\text{x}}{5}=\text{x}$
$\Rightarrow37+\text{x}=5\text{x}$
$\Rightarrow37=4\text{x}$
$\Rightarrow\text{x}=9.25,$ which is correct
View full question & answer→MCQ 71 Mark
Directions: In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R)$.Mark the correct choice as:
Assertion: The arithmetic mean of the following frequency distribution is $25$.
| Class Interval |
$0-10$ |
$10-20$ |
$20-30$ |
$30-40$ |
$40-50$ |
| Frequency |
$5$ |
$18$ |
$15$ |
$16$ |
$6$ |
Reason: $\text{Mean }(\overline{\text{x}})=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$ where $\text{x}_\text{i}=\frac{1}{2} ($Lower limit $+$ Upper limit$)$ of the $\text{i}^\text{th}$ class interval and $\text{f}_\text{i}$ is its frequency. - ✓
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: A. Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
Clearly, Reason is correct.
Now, the frequency distribution table from the given data can be drawn as:
| Class Interval |
Frequency $(f_i)$ |
$x_i$ |
$f_ix_i$ |
| $0-10$ |
$5$ |
$5$ |
$25$ |
| $10-20$ |
$18$ |
$15$ |
$270$ |
| $20-30$ |
$15$ |
$25$ |
$375$ |
| $30-40$ |
$16$ |
$35$ |
$560$ |
| $40-50$ |
$6$ |
$45$ |
$270$ |
| |
$\sum\text{f}_\text{i}=60$ |
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=1500$ |
$\therefore\text{Mean }=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow$ Mean $=\frac{1500}{60}$
$\Rightarrow$ Mean $= 25$, which is true. View full question & answer→MCQ 81 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as:
Assertion : If the number of runs scored by $11$ players of a cricket team of India are $\{5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27\}$ then median is $30.$
Reason : Median $=\big(\frac{\text{n}+1}{2}\big)^\text{th},$ value, if n is odd.
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true.
Arranging the terms in ascending order $, \{0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52\}$
Median value $=\big(\frac{11+1}{2}\big)^\text{th}$
$=6^\text{th}\text{ value}=27$
View full question & answer→MCQ 91 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : If for a certain frequency distribution, $\text{l}=24.5,\text{h}=4,\text{f}_0=14,\text{f}_1=14,\text{f}_2=15$ then the value of mode is $25.$
Reason : Mode of a frequency distribution is given by :
$\text{Mode}=\text{l}+\bigg(\frac{\text{f}_1-\text{f}_0}{2\text{f}_1-\text{f}_0-\text{f}_2}\bigg)\times\text{h}$
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true.
Now, it is given that $\text{l}=24.5,\text{h}=4,\text{f}_0=14,\text{f}_1=14,\text{f}_2=15$
$\therefore\text{Mode}=24.5+\bigg(\frac{14-14}{28-14-15}\bigg)\times4$
$\Rightarrow\text{Mode}=24.5+0$
$\Rightarrow\text{Mode}=24.5$
View full question & answer→MCQ 101 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R)$.Mark the correct choice as :
Assertion : The arithmetic mean of the following given frequency distribution table is $13.81.$
| $x$ |
$4$ |
$7$ |
$10$ |
$13$ |
$16$ |
$19$ |
| $f$ |
$7$ |
$10$ |
$15$ |
$20$ |
$25$ |
$30$ |
Reason : $\overline{\text{x}}=\frac{\sum\text{f}_i\text{x}_\text{i}}{\sum\text{f}_\text{i}}$ - ✓
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: A. Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
View full question & answer→MCQ 111 Mark
Statement $A ($Assertion$)$ : Consider the following frequency distribution:
| Class interval |
$3-6$ |
$6-9$ |
$9-12$ |
$12-15$ |
$15-18$ |
$18-21$ |
| Frequency |
$2$ |
$5$ |
$21$ |
$23$ |
$10$ |
$12$ |
The mode of the above data is $12.4 .$
Statement $R ($Reason$):$ The value of the variable which occurs most often is the mode. - A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- ✓
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is not the correct explanation of assertion $(A).$
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion $(A)$ is false but reason $(R)$ is true.
AnswerCorrect option: B. Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is not the correct explanation of assertion $(A).$
Clearly, Reason is true.
The maximum frequency is $23$ and the modal class is $12-15.$
$\therefore l=12, f_1=23, f_0=21, f_2=10$ and $h=3$
$\therefore \text { Mode }=12+\left(\frac{23-21}{2 \times 23-21-10}\right) \times 3$
$=\left(12+3 \times \frac{2}{15}\right)$
$=12.4$
$\therefore$ Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
| Class |
Class marks $\left(x_i\right)$ |
Frequency $\left(f_i\right)$ |
$f_i x_i$ |
| $1-3$ |
$2$ |
$12$ |
$24$ |
| $3-5$ |
$4$ |
$22$ |
$88$ |
| $5-7$ |
$6$ |
$27$ |
$165$ |
| $7-9$ |
$8$ |
$19$ |
$152$ |
| |
|
$\Sigma f_i=80$ |
$\Sigma f_i x_i=426$ |
$\therefore$ Mean $=\frac{\sum f_i x_i}{\sum f_i}=\frac{426}{80}=5.325$ View full question & answer→MCQ 121 Mark
Statement A (Assertion) : Consider the following frequency distribution:| Class interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |
| Frequency | 6 | 3 | 5 | 20 | 10 |
The median class is 12-16.
Statement R (Reason): Let $n=\sum f_i$. Then, the class whose cumulative frequency is just lesser than $\left(\frac{n}{2}\right)$ is the median class. - ✓
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is not the correct explanation of assertion (A).
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion (A) is false but reason $(R)$ is true.
AnswerCorrect option: A. Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is the correct explanation of assertion $(A)$.
(a) : Clearly, Reason is correct. Now, the frequency distribution table from the given data can be drawn as :| Class interval | Frequency $\left(f_i\right)$ | $x_i$ | $f_i x_i$ |
| 0-10 | 5 | 5 | 25 |
| 10-20 | 18 | 15 | 270 |
| 20-30 | 15 | 25 | 375 |
| 30-40 | 16 | 35 | 560 |
| 40-50 | 6 | 45 | 270 |
| | $\sum f_i=60$ | | $\sum f_i x_i=1500$ |
$\therefore \quad$ Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{1500}{60}=25$, which is true.
$\therefore \quad$ Both Assertion and Reason are true and Reason is the correct explanation of Assertion. View full question & answer→MCQ 131 Mark
Statement A (Assertion) : If the median and mode of a frequency distribution are 8.9 and 9.2 respectively, then its mean is 9 .
Statement R (Reason) : Mean, median and mode of a frequency distribution are related as:
Mode $=3$ median -2 mean
- A
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is not the correct explanation of assertion (A).
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion (A) is false but reason $(R)$ is true.
AnswerCorrect option: D. Assertion (A) is false but reason $(R)$ is true.
(d): Clearly, Reason is true.Using the relation,
$
\begin{aligned}
2 \text { Mean } & =3 \text { Median }- \text { Mode }=3 \times 8.9-9.2 \\
& =26.7-9.2=17.5
\end{aligned}
$
$\Rightarrow$ Mean $=8.75$
$\therefore$ Assertion is false but Reason is true.
View full question & answer→MCQ 141 Mark
Statement A (Assertion) : Consider the following frequency distribution:| Class interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |
| Frequency | 6 | 3 | 5 | 20 | 10 |
The median class is 12-16.
Statement R (Reason): Let $n=\sum f_i$. Then, the class whose cumulative frequency is just lesser than $\left(\frac{n}{2}\right)$ is the median class. - A
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is not the correct explanation of assertion (A).
- ✓
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion (A) is false but reason $(R)$ is true.
AnswerCorrect option: C. Assertion $(A)$ is true but reason $(R)$ is false.
(c) : We know that, if $n=\Sigma f i$, then the class whose cumulative frequency is just greater than $\left(\frac{n}{2}\right)$ is the median class. So, Reason is false.
The cumulative frequency distribution table from the given data can be drawn as :| Class interval | Frequency | Cumalative frequency |
| 0-4 | 6 | 6 |
| 4-8 | 3 | 9 |
| 8-12 | 5 | 14 |
| 12-16 | 20 | 34 |
| 16-20 | 10 | 44 |
Here, $n=44 \Rightarrow \frac{n}{2}=22$. So, the class whose cumulative frequency is just greater than 22 is 34 , which lies in the interval 12 - 16.
So, the median class is 12-16.
$\therefore$ Assertion is true but Reason is false. View full question & answer→MCQ 151 Mark
Statement A (Assertion) : If for a certain frequency distribution, $l=24.5, h=4, f_0=14, f_1=14$, $f_2=15$, then the value of mode is 25 .
Statement R (Reason) : Mode of a frequency distribution is given by :
$
\text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h
$
- A
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is not the correct explanation of assertion (A).
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion (A) is false but reason $(R)$ is true.
AnswerCorrect option: D. Assertion (A) is false but reason $(R)$ is true.
(d) : Clearly, Reason is true.
Now, it is given that $l=24.5, h=4, f_0=14, f_1=14, f_2=15$
$\therefore \quad$ Mode $=24.5+\left(\frac{14-14}{28-14-15}\right) \times 4=24.5+0=24.5$
$\therefore \quad$ Assertion is false but Reason is true.
View full question & answer→MCQ 161 Mark
Statement A (Assertion) : Consider the following frequency distribution:| Class interval | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
| Frequency | 5 | 9 | 12 | 6 | 5 |
The modal class is $10-15$.
Statement $K$ (Keason) : The class having maximum frequency is called the modal class. - A
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is not the correct explanation of assertion (A).
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion (A) is false but reason $(R)$ is true.
AnswerCorrect option: D. Assertion (A) is false but reason $(R)$ is true.
(d) : Clearly, Reason is true.
The maximum frequency is 12 , which lies in the interval $20-25$. So, the modal class is 20-25.
$\therefore \quad$ Assertion is false but Reason is true.
View full question & answer→MCQ 171 Mark
Statement A (Assertion): If the arithmetic mean of $5,7, x, 10,15$ is $x$, then $x=9.25$.
Statement $R$ (Reason): If $x_1, x_2, x_3, \ldots, x_n$ are $n$ values of a variable $X$, then the arithmetic mean of these values is given by $\frac{x_1+x_2+x_3+\ldots+x_n}{2 n}$.
- A
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is the correct explanation of assertion $(A)$.
- B
Both assertion (A) and reason ( $R$ ) are true and reason $(R)$ is not the correct explanation of assertion (A).
- ✓
Assertion $(A)$ is true but reason $(R)$ is false.
- D
Assertion (A) is false but reason $(R)$ is true.
AnswerCorrect option: C. Assertion $(A)$ is true but reason $(R)$ is false.
(c) : If $x_1, x_2, x_3, \ldots, x_n$ are $n$ values of a variable $X$, then the arithmetic mean of these values is given by $\frac{x_1+x_2+x_3+\ldots+x_n}{n}$.
$\therefore$ Reason is false.
Now, mean of given values $=x$
[Given]
$\Rightarrow \frac{5+7+x+10+15}{5}=x \quad \Rightarrow \quad \frac{37+x}{5}=x$
$\Rightarrow 37+x=5 x \Rightarrow 37=4 x$
$\Rightarrow x=9.25$
$\therefore \quad$ Assertion is true but Reason is false.
View full question & answer→