Mathematics Class XI

Question 1 of 25

1. Question
1 point(s)
If the mean of the distribution is $ 2.6 $, then the value of $ y $ is________
- $ 24 $
- $ 13 $
- $ 8 $
- $ 3 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Mean}\ (\overline{x})&=\dfrac{\sum_{i=1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}\\\Longrightarrow \dfrac{1\times4+2\times5+3\times y+4\times1+5\times2}{4+5+y+1+2}&=2.6\\\Longrightarrow \frac{28+3y}{12+y}&=2.6\\\Longrightarrow 31.2+2.6y&=28+3y\\\Longrightarrow 0.4y&=3.2\\\Longrightarrow y&=\dfrac{3.2}{0.4}=8 \end{aligned}$
Question 2 of 25

2. Question
1 point(s)
The harmonic mean of $ 4,8,16 $ is______
- $ 6.42 $
- $ 6.71 $
- $ 6.85 $
- $ 6.73 $
Correct

Incorrect

Hint

$\begin{aligned} HM&=\dfrac{3}{\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}}=\dfrac{3}{\dfrac{4+2+1}{16}}\\&=\dfrac{48}{7}=6.85 \end{aligned}$
Question 3 of 25

3. Question
1 point(s)
Mean of $ 100 $ items is $ 49 $. It was discovered that three items which should been $ 60,70,80 $ were wrongly readed as $ 40,20,50 $ respectively. Then, the correct mean is_______
- $ 50 $
- $ 52 $
- $ 58 $
- $ 45 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Sum of}\ 100\ \text{Items}&=49\times100=4900\\\text{Sum of items added}&=60+70+80=210\\\text{Sum of items replaced}&=40+20+50=110\\\text{New sum}&=4900+210-110=5000\\\text{Correct mean}&=\dfrac{5000}{100}=50 \end{aligned}$
Question 4 of 25

4. Question
1 point(s)
A school has four sections of Chemistry in class XI having $ 40,35,45 $ and $ 42 $ students. The mean marks obtained in chemistry test are $ 50,60,55 $ and $ 45 $ respectively for the four sections, the over all average of marks per students is______
- $ 53 $
- $ 45 $
- $ 55.3 $
- $ 52.25 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Total students}&=40+35+45+42=162\\\text{Total marks obtained}&=(40\times 50)+(35\times 60)+(45\times55)+(42\times45)=8465\\\text{Average}&=\dfrac{8465}{162}=52.25 \end{aligned}$
Question 5 of 25

5. Question
1 point(s)
The mean of $ 5 $ numbers is $ 18 $. If one number is excluded, their mean becomes $ 16 $. Then the excluded number is _______
- $ 18 $
- $ 25 $
- $ 26 $
- $ 30 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Sum of numbers}&=18\times 5=90\\\text{After one number excluded, sum of the rest numbers}&=16\times4=64\\\therefore\text{Excluded number}&=90-64=26 \end{aligned}$
Question 6 of 25

6. Question
1 point(s)
The class marks of a distribution are $ 6,10,14,18,22,26,30 $, then the class size is_______
- $ 4 $
- $ 5 $
- $ 6 $
- $ 7 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Class length}\ &(i)\ 10-6=4\\&(ii)\ 14-10=4\\&(iii)\ 18-14=4\\&(iv)\ 22-18=4\\&(v)\ 26-22=4\\&(vi)\ 30-26=4\\\therefore \text{Class Size}&=4 \end{aligned}$
Question 7 of 25

7. Question
1 point(s)
For a frequently distribution $ 7^{th} $ decide is computed by ______
- $ l+\dfrac{\left( \dfrac{n}{7}-c \right)}{f}\times i $
- $ l+\dfrac{\left( \dfrac{N}{10}-c \right)}{f}\times i $
- $ l+\dfrac{\left( \dfrac{7N}{10}-c \right)}{f}\times i $
- $ l+\dfrac{\left( \dfrac{10N}{7}-c \right)}{f}\times i $
Correct

Incorrect

Hint

$D_7=l+\dfrac{\left( \dfrac{7N}{10}-C \right)}{f}\times i$
Question 8 of 25

8. Question
1 point(s)
Which of the following, in case if a discrete data, is not equal to the median_______
- $ 50^{th} $ percentile
- $ 5^{th} $ decile
- $ 2^{nd} $ quartile
- Lower Quartile
Correct

Incorrect

Hint

Lower quartile is not equal to the median.
Question 9 of 25

9. Question
1 point(s)
The mean of $ 10,14,11,9,8,12,6 $ is_______
- $ 10 $
- $ 12 $
- $ 14 $
- $ 11 $
Correct

Incorrect

Hint

Arranging in ascending order that is $6,8,9,10,11,12,14$ Total term $=7$ Median $=\left( \dfrac{7+2}{2} \right)^{th}=4^{th}$ term $=10$
Question 10 of 25

10. Question
1 point(s)
The relation between the median $ M $, the $ 2^{nd} $ quartile $ Q_2 $, the $ 5^{th} $ decile $ D_5 $ and the $ {50}^{th} $ percentile $ P_{50} $, of a set of observations is _______
- $ M=Q_2=D_5=P_{50} $
- $ M $ < $ Q_2 $ <$ D_5 $ <$P_{50} $
- $M $ > $ Q_2 $ >$ D_5 $ >$P_{50} $
- None of these
Correct

Incorrect

Hint

$M=Q_2=D_5=P_{50}$ , as their coressponding observation
Question 11 of 25

11. Question
1 point(s)
For a symmetrical distribution, $ Q_1=27 $ and $ Q_3=43 $, the median is______
- $ 28 $
- $ 25 $
- $ 35 $
- $ 32 $
Correct

Incorrect

Hint

$Q_2(\text{Median})=\dfrac{Q_1+Q_3}{2}=\dfrac{27+43}{2}=\dfrac{70}{2}=35$
Question 12 of 25

12. Question
1 point(s)
A set of numbers consists of three $ 4’s $, five $ 5’s $, six $ 6’s $, eight $ 8’s $ and seven $ 10’s $. The mode of this set of numbers is______
- $ 6 $
- $ 7 $
- $ 8 $
- $ 10 $
Correct

Incorrect

Hint

Mode $=8$ , as it has maximum frequencuy.
Question 13 of 25

13. Question
1 point(s)
The mode of the following items is $ 0,1,6,7,2,3,7,6,6,2,6,0,5,6,0 $______
- $ 0 $
- $ 6 $
- $ 5 $
- $ 2 $
Correct

Incorrect

Hint

Since $6$ occurs most number of times that is $5$ times. $\therefore$ Mode $=6$
Question 14 of 25

14. Question
1 point(s)
For a continous series mode is ______
- $ l+\left( \dfrac{f_1}{f_m-f_1-f_2} \right)\times i $
- $ l+\left( \dfrac{f_m-f_1}{f_m-f_1-f_2} \right)\times i $
- $ l+\left( \dfrac{f_m-f_1}{2f_m-f_1-f_2} \right)\times i $
- $ l+\left( \dfrac{2f_m-f_1}{f_m-f_1-f_2} \right)\times i $
Correct

Incorrect

Hint

For a continous series, Mode $=l+\left( \dfrac{f_m-f_i}{2f_m-f_i-f_2} \right)\times i$
Question 15 of 25

15. Question
1 point(s)
The mode of the distribution is_________
- $ 5 $
- $ 6 $
- $ 7 $
- $ 9 $
Correct

Incorrect

Hint

Mode $=6$ Since, $6$ has maximum number of frequencies that is $10$
Question 16 of 25

16. Question
1 point(s)
In a moderately asymmetrical distribution the mode and mean are $ 7 \text{ and } 4 $ respectively. Then, the median is______
- $ 4 $
- $ 5 $
- $ 6 $
- $ 7 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Mode}&=3\text{Median}-2\text{Mean}\\\Longrightarrow 7&=3x-2\times4\\\Longrightarrow 3x&=7+8=15\\\Longrightarrow x&=\dfrac{15}{3}=5\\\therefore\text{Median}&=5 \end{aligned}$
Question 17 of 25

17. Question
1 point(s)
The total expenditure incurred by an industry under different heads is best presented as a ________
- Bar Diagram
- Pie Diagram
- Histogram
- Frequency Polygon
Correct

Incorrect

Hint

Pie diagram is the best feature to incur the total expenditure under different heads.
Question 18 of 25

18. Question
1 point(s)
The expenditure of a family for a certain month were as follows:
Food-:$ RS.560 $,
Rent-:$ RS.420 $,
Clothes-:$ RS.180 $,
Education-:$ RS.160 $,
Others-:$ RS.120 $
A pie chart representing this data would show the expenditure for clothes by a sector whose angle equals ______
- $ 180^o $
- $ 90^o $
- $ 45^o $
- $ 64^o $
Correct

Incorrect

Hint

Required angle $=\dfrac{180\times360}{1440}=45^o$
Question 19 of 25

19. Question
1 point(s)
In a frequently distribution, the mean and median are $ 21 $ and $ 22 $ respectively, then its mode is________
- $ 25.5 $
- $ 24 $
- $ 22 $
- $ 20.5 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Mode}&=3\text{Median}-2\text{Mean}\\&=(3\times 22)-(2\times 21)\\&=66-42=24 \end{aligned}$
Question 20 of 25

20. Question
1 point(s)
Mean of $ 100 $ observations is $ 45 $. It was later found that two observations $ 19 $ and $ 31 $ were incorrectly recorded as $ 91 $ and $ 13 $. The correct mean is______
- $ 44.12 $
- $ 44.46 $
- $ 45.01 $
- $ 45.54 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Sum of}\ 100\ \text{Items}&=45\times100=4500\\\text{Sum of items added}&=19+31=50\\\text{Sum of itemsreplaced}&=91+13=104\\\text{New sum}&=4500+50-104=4446\\\text{New mean}&=\dfrac{4446}{100}=44.46 \end{aligned}$
Question 21 of 25

21. Question
1 point(s)
The Pie chart is to be drawn for respecting the following data

The value of central angle for food and clothing would be______
- $ 90^o $
- $ 28^o $
- $ 150^o $
- $ 144^o $
Correct

Incorrect

Hint

$\begin{aligned} \text{Required angle}&=\dfrac{400}{1000}\times360^o\\&=144^o \end{aligned}$
Question 22 of 25

22. Question
1 point(s)
The standard deviation of $ 5 $ scores $ 1,2,3,4,5 $ is______
- $ \dfrac{2}{5} $
- $ \dfrac{3}{5} $
- $ \sqrt{2} $
- $ \sqrt{3} $
Correct

Incorrect

Hint

$\begin{aligned} \overline{x}&=\dfrac{1+2+3+4+5}{5}=3\\SD(\sigma)&=\sqrt{\dfrac{1}{n}\Sigma x_i^2-(\overline{x})^2}\\&=\sqrt{\dfrac{1}{5}(1+4+9+16+25)-9}\\&=\sqrt{11-9}=\sqrt{2} \end{aligned}$
Question 23 of 25

23. Question
1 point(s)
The varience of the data $ 2,4,6,8,10 $ is_______
- $ 6 $
- $ 7 $
- $ 9 $
- $ 8 $
Correct

Incorrect

Hint

$\begin{aligned} \overline{x}&=\dfrac{2+4+6+8+10}{5}=6\\\sigma^2&=\dfrac{1}{n}\Sigma(x_i-\overline{x})^2\\&=\dfrac{1}{5}\{ (2-6)^2+(4-6)^2+(6-6)^2+(6-8)^2+(10-6)^2 \}\\&=\dfrac{1}{5}\{ 16+4+0+4+16 \}\\&=\dfrac{40}{5}=8 \end{aligned}$
Question 24 of 25

24. Question
1 point(s)
Quartile deviation for a frequency distribution is_______
- $ Q=Q_3-Q_1 $
- $ Q=\dfrac{1}{2}(Q_3-Q_1) $
- $ Q=\dfrac{1}{3}(Q_3-Q_1) $
- $ Q=\dfrac{1}{4}(Q_3-Q_1) $
Correct

Incorrect

Hint

For a quartile deviation $Q=\dfrac{1}{2}(Q_3-Q_1)$
Question 25 of 25

25. Question
1 point(s)
The range of the following set of observations $ 2,3,5,9,8,7,6,5,7 $ is_______
- $ 11 $
- $ 7 $
- $ 5.5 $
- $ 6 $
Correct

Incorrect

Hint

$\begin{aligned} \text{Range}&=X_{\text{max}}-X_{\text{min}}\\&=9-2=7 \end{aligned}$