Rs Aggarwal 2021 2022 for Class 9 Maths Chapter 18 - Mean, Median And Mode Of Ungrouped Data

Answer:

We know:

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

(i) The first eight natural numbers are 1, 2, 3, 4, 5, 6, 7 and 8.
Mean of these numbers:

$$\begin{gathered} \frac{1+2+3+4+5+6+7+8}{8} \\ =\frac{36}{8} \\ =4.5 \end{gathered}$$

(ii) The first ten odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.
Mean of these numbers:

$$\begin{gathered} \frac{1+3+5+7+9+11+13+15+17+19}{10} \\ =\frac{100}{10} \\ =10 \end{gathered}$$

(iii) The first seven multiples of 5 are 5, 10, 15, 20, 25, 30 and 35.
Mean of these numbers:
 
$$\begin{gathered} \frac{5+10+15+20+25+30+35}{7} \\ =\frac{140}{7} \\ =20 \end{gathered}$$

(iv) The factors of 20 are 1, 2, 4, 5, 10 and 20.
Mean of these numbers:

$$\begin{gathered} \frac{1+2+4+5+10+20}{6} \\ = \frac{42}{6} \\ =7 \end{gathered}$$

(v) The prime numbers between 50 and 80 are 53, 59, 61, 67, 71, 73 and 79.
Mean of these numbers:
 
$$\begin{gathered} \frac{53+59+61+67+71+73+79}{7} \\ =\frac{463}{7} \\ =66.14 \end{gathered}$$

Answer:

Numbers of children in 10 families = 2, 4, 3, 4, 2, 0, 3, 5, 1 and 6.
Thus, we have:

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$
 
$= \frac{2+4+3+4+2+0+3+5+1+6}{10}$
$$\begin{gathered} =\frac{30}{10} \\ =3 \end{gathered}$$

Answer:

Numbers of books issued in the school library: 105, 216, 322, 167, 273, 405 and 346
Thus, we have:

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$
 
$$\begin{gathered} =\frac{105+216+322+167+273+405+346}{7} \\ = \frac{346}{7} \\ = 262 \end{gathered}$$

Answer:

Daily minimum temperatures = 35.5, 30.8, 27.3, 32.1, 23.8 and 29.9
Thus, we have:

$$\begin{gathered} \mathrm{Mean} \mathrm{temperature}=\frac{35.5 + 30.8+27.3+32.1+23.8+29.9}{6} \\ = \frac{179.4}{6} \\ = 29.9°\mathrm{F} \end{gathered}$$

Answer:

We know that,

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

The first five observations are x, x + 2, x + 4, x + 6 and x + 8.

Mean of these numbers = $\frac{x+\left(x+2\right)+\left(x+4\right)+\left(x+6\right)+\left(x+8\right)}{5}$
$$\begin{gathered} \Rightarrow 13=\frac{5x+20}{5} \\ \Rightarrow 13\times 5=5x+20 \\ \Rightarrow 65=5x+20 \\ \Rightarrow 5x=65-20 \\ \Rightarrow 5x=45 \\ \Rightarrow x=\frac{45}{5} \\ \Rightarrow x=9 \end{gathered}$$

Hence, the value of x is 9.

Now, the last three observations are 13, 15 and 17.
Mean of these observations = $\frac{13+15+17}{3}$
                                             = $\frac{45}{3}$
                                             = 15

Hence, the mean of the last three observations is 15.

Answer:

The individual weights of five boys are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg.
Now,
Let the weight of the sixth boy be x kg.
We know:
$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Also,
Given mean = 48 kg
Thus, we have:

$$\begin{gathered} \Rightarrow 48 = \frac{51+45+49+46+44+x}{6} \\ \Rightarrow 288 = 235+x \\ \Rightarrow x = 53 \end{gathered}$$

Therefore, the sixth boy weighs 53 kg.

Answer:

Let the marks scored by 50 students be x₁, x₂,...x₅₀.
Mean = 39
We know:
 $\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Thus, we have:

$$\begin{gathered} 39 = \frac{x_1+x_2+...+x_{50}}{50} \\ \Rightarrow x \end{gathered}$$
_{$\Rightarrow x_1+x_2+ ....+x_{50}=1950 ......\left(\mathrm{i}\right)$}

Also, a score of 43 was misread as 23.
$$\begin{gathered} \therefore \mathrm{New} \mathrm{Mean} = \frac{\left(x_1+x_2+...+x_{50}\right)-23 +43}{50} \\ = \frac{1950-23 +43}{50} \left[ \mathrm{using} \left(\mathrm{i}\right)\right] \\ = \frac{1970}{50} \\ = 39.4 \end{gathered}$$

Answer:

Let the numbers be x₁, x₂,...x₂₄.

We know:

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Thus, we have:

$35= \frac{x_1+x_2+.........+x_{24}}{24}$

_{$\Rightarrow 840=x_1+x_2+......+x_{24} .......\left(\mathrm{i}\right)$}


After addition, the new numbers become (x₁+3), (x₂+3),...(x₂₄+3).
New mean:

$$\begin{gathered} =\frac{\left(x_1+3\right)+\left(x_2+3\right)+..........+\left(x_{24}+3\right)}{24} \\ =\frac{(x_1+x_2+.........+x_{24}) + \left(24\times 3\right)}{24} \\ = \frac{840+72}{24} [\mathrm{From} (\mathrm{i}\left)\right] \\ = \frac{912}{24} \\ = 38 \end{gathered}$$

Answer:

Let the numbers be x₁, x₂,...x₂₀.
We know:
$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Thus, we have:

 $43 = \frac{x_1+x_2+......+x_{20}}{20}$

_{$\Rightarrow 860=x_1+x_2+......+x_{20} ......\left(\mathrm{i}\right)$}

Numbers after subtraction: (x₁$-$6), (x₂$-$6),...(x₂₀$-$6)

∴ New Mean = $\frac{(x_1-6)+(x_2-6)+........+(x_{20}-6)}{20}$

$$\begin{gathered} =\frac{(x_1+x_2+..........+x_{20})-(20\times 6)}{20} \\ = \frac{860-120}{20} [\mathrm{From} (\mathrm{i}\left)\right] \\ = 37 \end{gathered}$$

Answer:

Let the numbers be x₁, x₂,...x₁₅
We know:
 
$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Thus, we have:

$27 = \frac{x_1+x_2+...........+x_{15}}{15}$

$\Rightarrow x_1+x_2+.........+x_{15}=405.........\left(\mathrm{i}\right)$

After multiplication, the numbers become 4x₁, 4x₂,...4x₁₅
∴ New Mean = $\frac{4x_1+4x_2+......+4x_{15}}{15}$

$$\begin{gathered} =\frac{4(x_1+x_2+.......+x_{15})}{15} \\ = \frac{4\times 405}{15} [\mathrm{From} (\mathrm{i}\left)\right] \\ =\frac{1620}{15} \\ = 108 \end{gathered}$$

Answer:

Let the numbers be x₁, x₂,...x₁₂.
We know:

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Thus, we have:

$40 = \frac{x_1+x_2+....+x_{12}}{12}$


$\Rightarrow x_1+x_2+....+x_{12}=480 .....\left(\mathrm{i}\right)$

After division, the numbers become:

$\frac{x_1}{8} , \frac{x_2}{8} , ........ , \frac{x_{12}}{8}$

$$\begin{gathered} \therefore \mathrm{New} \mathrm{mean} = \frac{\left({\displaystyle \raisebox{1ex}{$x_1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)+\left({\displaystyle \raisebox{1ex}{$x_2$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)+....+\left({\displaystyle \raisebox{1ex}{$x_{12}$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)}{12} \\ = \frac{x_1+x_2+...+x_{12}}{12\times 8} \\ = \frac{x_1+x_2+...+x_{12}}{96} \\ = \frac{480}{96} \left[\mathrm{From} \left(\mathrm{i}\right)\right] \\ =5 \end{gathered}$$

Answer:

Let the numbers be x₁, x_2,...x_20.
We know:
$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Thus, we have:

$18 = \frac{x_1+x_2+.......+x_{20}}{20}$

_{$\Rightarrow x_1+x_2+ ..... +\hspace{0.17em}{x}_{20}=360 .....\left(\mathrm{i}\right)$}


New numbers are:

(x₁ + 3), (x₂ + 3),...(x₁₀ + 3), x₁₁,...x₂₀

New Mean:

$$\begin{gathered} =\frac{\left(x_1+3\right)+\left(x_2+3\right)+........+\left(x_{10}+3\right)+x_{11}+........+x_{20}}{20} \\ =\frac{(x_1+x_2+......+x_{10})+\left(3\times 10\right)+x_{11}+........+x_{20}}{20} \\ = \frac{(x_1+x_2+.......+x_{20})+ 30}{20} \\ = \frac{360+30}{20} \left[ \mathrm{From} \left(\mathrm{i}\right)\right] \\ =19.5 \end{gathered}$$

Answer:

Let the numbers be x₁, x₂,..., x₆.
Mean = 23
We know:
$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

Thus, we have:

$23 = \frac{x_1+x_2+...+x_6}{6}$
$x_1+x_2.......x_6=138$.....................(i)

If one number, say, x₆, is excluded, then we have:

$20 = \frac{x_1+x_2+...+x_5}{5}$
$x_1+x_2......+x_5=100....................\left(\mathrm{ii}\right)$

Using (i) and (ii), we get:

$$\begin{gathered} 138 = x_1+x_2+...+x_5+x_6 \\ \Rightarrow 138 = 100+x_6 ...\left(\mathrm{i}\right) \\ \Rightarrow x_6=38 \end{gathered}$$

Thus, the excluded number is 38

Answer:

We know that,

$\mathrm{Mean} =\frac{\mathrm{Sum} \mathrm{of} \mathrm{observa}\mathrm{tions}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$
Mean of height of 30 boys = $\frac{{\displaystyle \sum _{i=1}^{30}{x}_i}}{30}$
$$\begin{gathered} \Rightarrow 150=\frac{{\displaystyle \sum _{i=1}^{30}}{x}_i}{30} \\ \Rightarrow \sum _{i=1}^{30}{x}_i=150\times 30 \\ \Rightarrow \sum _{i=1}^{30}{x}_i=4500 ...\left(1\right) \end{gathered}$$
It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean.
 $$\begin{gathered} \therefore \mathrm{The} \mathrm{correct} \mathrm{mean}=\frac{{\displaystyle \underset{i=1}{\overset{30}{\sum x_i}}-135+165}}{30} \\ =\frac{{\displaystyle \underset{i=1}{\overset{30}{\sum x_i}}+30}}{30} \\ =\frac{{\displaystyle 4500+30}}{30} \left(\mathrm{from} \left(1\right)\right) \\ =\frac{{\displaystyle 4530}}{30} \\ =151 \end{gathered}$$
Hence, the correct mean is 151.

Answer:

Mean weight of 34 students = 46.5 kg
Sum of the weights of 34 students = $(46.5\times 34) \mathrm{kg}=1581$ kg
Increase in the mean weight when the weight of the teacher is included = 500 g = 0.5 kg
∴ New mean weight = (46.5 + 0.5) kg = 47 kg
Now,
Let the weight of the teacher be x kg.
Thus, we have:

$$\begin{gathered} \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{weights} \mathrm{of} 34 \mathrm{students}+\mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{teacher}}{35}=47 \\ \Rightarrow \frac{1581+x}{35}=47 \\ \Rightarrow 1581+x=1645 \\ \Rightarrow x=64 \end{gathered}$$

Therefore, the weight of the teacher is 64 kg.

Answer:

Mean weight of 36 students = 41 kg
Sum of the weights of 36 students = $\left(41\times 36\right) \mathrm{kg}=1476 \mathrm{kg}$
Decrease in the mean when one of the students left the class = 200 g = 0.2 kg
Mean weight of 35 students = (41 $-$ 0.2) kg = 40.8 kg
Now,
Let the weight of the student who left the class be x kg.

$$\begin{gathered} \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{weights} \mathrm{of} 36 \mathrm{students}-x}{35}=40.8 \\ \Rightarrow \frac{1476-x}{35}=40.8 \\ \Rightarrow 1476-x=1428 \\ \Rightarrow x=48 \end{gathered}$$

Hence, the weight of the student who left the class is 48 kg.

Answer:

Average weight of 39 students = 40 kg
Sum of the weights of 39 students = $\left(40\times 39\right) \mathrm{kg}=1560 \mathrm{kg}$
Decrease in the average when new student is admitted in the class = 200 g = 0.2 kg
∴ New average weight = (40 $-$ 0.2) kg = 39.8 kg
Now,
Let the weight of the new student be x kg.
Thus, we have:

$$\begin{gathered} \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{weights} \mathrm{of} 39 \mathrm{students}+x}{40}=39.8 \\ \Rightarrow \frac{1560+x}{40}=39.8 \\ \Rightarrow 1560+x=1592 \\ \Rightarrow x=32 \end{gathered}$$

Therefore, the weight of the new student is 32 kg.

Answer:

Let the average weight of 10 oarsmen be x kg.
Sum of the weights of 10 oarsmen = 10x kg
∴ New average weight = (x + 1.5) kg
Now, we have:
$$\begin{gathered} \mathrm{New} \mathrm{average} \mathrm{weight}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{weights} \mathrm{of} \mathrm{initial} 10 \mathrm{oarsmen}-58+\mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{man}}{10} \\ \Rightarrow x+1.5=\frac{10x-58+\mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{man}}{10} \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{man} +10x-58=10x+15 \\ \Rightarrow \mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{man} -58=15 \\ \Rightarrow \mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{man} =15+58 \\ =73 \mathrm{kg} \end{gathered}$$

Answer:

Mean of 8 numbers = 35
Sum of 8 numbers = $35\times 8=280$
Let the excluded number be x.
Now,
New mean = 35 $-$ 3 = 32
Thus, we have:

$$\begin{gathered} \frac{\mathrm{Sum} \mathrm{of} 8 \mathrm{numbers}-x}{7}=32 \\ \Rightarrow \frac{280-x}{7}=32 \\ \Rightarrow 280-x=224 \\ \Rightarrow x=56 \end{gathered}$$

Therefore, the excluded number is 56.

Answer:

Mean of 150 items = 60
Sum of 150 items = $(150\times 60)=9000$
New sum = [9000 $-$ (52 + 8) + (152 + 88)] = 9180

Correct mean = $\frac{\mathrm{New} \mathrm{sum}}{\mathrm{Total} \mathrm{items}}=\frac{9180}{150}=61.2$

Therefore, the correct mean is 61.2.

Answer:

Mean of 31 results = 60
Sum of 31 results = $31\times 60=1860$
Mean of the first 16 results = 58
Sum of the first 16 results = $58\times 16=928$
Mean of the last 16 results = 62
Sum of the last 16 results = $62\times 16=992$
Value of the 16th result = (Sum of the first 16 results + Sum of the last 16 results) $-$ Sum of 31 results
= (928 + 992) $-$ 1860
= 1920 $-$ 1860
= 60

Answer:

Mean of 11 numbers = 42
Sum of 11 numbers = 42$\times$11 = 462
Mean of the first 6 numbers = 37
Sum of the first 6 numbers = 37$\times$6 = 222
Mean of the last 6 numbers = 46
Sum of the last 6 numbers = 46$\times$6 = 276
∴ 6th number = [(Sum of the first 6 numbers + Sum of the last 6 numbers) $-$ Sum of 11 numbers]
= [(222 + 276) $-$ 462]
= [498 $-$ 462]
= 36
Hence, the 6th number is 36.

Answer:

Mean weight of 25 students = 52 kg
Sum of the weights of 25 students = (52$\times$25) kg = 1300 kg
Mean weight of the first 13 students = 48 kg
Sum of the weights of the first 13 students = (48$\times$13) kg = 624 kg
Mean weight of the last 13 students = 55 kg
Sum of the weights of the last 13 students = (55$\times$13) kg =715 kg
Weight of the 13th student = (Sum of the weights of the first 13 students + Sum of the weights of the last 13 students) $-$ Sum of the weights of 25 students
                                        = [(624+715)$-$1300] kg
                                        = 39 kg
Therefore, the weight of the 13th student is 39 kg.

Answer:

Mean score of 25 observations = 80
Sum of the scores of 25 observations = 80$\times$25 = 2000
Mean score of another 55 observations = 60
Sum of the scores of another 55 observations = 60$\times$55 = 3300

$\mathrm{Mean} \mathrm{score} \mathrm{of} \mathrm{the} \mathrm{whole} \mathrm{set} \mathrm{of} \mathrm{observations}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{scores} \mathrm{of} 25 \mathrm{observations}+\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{scores} \mathrm{of} \mathrm{another} 55 \mathrm{observations}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{observations}}$
     
                                                       $$\begin{gathered} =\frac{2000+3300}{80} \\ =\frac{5300}{80} \\ =66.25 \end{gathered}$$

Therefore, the mean score of the whole set of observations is 66.25.

Answer:

Marks scored by Arun in English = 36
Marks scored by Arun in Hindi = 44
Marks scored by Arun in mathematics = 75
Marks scored by Arun in science = x
Average marks = 50
Thus, we have:
$$\begin{gathered} \mathrm{Average} \mathrm{marks}=\frac{36+44+75+x}{4} \\ \Rightarrow 50=\frac{155+x}{4} \end{gathered}$$
$$\begin{gathered} \Rightarrow 155+x=200 \\ \Rightarrow x=200-155=45 \end{gathered}$$

∴ Marks scored by Arun in science = 45

Answer:

Let the distance from the starting point to the island be x km.
Speed of the ship sailing out to the island = 15 km/h
Speed of the ship sailing back to the starting point = 10 km/h
We know:
$$\begin{gathered} \mathrm{Time} =\frac{\mathrm{Distance} }{\mathrm{Speed}} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{ship} \mathrm{to} \mathrm{travel} \mathrm{from} \mathrm{the} \mathrm{starting} \mathrm{point} \mathrm{to} \mathrm{the} \mathrm{island} =\frac{x}{15} \mathrm{h} \\ \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{ship} \mathrm{to} \mathrm{travel} \mathrm{from} \mathrm{the} \mathrm{island} \mathrm{to} \mathrm{the} \mathrm{starting} \mathrm{point}=\frac{x}{10} \mathrm{h} \\ \mathrm{Average} \mathrm{speed}=\frac{\mathrm{Total} \mathrm{distance} \mathrm{travelled}}{\mathrm{Total} \mathrm{time} \mathrm{taken}} \\ =\frac{x+x}{{\displaystyle \frac{x}{15}}+{\displaystyle \frac{x}{10}}} \\ =\frac{2x}{{\displaystyle \frac{2x+3x}{30}}} \\ =\frac{2x}{{\displaystyle \frac{5x}{30}}} \\ =\frac{60}{5} \\ =12 \mathrm{km}/\mathrm{h} \end{gathered}$$

Therefore, the average speed of the ship in the whole journey was 12 km/h.

Answer:

Total students in the class = 50
Number of boys = 40
∴ Number of girls = (50 $-$ 40) = 10
Average weight of students in the class = 44 kg
Average weight of girls in the class = 40 kg
Sum of the weights of girls in the class = (40 $\times$10) kg = 400 kg
Thus, we have:

$$\begin{gathered} \mathrm{Average} \mathrm{weight} \mathrm{of} \mathrm{students} \mathrm{in} \mathrm{the} \mathrm{class}=\frac{\mathrm{Total} \mathrm{weight} \mathrm{of} \mathrm{girls} \mathrm{in} \mathrm{the} \mathrm{class}+\mathrm{Total} \mathrm{weight} \mathrm{of} \mathrm{boys} \mathrm{in} \mathrm{the} \mathrm{class}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students} \mathrm{in} \mathrm{the} \mathrm{class}} \\ \Rightarrow 44=\frac{400+\mathrm{Total} \mathrm{weight} \mathrm{of} \mathrm{boys} \mathrm{in} \mathrm{the} \mathrm{class}}{50} \end{gathered}$$
$$\begin{gathered} \Rightarrow \mathrm{Total} \mathrm{weight} \mathrm{of} \mathrm{boys}+400 =2200 \\ \Rightarrow \mathrm{Total} \mathrm{weight} \mathrm{of} \mathrm{boys} =1800 \mathrm{kg} \end{gathered}$$
$\mathrm{Average} \mathrm{weight} \mathrm{of} \mathrm{boys} =\frac{1800}{40}=45 \mathrm{kg}$

Answer:

The aggregate yearly expenditure of a family = Rs (18720 × 3) + Rs (20340 × 4) + Rs (21708 × 5)
                                                                            = Rs (56160 + 81360 + 108540)
                                                                            = Rs 246060

The total savings during the year = Rs 35340

The average yearly income = Rs 246060 + Rs 35340
                                            = Rs 281400

∴ The average monthly income of the family = $\frac{281400}{12} =\mathrm{Rs} 23450$

Hence, The average monthly income of the family is Rs 23450.

Answer:

Total salary of 75 workers = ₹ 5680 × 75
                                           = ₹ 426000

Total salary of 25 workers = ₹ 5400 × 25
                                           = ₹ 135000

Total salary of 30 workers = ₹ 5700 × 30
                                           = ₹ 171000

No. of remaining workers = 75 − (25 +30)
                                          = 20

Total salary of 20 workers = ₹ (426000 − 135000 − 171000)
                                           = ₹ 120000

∴ The mean weekly payment of the 20 workers = $\frac{120000}{20}$
                                                                             = ₹ 6000

Hence, the mean weekly payment of the remaining workers is ₹ 6000.

Answer:

Let the number of girls be x and the number of boys be y.

Mean marks of boys = $\frac{\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{boys}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{boys}}$
$$\begin{gathered} \Rightarrow 70=\frac{\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{boys}}{y} \\ \Rightarrow 70y=\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{boys} ...\left(1\right) \end{gathered}$$

Mean marks of girls = $\frac{\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{girls}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{girls}}$
$$\begin{gathered} \Rightarrow 73=\frac{\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{girls}}{x} \\ \Rightarrow 73x=\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{girls} ...\left(2\right) \end{gathered}$$

Mean marks of all the students = $\frac{\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{all}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}$
$$\begin{gathered} \Rightarrow 71=\frac{\mathrm{sum} \mathrm{of} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{boys} \mathrm{and} \mathrm{girls}}{x+y} \\ \Rightarrow 71\left(x+y\right)=70y+73x \left(\mathrm{from} \left(1\right) \mathrm{and} \left(2\right)\right) \\ \Rightarrow 71x+71y=70y+73x \\ \Rightarrow 71y-70y=73x-71x \\ \Rightarrow y=2x \\ \Rightarrow \frac{y}{x}=\frac{2}{1} \end{gathered}$$
Hence, the ratio of the number of boys to the number of girls is 2:1.

Answer:

Average monthly salary of 20 workers = Rs 45900
Sum of the monthly salaries of 20 workers = $\mathrm{Rs} \left(45900\times 20\right)=\mathrm{Rs} 918000$
By adding the manager's monthly salary, we get:
Average salary = Rs 49200
Now,
Let the manager's monthly salary be Rs x.
Thus, we have:

$$\begin{gathered} \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{monthly} \mathrm{salaries} \mathrm{of} 20 \mathrm{workers}+x}{21}=49200 \\ \Rightarrow \frac{918000+x}{21}=49200 \\ \Rightarrow 918000+x=1033200 \\ \Rightarrow x=115200 \end{gathered}$$

Therefore, the manager's monthly salary is Rs 115200.

Answer:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Answer:

We will make the following table:
 

Thus, we have:

$\mathrm{Mean} = \frac{{\displaystyle \underset{}{\overset{}{\sum f_i{x}_i}}}}{{\displaystyle \sum _{}^{}}{x}_i}$
= $\frac{771}{12}= 64.25 \mathrm{kg}$

Answer:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Answer:

We will make the following table:
 

Answer:

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Answer:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Answer:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Answer:

We will make the following table:
 

Answer:

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Answer:

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Answer:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Answer:

We will prepare the following table:
 

Answer:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Answer:

We know that,
Mean = $\frac{{\displaystyle \sum _{}{x}_i{f}_i}}{\sum _{}{f}_i}$

For the following data:

Answer:

(i) Arranging the numbers in ascending order, we get:
      2, 2, 3, 5, 7, 9, 9, 10, 11
Here, n is 9, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{9+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 5\mathrm{th} \mathrm{observation} \\ =7 \end{gathered}$$

(ii) Arranging the numbers in ascending order, we get:
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, n is 9, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{9+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 5\mathrm{th} \mathrm{observation} \\ =16 \end{gathered}$$

(iii) Arranging the numbers in ascending order, we get:
    6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, n is 11, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{11+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 6\mathrm{th} \mathrm{observation} \\ =16 \end{gathered}$$

(iv) Arranging the numbers in ascending order, we get:
   0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, n is 13, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{13+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 7\mathrm{th} \mathrm{observation} \\ =4 \end{gathered}$$

Answer:

(i) Arranging the numbers in ascending order, we get:
9, 10, 17, 19, 21, 22, 32, 35
Here, n is 8, which is an even number.
If n is an even number, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
Now,
$$\begin{gathered} \mathrm{Median} =\mathrm{Mean} \mathrm{of} \left(\frac{8}{2}\right)\mathrm{th} \& \left(\frac{8}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} \mathrm{the} 4\mathrm{th} \& 5\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(19+21\right) \\ = 20 \end{gathered}$$

(ii) Arranging the numbers in ascending order, we get:
   29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here, n is 10, which is an even number.
If n is an even number, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
Now,
$$\begin{gathered} \mathrm{Median} =\mathrm{Mean} \mathrm{of} \left(\frac{10}{2}\right)\mathrm{th} \& \left(\frac{10}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} \mathrm{the} 5\mathrm{th} \& 6\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(60+63\right) \\ =61.5 \end{gathered}$$

(iii) Arranging the numbers in ascending order, we get:
    3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here, n is 12, which is an even number.
If n is an even number, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
Now,
$$\begin{gathered} \mathrm{Median} =\mathrm{Mean} \mathrm{of} \left(\frac{12}{2}\right)\mathrm{th} \& \left(\frac{12}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} \mathrm{the} 6\mathrm{th} \& 7\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(15+17\right) \\ =16 \end{gathered}$$

Answer:

Arranging the marks of 15 students in ascending order, we get:
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Here, n is 15, which is an odd number.
We know:
 $\mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Thus, we have:
$$\begin{gathered} \mathrm{Median} \mathrm{score}=\mathrm{Value} \mathrm{of} \left(\frac{15+1}{2}\right) \mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 8\mathrm{th} \mathrm{observation} \\ =23 \end{gathered}$$

Answer:

Arranging the given data in ascending order:
144, 145, 147, 148, 149, 150, 152, 155, 160

Number of terms = 9 (odd)
$$\begin{gathered} \therefore \mathrm{Median}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ ={\left(\frac{9+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ =5^{\mathrm{th}} \mathrm{term} \\ =149 \end{gathered}$$

Hence, the median height is 149.

Answer:

Arranging the weights (in kg) in ascending order, we have:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here, n is 8, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
$$\begin{gathered} \mathrm{Median} \mathrm{weight}=\mathrm{Mean} \mathrm{of} \left(\frac{8}{2}\right)\mathrm{th} \& \left(\frac{8}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} 4\mathrm{th} \& 5\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(13.4+14.3\right) \\ =13.85 \end{gathered}$$

Hence, the median weight is 13.85 kg.

Answer:

Arranging the ages (in years) in ascending order, we have:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, n is 10, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
$$\begin{gathered} \mathrm{Median} \mathrm{age} =\mathrm{Mean} \mathrm{of} \left(\frac{10}{2}\right)\mathrm{th} \& \left(\frac{10}{2}+1\right)\mathrm{th} \mathrm{observations} \\ = \mathrm{Mean} \mathrm{of} 5\mathrm{th} \& 6\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(40+44\right) \\ =42 \\ \mathrm{Hence}, \mathrm{the} \mathrm{median} \mathrm{age} \mathrm{is} 42 \mathrm{years}. \end{gathered}$$