RS AGGARWAL CLASS 9 CHAPTER 18 MEAN, MEDIAN AND MODE OF UNGROUPED DATA EXERCISE 18A

 EXERCISE 18A

PAGE NO-669

Question 1:

Find the mean of:
(i) the first eight natural numbers
(ii) the first ten odd numbers
(iii) the first seven multiples of 5
(iv) all the factors of 20
(v) all prime numbers between 50 and 80.

Answer 1:

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:
 
$$\begin{gathered} \frac{53+59+61+67+71+73+79}{7} \\ =\frac{463}{7} \\ =66.14 \end{gathered}$$

Question 2:

The number of children in 10 families of a locality are
2, 4, 3, 4, 2, 0, 3, 5, 1, 6.
Find the mean number of children per family.

Answer 2:

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

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

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Question 3:

The following are the number of books issued in a school library during a week:
105, 216, 322, 167, 273, 405 and 346.
Find the average number of books issued per day.

Answer 3:

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

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

Question 4:

The daily minimum temperature recorded (in degree F) at a place during a week was as under:

Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
35.5	30.8	27.3	32.1	23.8	29.9

Find the mean temperature.

Answer 4:

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

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

Question 5:

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 13, find the value of x and hence find the mean of the last three observations.

Answer 5:

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.

Question 6:

The mean weight of 6 boys in a group is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. Find the weight of the sixth boy.

Answer 6:

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.

Question 7:

The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.

Answer 7:

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

Question 8:

The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?

Answer 8:

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:

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

Question 9:

The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean?

Answer 9:

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

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

Question 10:

The mean of 15 numbers is 27. If each number is multiplied by 4, what will be the mean of the new numbers?

Answer 10:

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 = $\frac{4x_1+4x_2+......+4x_{15}}{15}$

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

Question 11:

The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?

Answer 11:

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:

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

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

Question 12:

The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of the new set of 20 numbers.

Answer 12:

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:

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

Question 13:

The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.

Answer 13:

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

Question 14:

The average height of 30 boys was calculated to be 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.

Answer 14:

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.

Question 15:

The mean weight of a class of 34 students is 46.5 kg. If the weight of the teacher is included, the mean is rises by 500 g. Find the weight of the teacher.

Answer 15:

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.

Question 16:

The mean weight of a class of 36 students is 41 kg. If one of the students leaves the class then the mean is decreased by  200 g. Find the weight of the student who left.

Answer 16:

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.

PAGE NO-671

Question 17:

The average weight of a class of 39 students is 40 kg. When a new student is admitted to the class, the average decreases by 200 g. Find the weight of the new student.

Answer 17:

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.

Question 18:

The average weight of 10 oarsmen in a boat is increased by 1.5 kg when one of the crew who weighs 58 kg is replaced by a new man. Find the weight of the new man.

Answer 18:

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:
$$\begin{gathered} \mathrm{New} \mathrm{average} \mathrm{weight}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{weights} \mathrm{of} \mathrm{initial} 10 \mathrm{oarsmen}-58+\mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{man}}{10} \\ \Rightarrow x+1.5=\frac{10x-58+\mathrm{Weight} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{man}}{10} \end{gathered}$$

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

Question 19:

The mean of 8 numbers is 35. If a number is excluded then the mean is reduced by 3. Find the excluded number.

Answer 19:

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.

Question 20:

The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct men.

Answer 20:

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.

Question 21:

The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result.

Answer 21:

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

Question 22:

The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46, find the 6th number.

Answer 22:

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.

Question 23:

The mean weight of 25 students of a class is 52 kg. If the mean weight of the first 13 students of the class is 48 kg and that of the last 13 students is 55 kg, find the weight of the 13th student.

Answer 23:

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.

Question 24:

The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations.

Answer 24:

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.

Question 25:

Arun scored 36 marks in English, 44 marks in Hindi, 75 marks in mathematics and x marks in science. If he has secured an average of 50 marks, find the value of x.

Answer 25:

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

Question 26:

A ship sails out to an island at the rate of 15 km/h and sails back to the starting point at 10 km/h. Find the average sailing speed for the whole journey.

Answer 26:

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.

Question 27:

There are 50 students in a class, of which 40 are boys. The average weight of the class is 44 kg an that of the girls is 40 kg. Find the average weight of the boys.

Answer 27:

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:

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

Question 28:

The aggregate monthly expenditure of a family was ₹ 18720 during the first 3 months, ₹ 20340 during the next 4 months and ₹ 21708 during the last 5 months of a year. If the total savings during the year be ₹ 35340 find the average monthly income of the family.

Answer 28:

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.

Question 29:

The average weekly payment to 75 workers in a factory is ₹ 5680. The mean weekly payment to 25 of them is ₹ 5400 and that of 30 others is ₹ 5700. Find the mean weekly payment of the remaining workers.

Answer 29:

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.

Question 30:

The mean marks (out of 100) of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.

Answer 30:

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.

PAGE NO-672

Question 31:

The average monthly salary of 20 workers in an office is ₹ 45900. If the manager's salary is added, the average salary becomes ₹ 49200 per month. What's manager's monthly salary?

Answer 31:

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.