RD Sharma solution class 7 chapter 23 Data Handling II(Central values) Exercise 23.1

Exercise 23.1

Page-23.6

Question 1:

Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days. How many hours does he study daily on an average?

Answer 1:

Average number of  study hours =  ( 4  + 5 + 3) $÷$ 3
                                                 =  12 $÷$ 3

                                                 = 4 hours          
           Thus, Ashish studies for 4 hours on an average.

Question 2:

A cricketer scores the following runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100. Find the mean score.

Answer 2:

We have:
The mean score = $\frac{(58 + 76 +40 +35 +48 +45 + 0+ 100)}{8} =\frac{402}{8} = 50.25$ runs.

Thus, the mean score of the cricketer is 50.25 runs.

Question 3:

The marks (out of 100) obtained by a group of students in science test are 85, 76, 90, 84, 39, 48, 56, 95, 81 and 75. Find the
(i) highest and the lowest marks obtained by the students.
(ii) range of marks obtained.
(iii) mean marks obtained by the group.

Answer 3:

In order to find the highest and lowest marks, let us arrange the marks in ascending order as follows:
39, 48, 56, 75, 76, 81, 84, 85, 90, 95
(i) Clearly, the highest mark is 95 and the lowest is 39.
(ii) The range of the marks obtained is: ( 95 - 39) = 56.
(iii) We have:
      Mean marks = Sum of the marks ​$÷$ Total number of students

⇒ Mean marks = $\frac{(39 +48 + 56 +75 +76 +81 +84 +85 +90 +95)}{10} = \frac{729}{10} = 72.9.$

Hence, the mean marks of the students is 72.9.

Question 4:

The enrolment of a school during six consecutive years was as follows:
1555, 1670, 1750, 2019, 2540, 2820
Find the mean enrollment of the school for this period.

Answer 4:

The mean enrolment = Sum of the enrolments in each year $÷$ Total number of years

The mean enrolment = $\frac{(1555 +1670 +1750 +2019 +2540 +2820) }{6} = \frac{12354}{6} = 2059$.

Thus, the mean enrolment of the school for the given period is 2059.

Question 5:

The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun
Rainfall (in mm)	0.0	12.2	2.1	0.0	20.5	5.3	1.0

(i) Find the range of the rainfall from the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.

Answer 5:

(i) The range of the rainfall = Maximum rainfall - Minimum rainfall

                                    =    20.5  - 0.0

                                    =    20.5 mm .            

(ii) The mean rainfall = $\frac{(0.0+12.2 +2.1 +0.0+20.5 + 5.3 +1.0)}{7} = \frac{41.1}{7} = 5.87 \mathrm{mm}$.

(iii) Clearly, there are 5 days (Mon, Wed, Thu, Sat, and Sun), when the rainfall was less than the mean, i.e., 5.87 mm.

Question 6:

If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.

Answer 6:

The mean height = Sum of the heights $÷$ Total number of persons

                        = $\frac{(140 + 150 +152 +158 +161)}{5} = \frac{761}{5}= 152.2 \mathrm{cm}$

Thus, the mean height of 5 persons is 152.2 cm.

Page-23.7

Question 7:

Find the mean of 994, 996, 998, 1002 and 1000.

Answer 7:

Mean = Sum of the observations $÷$ Total number of observations

 Mean  = $\frac{(994 + 996 + 998 + 1002 +1000)}{5} = \frac{4990}{5} = 998$.

Question 8:

Find the mean of first five natural numbers.

Answer 8:

The first five natural numbers are 1, 2, 3, 4 and 5. Let $\overline{X}$ denote their arithmetic mean. Then,

$\overline{X}$ = $\frac{1+2+3+4+5}{5} = \frac{15}{5} = 3$.

Question 9:

Find the mean of all factors of 10.

Answer 9:

The factors of 10 are 1, 2, 5 and 10 itself. Let ​$\overline{X}$ denote their arithmetic mean. Then,

$\overline{X}$ = $\frac{1+2+5 +10}{4} = \frac{18}{4} = 4.5$.

Question 10:

Find the mean of first 10 even natural numbers.

Answer 10:

The first 10 even natural numbers are 2,4, 6, 8,10,12,14,16,18 and 20. Let ​$\overline{X}$ denote their arithmetic mean. Then,

$\overline{X}$ = $\frac{2+4+6+8+10+12+14+16+ 18+20}{10} = \frac{110}{10} = 11$.

Question 11:

Find the mean of x, x + 2, x + 4, x + 6, x + 8.

Answer 11:

Mean =  $\frac{\mathrm{Sum} \mathrm{of} \mathrm{o}\mathrm{bservations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

$$\begin{gathered} \Rightarrow \mathrm{Mean} = \frac{x +x+2 +x+4 +x+6 +x+8}{5} \\ \Rightarrow \mathrm{Mean}= \frac{5x + 20}{5} = \frac{5(x+4)}{5} \\ \Rightarrow \mathrm{Mean}= x+4. \end{gathered}$$

Question 12:

Find the mean of first five multiples of 3.

Answer 12:

The first five multiples of 3 are 3,6,9,12 and 15. Let $\overline{X}$ denote their arithmetic mean. Then,

$\overline{X}$ = $\frac{3+6+9+12+15}{5} = \frac{45}{5} = 9$.

Question 13:

Following are the weights (in kg) of 10 new born babies in a hospital on a particular day:
3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean $\overline{X}$.

Answer 13:

We have:
 $$\begin{gathered} \overline{X} = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}} \\ \Rightarrow \overline{X} = \frac{3.4 +3.6 +4.2 +4.5 + 3.9 +4.1 +3.8 +4.5 +4.4 +3.6}{10} \\ \Rightarrow \overline{X} = \frac{40}{10} = 4 \mathrm{kg}. \end{gathered}$$

Question 14:

The percentage of marks obtained by students of a class in mathematics are:
64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.

Answer 14:

We have :
 $$\begin{gathered} \mathrm{Mean} = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{marks} \mathrm{obtained} \mathrm{by} \mathrm{students}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}.} \\ \Rightarrow \mathrm{Mean} =\frac{64 +36 +47 +23+ 19+81+93 +72 +35 +3+1}{12} \\ \Rightarrow \mathrm{Mean} =\frac{474}{12} = 39.5 \%. \end{gathered}$$

Question 15:

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

Answer 15:

The mean number of children per family = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{children}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{families}}$
 

Mean = $\frac{2+4+3+4+2+3+5+1+1+5}{10} = \frac{30}{10} = 3$.

Thus, on an average there are 3 children per family in the locality.

Question 16:

The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.

Answer 16:

We have:
 

n = The number of observations = 100,  Mean = 40

 $$\begin{gathered} \mathrm{Mean} = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{observations}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{observations}} \\ \Rightarrow 40 = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{observations}}{100} \\ \Rightarrow 40 \times 100 = \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{observations} \end{gathered}$$

Thus, the incorrect sum of the observations = 40 x 100 = 4000.

Now,

The correct sum of the observations = Incorrect sum of the observations - Incorrect observation + Correct observation

⇒ The correct sum of the observations = 4000 - 83 + 53
⇒ The correct sum of the observations = 4000 - 30 =  3970

∴ $\mathrm{Correct} \mathrm{mean} = \frac{\mathrm{Correct} \mathrm{sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}} = \frac{3970}{100} = 39.7$

Question 17:

The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Answer 17:

We have:

 $$\begin{gathered} \mathrm{Mean} = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{five} \mathrm{numbers}}{5} = 27 \\ \mathrm{So}, \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{five} \mathrm{numbers} = 5 \times 27 = 135. \\ \mathrm{Now}, \\ \mathrm{The} \mathrm{mean} \mathrm{of} \mathrm{four} \mathrm{numbers} = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{four} \mathrm{numbers}}{4} = 25 \\ \mathrm{So}, \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{four} \mathrm{numbers} = 4 \times 25 = 100. \end{gathered}$$

Therefore, the excluded number  = Sum of the five numbers - sum of the four numbers

⇒ The excluded number = 135 - 100 = 35.

Question 18:

The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Answer 18:

We have:

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{weight}s \mathrm{of} the \mathrm{students}}{\mathrm{Number} \mathrm{of} \mathrm{students}}$

Let the weight of the seventh student be x kg.

 $$\begin{gathered} \mathrm{Mean} = \frac{52+54 + 55 +53 +56 +54 +x}{7} = 55 \\ \Rightarrow \frac{324 +x}{7} = 55 \\ \Rightarrow 324 +x = 385 \\ \Rightarrow x = 385 - 324 \\ \Rightarrow x = 61 \mathrm{kg}. \end{gathered}$$
Thus, the weight of the seventh student is 61 kg.

Question 19:

The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean?

Answer 19:

Let x₁, x₂, x₃...x₈ be the eight numbers whose mean is 15 kg. Then,

 $15= \frac{x_1+ x_2+ x_3 + ...+ x_8}{8}$

    x₁ + x₂ + x₃ +...+ x₈ = 15×8 
⇒x₁ + x₂ + x₃ +...+ x₈ = 120.

Let the new numbers be 2x₁ , 2x₂, 2x₃, ...2x_8. Let M be the arithmetic mean of the new numbers.
Then,

​$$\begin{gathered} M= \frac{2x_1+ 2x_2+ 2x_3 + ....+ 2x_8}{8} \\ =>M = \frac{2(x_1+ x_2+ x_3 + ....x_8)}{8}=>M=\frac{2\times 120}{8}=>M=30 \end{gathered}$$ 

Question 20:

The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Answer 20:

Let $x_1,x_2,x_3,x_{4 }\& x_5$ be five numbers whose mean is 18. Then,
18 = Sum of five numbers $÷$ 5
∴ Sum of five numbers = 18 $\times$ 5 = 90.

Now, if one number is excluded, then their mean is 16.
So,
​16= Sum of four numbers ​$÷$ 4
∴ Sum of four numbers = 16 $\times$ 4 = 64.

 The excluded number =  Sum of the five observations - Sum of the four observations

∴ The excluded number = 90 - 64
 ∴ The excluded number = 26.

Question 21:

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Answer 21:

n = Number of observations = 200

$$\begin{gathered} \mathrm{Mean} = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{o}\mathrm{bservations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}} \\ \Rightarrow 50 = \frac{\mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{o}\mathrm{bservations}}{200} \\ \Rightarrow \mathrm{Sum} \mathrm{of} \mathrm{the} \mathrm{observations} = 50 \times 200 = 10,000 \end{gathered}$$.

Thus, the incorrect sum of the observations = 50 x 200
Now,
The correct sum of the observations = Incorrect sum of the observations - Incorrect observations + Correct observations
⇒​Correct sum of the observations =  10,000 - (92+ 8) + (192 + 88)

⇒ Correct sum of the observations = 10,000 - 100 + 280
⇒ Correct sum of the observations = 9900 +280
⇒ Correct sum of the observations = 10180.

∴ Correct Mean = $\frac{\mathrm{Correct} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}} = \frac{10180}{200} = 50.9$

Question 22:

The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.

Answer 22:

We have:
Mean =  Sum of  five numbers $÷$ 5
⇒ Sum of the five numbers = 27 x 5 = 135.

Now, New mean = 25
25 = Sum of six numbers ​$÷$ 6
⇒ Sum of the six numbers = 25 x 6 = 150.

The included number = Sum of the six numbers - Sum of the five numbers
⇒​The included number = 150 - 135
⇒The included number =  15.

Question 23:

The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.

Answer 23:

Let x₁, x₂, x₃, ... x₇₅ be 75 numbers with their mean equal to 35. Then,

$\Rightarrow 35 = \frac{x_1 +x_2 +x_3+... + x_{75}}{75}$

x₁+ x₂ + x₃ +... + x₇₅ = 35 ×75 

⇒ x₁ + x₂ + x₃ +... + x₇₅  = 2625 

The new numbers are 4x₁, 4x₂, 4x₃, ...4x_75. Let M be the arithmetic mean of the new numbers. Then,

$$\begin{gathered} \Rightarrow M = \frac{4x_1 +4x_2 +4x_3+... +4 x_{75}}{75} \\ \Rightarrow M = \frac{4 (x_1 +x_2 +x_3+...+ x_{75})}{75} \\ \Rightarrow M =\frac{4\times 2625}{75}\Rightarrow M =35\times 4\Rightarrow M =140. \end{gathered}$$