Problem Set 7 : Textbook Page 126

Write the correct alternative answer for each of the following questions:

(Click on the question to view the answer)

(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

(B) 35

(D) 30

(B) 10 – 20

(A) \(\displaystyle \frac {\overline{x} + \overline{y}}{2}\)

The mean of 5 numbers is 50.
∴ The sum of 5 numbers
= 5 × 50
= 250 ... (i)

Also, the mean of four numbers is 46.
∴ The sum of 4 numbers
= 4 × 46
= 184 ... (ii)

∴ The fifth number
= 250 − 184 ... [From (i) and (ii)]
= 66 ... (iii)

∴ The correct option is: (D) 66.

The 9th observation is replaced by 70 instead of 30.
∴ The total will increase by
70 − 30 = 40

∴ The mean will increase by

\(\displaystyle \frac {40}{100} = 0.4\)

∴ New mean = 40 + 0.4 = 40.4

∴ The correct option is: (B) 40.4.

(A) 15

Let’s arrange the given data in ascending order:
5, 7, 7, 9, 10, 10
Here, N = 6, i.e. Even number
∴ Median is the average of middle two scores.
∴ Median is the average of 7 and 9.

∴ Median = \(\displaystyle \frac {7 + 9}{2}\)

∴ Median = 8

∴ The correct option is: (C) 8.

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Frequency	7	3	12	13	2

7 + 3 + 12 + 13 = 35

∴ The correct option is: (C) 35.

2. The mean salary of 20 workers is ₹ 10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.

Solution:

The mean salary of 20 workers is ₹ 10,250.
∴ The sum of their salary
= 20 × 10250
= 205000 ... (i)

If the salary of the office superintendent is added, then the mean increases by ₹ 750/-
∴ The new mean
= ₹ 10250 + ₹ 750
= ₹ 11000

But, this is the mean for 21 persons.
∴ The sum of the salary of 21 persons
= 21 × 11000
= ₹ 231000 ... (ii)

∴ The salary of the office superitendent
= 231000 − 205000 ... [From (i) and (ii)]
= 26000

∴ The salary of the office superitendent is ₹ 26,000/-.

3. The mean of nine numbers is 77. If one more number is added to it then the mean increases by 5. Find the number added in the data.

Solution:

The mean of nine numbers is 77.
∴ The sum of nine numbers
= 9 × 77
= 693 ... (i)

Now, if one more number is added, the mean increases by 5.
∴ The mean becomes
= 77 + 5
= 82

But, this is the mean for 10 numbers.
∴ The sum of 10 numbers
= 10 × 82
= 820 ... (ii)

∴ The number added in the data
= 820 − 693 ... [From (i) and (ii)]
= 127 ... (iii)

∴ The number added in the data is 127.

4. The monthly maximum temperature of a city is given in degree Celsius in the following data.

29.2 29.0 28.1 28.5 32.9 29.2 34.2 36.8 32.0 31.0

30.5 30.0 33 32.5 35.5 34.0 32.9 31.5 30.3 31.4

30.3 34.7 35.0 32.5 33.5 29.0 29.5 29.9 33.2 30.2

By taking suitable classes, prepare the grouped frequency distribution table.

Solution:

First, let’s find out the lowest and the highest scores.

The lowest score is 28.1 and the highest score is 36.8.

Let’s take the classes as 28 – 30, 30 – 32, 32 – 34, 34 – 36, 36 – 38 and prepare a grouped frequency distribution table.

Class (Temp °C)	Tally Marks	Frequency (f)
28 – 30	\(\enclose{downdiagonalstrike}{\|\|\|\|}{\|\|\|}\)	8
30 – 32	\(\enclose{downdiagonalstrike}{\|\|\|\|}{\|\|\|}\)	8
32 – 34	\(\enclose{downdiagonalstrike}{\|\|\|\|}{\|\|\|}\)	8
34 – 36	\(\enclose{downdiagonalstrike}{\|\|\|\|}\)	5
36 – 38	\(\|\)	1
\(\displaystyle \text N = \sum{f} = 30\)

From the table, answer the following questions:

For how many days the maximum temperature was less than 34°C?

For how many days the maximum temperature was 34°C or more than 34°C?

Solution:

8 + 8 + 8 = 24 days
5 + 1 = 6 days

5. If the mean of the following data is 20.2, then find the value of p.

x_i 10 15 20 25 30

f_i 6 8 p 10 6

Solution:

Let’s prepare a table as shown:

x_i	f_i	f_ix_i
10	6	60
15	8	120
20	p	20p
25	10	250
30	6	180
	\(\displaystyle \sum{{f}_i} = 30 + p\)	\(\displaystyle \sum{{f}_i{x}_i} = 610 + 20p\)

Now, \(\displaystyle \overline{x}\) = \(\displaystyle \frac {\sum{{f}_i{x}_i}}{\sum{{f}_i}}\)

∴ \(\displaystyle 20.2 = \displaystyle \frac {610 + 20p}{30 + p}\)

∴ 20.2 × (30 + p) = 610 + 20p

∴ 606 + 20.2p = 610 + 20p

∴ 20.2p − 20p = 610 − 606

∴ 0.2p = 4

∴ p = \(\displaystyle \frac {4}{0.2}\)

∴ p = \(\displaystyle \frac {40}{2}\)

∴ p = 20

∴ The value of p is 20.

6. There are 68 students of 9^th standard from Model High School, Nandpur. They have scored following marks out of 80, in written exam of mathematics.

70 50 60 66 45 46 38 30 40 47 56 68

80 79 39 43 57 61 51 32 42 43 75 43

36 37 61 71 32 40 45 32 36 42 43 55

56 62 66 72 73 78 36 46 47 52 68 78

80 49 59 69 65 35 46 56 57 60 36 37

45 42 70 37 45 66 56 47

By taking classes 30 – 40, 40 – 50, .... prepare a less than type cumulative frequency table.

Solution:

The highest score in the given data is 80.
∵ We are following a convention of excluding the upper class limit, we will have to take classes upto 80 – 90.

The completed table is given below:

Class (Marks)	Tally Marks	Frequency	Cumulative Frequency (Less Than Type)
30 – 40	\(\enclose{downdiagonalstrike}{\|\|\|\|}\enclose{downdiagonalstrike}{\|\|\|\|} \|\|\|\|\)	14	14
40 – 50	\(\enclose{downdiagonalstrike}{\|\|\|\|}\enclose{downdiagonalstrike}{\|\|\|\|}\enclose{downdiagonalstrike}{\|\|\|\|}\enclose{downdiagonalstrike}{\|\|\|\|}\)	20	14 + 20 = 34
50 – 60	\(\enclose{downdiagonalstrike}{\|\|\|\|}\enclose{downdiagonalstrike}{\|\|\|\|} \|\)	11	34 + 11 = 45
60 – 70	\(\enclose{downdiagonalstrike}{\|\|\|\|}\enclose{downdiagonalstrike}{\|\|\|\|} \|\|\)	12	45 + 12 = 57
70 – 80	\(\enclose{downdiagonalstrike}{\|\|\|\|} \|\|\|\|\)	9	57 + 9 = 66
80 – 90	\(\|\|\)	2	66 + 2 = 68
		N = 68

From the table, answer the following questions:

How many students have scored marks less than 80?
Answer: 66 students
How many students have scored marks less than 40?
Answer: 14 students
How many students have scored marks less than 60?
Answer: 45 students

7. By using data in example (6), and taking classes 30 – 40, 40 – 50, ... prepare equal to or more than type cumulative frequency table.

Solution:

The completed table is given below:

Class (Marks)	Frequency	Cumulative Frequency (More Than Type)
30 – 40	14	68
40 – 50	20	68 − 14 = 54
50 – 60	11	54 − 20 = 34
60 – 70	12	34 − 11 = 23
70 – 80	9	23 − 12 = 11
80 – 90	2	11 − 9 = 2
	N = 68

Based on the table above, answer the following questions:

How many students have scored marks 70 or more than 70?
Answer: 11 students
How many students have scored marks 30 or more than 30?
Answer: 68 students

8. There are 10 observations arranged in ascending order as given below:

45 47 50 52 x x + 2 60 62 63 74

The median of these observations is 53. Find the value of x. Also find the mean and the mode of the data.

Solution:

Here, N = 10 (Even number)
∴ Median is the average of the middle two scores.
∴ Median is the average of x and x + 2.

∴ Median = \(\displaystyle \frac {x + x + 2}{2}\)

∴ 53 = \(\displaystyle \frac {2x + 2}{2}\)

∴ 53 × 2 = 2x + 2
106 = 2x + 2
i.e. 2x + 2 = 106
∴ 2x = 106 − 2
∴ 2x = 104

∴ x = \(\displaystyle \frac {104}{2}\)

∴ x = 52 ... (i)
∴ The value of x is 52.

Also, x + 2 = 52 + 2 = 54

Thus, the given observations are:

Now, \(\displaystyle \overline{x}\) = \(\displaystyle \frac {\sum{{x}_i}}{\text N}\)

∴ \(\displaystyle \overline{x}\) = \(\displaystyle \frac {45 + 47 + 50 + 52 + 52 + 54 + 60 + 62 + 63 + 74}{10}\)

∴ \(\displaystyle \overline{x}\) = \(\displaystyle \frac {559}{10}\)

∴ \(\displaystyle \overline{x}\) = 55.9 ... (ii)
∴The mean is 55.9.

Also, the maximum frequency is of the score 52.
∴ The mode is 52. ... (iii)

∴ The value of x is 52, the mean is 55.9 and the mode is 52.

Lesson 9 : Main Page

Standard 9 : Main Page

This page was last modified on
02 March 2026 at 22:30

Write the correct alternative answer for each of the following questions: (Click on the question to view the answer)

What is the upper class limit for the class 25 – 35? 25 35 60 30

What is the class-mark of class 25 – 35? 25 35 60 30

If the classes are 0 – 10, 10 – 20, 20 – 30 ... then in which class should the observation 10 be included? 0 – 10 10 – 20 20 – 30 30 – 40

The mean of five numbers is 80, out of which mean of 4 numbers is 46, find the 5th number. 46 20 434 66

Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean. 40.6 40.4 40.3 40.7

What is the mode of 19, 19, 15, 20, 25, 15, 20, 15? 15 20 19 25

What is the median of 7, 10, 7, 5, 9, 10? 7 9 8 10

From the following table, what is the cumulative frequency of less than type for the class 30 – 40? Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequency 7 3 12 13 2 13 15 35 22

2. The mean salary of 20 workers is ₹ 10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.

Solution:

3. The mean of nine numbers is 77. If one more number is added to it then the mean increases by 5. Find the number added in the data.

Solution:

Solution:

From the table, answer the following questions: For how many days the maximum temperature was less than 34°C? For how many days the maximum temperature was 34°C or more than 34°C?

Solution:

5. If the mean of the following data is 20.2, then find the value of p. xi 10 15 20 25 30 fi 6 8 p 10 6

Solution:

Solution:

From the table, answer the following questions:

7. By using data in example (6), and taking classes 30 – 40, 40 – 50, ... prepare equal to or more than type cumulative frequency table.

Solution:

Based on the table above, answer the following questions:

8. There are 10 observations arranged in ascending order as given below: 45 47 50 52 x x + 2 60 62 63 74 The median of these observations is 53. Find the value of x. Also find the mean and the mode of the data.

Solution:

Write the correct alternative answer for each of the following questions:

(Click on the question to view the answer)

From the table, answer the following questions:

For how many days the maximum temperature was less than 34°C?

For how many days the maximum temperature was 34°C or more than 34°C?

5. If the mean of the following data is 20.2, then find the value of p.

x_i 10 15 20 25 30

f_i 6 8 p 10 6

8. There are 10 observations arranged in ascending order as given below:

45 47 50 52 x x + 2 60 62 63 74

The median of these observations is 53. Find the value of x. Also find the mean and the mode of the data.