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ICSE Class 10 Maths Selina Solutions Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles and Mode)

ICSE Class 10 Maths Selina Solutions Chapter 24 covers key concepts, formulas, and problem types to help students strengthen understanding and problem-solving skills. By revising solved examples and practice questions, students can improve accuracy and confidence for the Mathematics exam on 2 March 2026.
authorImageAnanya Gupta17 Feb, 2026
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ICSE Class 10 Maths Selina Solutions Chapter 24

ICSE Class 10 Maths Selina Solutions Chapter 24 is an essential resource for students preparing for the Mathematics exam on 2 March 2026, especially as the ICSE Class 10 board exams have started from 17 February 2026. This chapter provides detailed step-by-step solutions, worked examples, and a variety of practice problems to help students understand key concepts clearly.

By revising Chapter 24 thoroughly, students can strengthen their problem-solving skills, learn how to apply formulas accurately, and tackle both simple and complex questions with confidence. The solutions are explained in a simple and easy-to-follow language, making last-minute revision effective and helping students maximise their marks in the board exam.

Regular practice with these solutions also helps students identify recurring question patterns, improve speed and accuracy, and build the confidence needed to perform well under exam conditions.

ICSE Class 10 Maths Selina Solutions Chapter 24 Exercises

Below we have provided the ICSE Class 10 Maths Selina Solutions for Chapter 24 Measures of Central Tendency (Mean, Median, Quartiles, and Mode) to assist students. These solutions are designed to help students grasp the concepts more effectively and prepare better for their exams.

ICSE Class 10 Maths Selina Solutions Chapter 24 Exercise 24(A) Page No: 356

1. Find the mean of the following set of numbers:

(i) 6, 9, 11, 12 and 7

(ii) 11, 14, 23, 26, 10, 12, 18 and 6

Solution:

(i) By definition, we know Mean = ∑x/ n Here, n = 5 Thus, Mean = (6 + 9 + 11 + 12 + 7)/ 5 = 45/5 = 9 (ii) By definition, we know Mean = ∑x/ n Here, n = 8 Thus, Mean = (11 + 14 + 23 + 26 + 10 + 12 + 18 + 6)/ 8 = 120/8 = 15

2. Marks obtained (in mathematics) by 9 student are given below:

60, 67, 52, 76, 50, 51, 74, 45 and 56

(a) find the arithmetic mean

(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.

Solution:

(a) Mean = ∑x/ n Here, n = 9 Thus, Mean = (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/ 9 = 531/9 = 59 (b) If the marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

3. Find the mean of the natural numbers from 3 to 12.

Solution:

The numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Here n = 10 Mean = ∑x/ n = (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/ 10 = 75/10 = 7.5

4. (a) Find the mean of 7, 11, 6, 5, and 6

(b) If each number given in (a) is diminished by 2, find the new value of mean.

Solution:

(a) Mean = ∑x/ n , here n = 5 = (7 + 11 + 6 + 5 + 6)/ 5 = 35/5 = 7 (b) If 2 is subtracted from each number, then the mean will he changed as 7 – 2 = 5

5. If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’

Solution:

Given, No. of terms (n) = 5 Mean = 8 Sum of all terms = 8 x 5 = 40 …… (i) But, sum of numbers = 6 + 4 + 7 + a + 10 = 27 + a ….. (ii) On equating (i) and (ii), we get 27 + a = 40 Thus, a = 13

6. The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.

Solution:

Given, No. of terms (n) = 5 and mean = 8 So, the sum of all terms = 5 x 8 = 40 ……. (i) but sum of numbers = 6 + y + 7 + x + 14 = 27 + y + x ……. (ii) On equating (i) and (ii), we get 27 + y + x = 40 x + y = 13 Hence, y = 13 – x

7. The ages of 40 students are given in the following table:

Age( in yrs) 12 13 14 15 16 17 18
Frequency 2 4 6 9 8 7 4

Find the arithmetic mean.

Solution:

Age in yrs x i Frequency (f i ) f i x i
12 2 24
13 4 52
14 6 84
15 9 135
16 8 128
17 7 119
18 4 72
Total 40 614
Mean = ∑f i x i / ∑f i = 614/40 = 15.35
Exercise 24(B) Page No: 361

1. The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.

Age – Years 16 – 18 18 – 20 20 – 22 22- 24 24-26
No. of Students 2 7 21 17 3

Solution:

Age in years C.I. x i Number of students (f i ) x i f i
16 – 18 17 2 34
18 – 20 19 7 133
20 – 22 21 21 441
22 – 24 23 17 391
24 – 26 25 3 75
Total   50 1074
Mean = ∑f i x i / ∑f i = 1074/50 = 21.48

2. The following table gives the weekly wages of workers in a factory.

Weekly Wages (Rs) No. of Workers
50-55 5
55-60 20
60-65 10
65-70 10
70-75 9
75-80 6
80-85 12
85-90 8

Calculate the mean by using:

(i) Direct Method

(ii) Short – Cut Method

Solution:

(i) Direct Method
Weekly Wages (Rs) Mid-Value x i No. of Workers (f i ) f i x i
50-55 52.5 5 262.5
55-60 57.5 20 1150.0
60-65 62.5 10 625.0
65-70 67.5 10 675.0
70-75 72.5 9 652.5
75-80 77.5 6 465.0
80-85 82.5 12 990.0
85-90 87.5 8 700.0
Total   80 5520.00
Mean = ∑f i x i / ∑f i = 5520/80 = 69 (ii) Short – cut method
Weekly wages (Rs) No. of workers (f i ) Mid-value x i A = 72.5 d i = x – A f i d i
50-55 5 52.5 -20 -100
55-60 20 57.5 -15 -300
60-65 10 62.5 -10 -100
65-70 10 67.5 -5 -50
70-75 9 A = 72.5 0 0
75-80 6 77.5 5 30
80-85 12 82.5 10 120
85-90 8 87.5 15 120
Total 80     -280
Here, A = 72.5 Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(B) - 1

3. The following are the marks obtained by 70 boys in a class test:

Marks No. of boys
30 – 40 10
40 – 50 12
50 – 60 14
60 – 70 12
70 – 80 9
80 – 90 7
90 – 100 6

Calculate the mean by:

(i) Short – cut method

(ii) Step – deviation method

Solution:

(i) Short – cut method
Marks No. of boys (f i ) Mid-value x i A = 65 d i = x – A f i d i
30 – 40 10 35 -30 -300
40 – 50 12 45 -20 -240
50 – 60 14 55 -10 -140
60 – 70 12 A = 65 0 0
70 – 80 9 75 10 90
80 – 90 7 85 20 140
90 – 100 6 95 30 180
Total 70     -270
Here, A = 65 Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(B) - 2 (ii) Step – deviation method
Marks No. of boys (f i ) Mid-value x i A = 65 u i = (x i – A)/ h f i u i
30 – 40 10 35 -3 -30
40 – 50 12 45 -2 -24
50 – 60 14 55 -1 -14
60 – 70 12 A = 65 0 0
70 – 80 9 75 1 9
80 – 90 7 85 2 14
90 – 100 6 95 3 18
Total 70     -27
Here, A = 65 and h = 10 Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(B) - 3

4. Find mean by step – deviation method:

C. I. 63-70 70-77 77-84 84-91 91-98 98-105 105-112
Freq 9 13 27 38 32 16 15

Solution:

C. I. Frequency (f i ) Mid-value x i A = 87.50 u i = (x i – A)/ h f i u i
63 – 70 9 66.50 -3 -27
70 – 77 13 73.50 -2 -26
77 – 84 27 80.50 -1 -27
84 – 91 38 A = 87.50 0 0
91 – 98 32 94.50 1 32
98 – 105 16 101.50 2 32
105 – 112 15 108.50 3 45
Total 150     29
Here, A = 87.50 and h = 7 Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(B) - 4

5. The mean of the following frequency distribution is Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(B) - 5 . Find the value of ‘f’.

C. I. 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
freq 8 22 31 f 2

Solution:

Given, Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(B) - 6
C. I. frequency Mid-value (x i ) f i x i
0-10 8 5 40
10-20 22 15 330
20-30 31 25 775
30-40 f 35 35f
40-50 2 45 90
Total 63 + f   1235 + 35f
Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(B) - 7 9324 + 148f = 8645 + 245f 245f – 148f = 9324 – 8645 f = 679/97 Thus, f = 7

ICSE Class 10 Maths Selina Solutions Chapter 24 Exercise 24(C) Page No: 372

1. A student got the following marks in 9 questions of a question paper.

3, 5, 7, 3, 8, 0, 1, 4 and 6.

Find the median of these marks.

Solution:

Arranging the given data in descending order: 8, 7, 6, 5, 4, 3, 3, 1, 0 Clearly, the middle term is 4 which is the 5 th term. Hence, median = 4

2. The weights (in kg) of 10 students of a class are given below:

21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.

Find the median of their weights.

Solution:

Arranging the given data in descending order: 28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5 It’s seen that, The middle terms are 24 and 24, 5 th and 6 th terms Thus, Median = (24 + 24)/ 2 = 48/2 = 24

3. The marks obtained by 19 students of a class are given below:

27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find:

(i) median (ii) lower quartile

(iii) upper quartile (iv) interquartile range

Solution:

Arranging in ascending order: 22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37 (i) The middle term is 10 th term i.e. 29 Hence, median = 29 (ii) Lower quartile Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(C) - 1 (iii) Upper quartile = Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(C) - 2 (iv) Interquartile range = q 3 – q 1 =35 – 26 = 9

4. From the following data, find:

(i) Median

(ii) Upper quartile

(iii) Inter-quartile range

25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83

Solution:

Arranging the given data in ascending order, we have: 0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95 (i) Median is the mean of 8 th and 9 th term Thus, median = (40 + 45)/ 2 = 85/2 = 42.5 (ii) Upper quartile = Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(C) - 3 (iii) Interquartile range is given by, q 1 = 16 th /4 term = 18; q 3 = 65 Interquartile range = q 3 – q 1 Thus, q 3 – q 1 = 65 – 18 = 47

5. The ages of 37 students in a class are given in the following table:

Age (in years) 11 12 13 14 15 16
Frequency 2 4 6 10 8 7

Find the median.

Solution:

Age (in years) Frequency Cumulative Frequency
11 2 2
12 4 6
13 6 12
14 10 22
15 8 30
16 7 37
Number of terms (n) = 37 Median = Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(C) - 4 And, the 19 th term is 14 Therefore, the median = 14

ICSE Class 10 Maths Selina Solutions Chapter 24 Exercise 24(D) Page No: 374

1. Find the mode of the following data:

(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6

(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8

Solution:

(i) It’s seen that 7 occurs 4 times in the given data. Hence, mode = 7 (ii) Mode = 11 As 11 occurs 4 times in the given data.

2. The following table shows the frequency distribution of heights of 50 boys:

Height (cm) 120 121 122 123 124
Frequency 5 8 18 10 9

Find the mode of heights.

Solution:

Clearly, Mode is 122 cm because it has occurred the maximum number of times. i.e. frequency is 18.

3. Find the mode of following data, using a histogram:

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 12 20 9 4

Solution:

Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(D) - 1 Clearly, Mode is in 20-30, because in this class there are 20 frequencies.

4. The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:

Expenditure

(Rs)

No. of

students

20-25 4
25-30 7
30-35 23
35-40 18
40-45 6
45-50 2

Solution:

Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(D) - 2 Clearly, Mode is in 30-35 because it has the maximum frequency.

ICSE Class 10 Maths Selina Solutions Chapter 24 Exercise 24(E) Page No: 375

1. The following distribution represents the height of 160 students of a school.

Height (in cm) No. of Students
140 – 145 12
145 – 150 20
150 – 155 30
155 – 160 38
160 – 165 24
165 – 170 16
170 – 175 12
175 – 180 8

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:

i. The median height.

ii. The interquartile range.

iii. The number of students whose height is above 172 cm.

Solution:

Height (in cm) No. of Students Cumulative frequency
140 – 145 12 12
145 – 150 20 32
150 – 155 30 62
155 – 160 38 100
160 – 165 24 124
165 – 170 16 140
170 – 175 12 152
175 – 180 8 160
  N = 160  
Now, let’s draw an ogive taking height of student along x-axis and cumulative frequency along y-axis. Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(E) - 1 (i) So, Median = 160/2 = 80 th term Through mark for 80, draw a parallel line to x-axis which meets the curve; then from the curve draw a vertical line which meets the x-axis at the mark of 157.5. (ii) As, the number of terms = 160 Lower quartile (Q 1 ) = (160/4) = 40 th term = 152 Upper quartile (Q 3 ) = (3 x 160/4) = 120 th term = 164 Inner Quartile range = Q 3 – Q 1 = 164 – 152 = 12 (iii) Through mark for 172 on x-axis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the y-axis at the mark of 145. Now, The number of students whose height is above 172 cm = 160 – 144 = 16

2. Draw ogive for the data given below and from the graph determine: (i) the median marks.

(ii) the number of students who obtained more than 75% marks.

Marks 10 – 19 20 -29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99
No. of students 14 16 22 26 18 11 6 4 3

Solution:

Marks No. of students Cumulative frequency
9.5 – 19.5 14 14
19.5 – 29.5 16 30
29.5 – 39.5 22 52
39.5 – 49.5 26 78
49.5 – 59.5 18 96
59.5 – 69.5 11 107
69.5 – 79.5 6 113
79.5 – 89.5 4 117
89.5 – 99.5 3 120
Scale: 1cm = 10 marks on X axis 1cm = 20 students on Y axis Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(E) - 2 (i) So, the median = 120/ 2 = 60 th term Through mark 60, draw a parallel line to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B. The value of point B is the median = 43 (ii) Total marks = 100 75% of total marks = 75/100 x 100 = 75 marks Hence, the number of students getting more than 75% marks = 120 – 111 = 9 students.

3. The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m – 1 and median q. Find p and q.

Solution:

Mean of 1, 7, 5, 3, 4 and 4 = (1 + 7 + 5 + 3 + 4 + 4)/ 6 = 24/6 = 4 So, m = 4 Now, given that The mean of 3, 2, 4, 2, 3, 3 and p = m -1 = 4 – 1 = 3 Thus, 17 + p = 3 x n …. ,where n = 7 17 + p = 21 p = 4 Arranging the terms in ascending order, we have: 2, 2, 3, 3, 3, 3, 4, 4 Mean = 4 th term = 3 Hence, q = 3

4. In a malaria epidemic, the number of cases diagnosed were as follows:

Date (July) 1 2 3 4 5 6 7 8 9 10 11 12
Number 5 12 20 27 46 30 31 18 11 5 0 1

On what days do the mode and upper and lower quartiles occur?

Solution:

Date Number C.f.
1 5 5
2 12 17
3 20 37
4 27 64
5 46 110
6 30 140
7 31 171
8 18 189
9 11 200
10 5 205
11 0 205
12 1 206
(i) Mode = 5 th July as it has maximum frequencies. (ii) Total number of terms = 206 Upper quartile = 206 x (3/4) = 154.5 th = 7 th July Lower quartile = 206 x (1/4) = 51.5 th = 4 th July

5. The income of the parents of 100 students in a class in a certain university are tabulated below.

Income (in thousand Rs) 0 – 8 8 – 16 16 – 24 24 – 32 32 – 40
No. of students 8 35 35 14 8

(i) Draw a cumulative frequency curve to estimate the median income.

(ii) If 15% of the students are given freeships on the basis of the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.

(iii) Calculate the Arithmetic mean.

Solution:

Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(E) - 3 (i) Cumulative Frequency Curve
Income (in thousand Rs.) No. of students f Cumulative Frequency Class mark x fx
0 – 8 8 8 4 32
8 – 16 35 43 12 420
16 – 24 35 78 20 700
24 – 32 14 92 28 392
32 – 40 8 100 36 288
  ∑f = 100     ∑ fx = 1832
We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows: Here, N = 100 N/2 = 50 At y = 50, affix A. Through A, draw a horizontal line meeting the curve at B. Through B, a vertical line is drawn which meets OX at M. OM = 17.6 units Hence, median income = 17.6 thousands (ii) 15% of 100 students = (15 x 100)/ 100 = 15 From c.f. 15, draw a horizontal line which intersects the curve at P. From P, draw a perpendicular to x – axis meeting it at Q which is equal to 9.6 Thus, freeship will be awarded to students provided annual income of their parents is upto 9.6 thousands. (ii) Mean = ∑ fx/ ∑ f = 1832/100 = 18.32

6. The marks of 20 students in a test were as follows:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.

Calculate:

(i) the mean (ii) the median (iii) the mode

Solution:

Arranging the terms in ascending order: 2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20 Number of terms = 20 ∑ x = 2 + 6 + 8 + 9 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20 = 257 (i) Mean = ∑x/ ∑n = 257/ 20 = 12.85 (ii) Median = (10 th term + 11 th term)/ 2 = (13 + 14)/ 2 = 27/ 2 = 13.5 (iii) Mode = 15 since it has maximum frequencies i.e. 3

7. The marks obtained by 120 students in a mathematics test is given below:

Marks No. of students
0-10 5
10-20 9
20-30 16
30-40 22
40-50 26
50-60 18
60-70 11
70-80 6
80-90 4
90-100 3

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:

(i) the median

(ii) the number of students who obtained more than 75% in test.

(iii) the number of students who did not pass in the test if the pass percentage was 40.

(iv) the lower quartile

Solution:

Marks No. of students c.f.
0-10 5 5
10-20 9 14
20-30 16 30
30-40 22 52
40-50 26 78
50-60 18 96
60-70 11 107
70-80 6 113
80-90 4 117
90-100 3 120
Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(E) - 4 (i)  Median = (120 + 1)/ 2 = 60.5 th term Through mark 60.5, draw a parallel line to x-axis which meets the curve at A. From A draw a perpendicular to x-axis meeting it at B. Then, the value of point B is the median = 43 (ii) Number of students who obtained up to 75% marks in the test = 110 Number of students who obtained more than 75% marks in the test = 120 – 110 = 10 (iii) Number of students who obtained less than 40% marks in the test = 52 (from the graph; x = 40, y = 52) (iv) Lower quartile = Q 1 = 120 x (1/4) = 30 th term = 30

8. Using a graph paper, draw an ogive for the following distribution which shows a record of the width in kilograms of 200 students.

Weight Frequency
40 – 45 5
45 – 50 17
50 – 55 22
55 – 60 45
60 – 65 51
65 – 70 31
70 – 75 20
75 – 80 9

Use your ogive to estimate the following:

(i) The percentage of students weighing 55 kg or more

(ii) The weight above which the heaviest 30% of the student fall

(iii) The number of students who are (a) underweight (b) overweight, if 55.70 kg is considered as standard weight.

Solution:

Weight Frequency c. f.
40-45 5 5
45-50 17 22
50-55 22 44
55-60 45 89
60-65 51 140
65-70 31 171
70-75 20 191
75-80 9 200
Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(E) - 5 (i) The number of students weighing more than 55 kg = 200 – 44 = 156 Thus, the percentage of students weighing 55 kg or more = (156/200) x 100 = 78 % (ii) 30% of students = (30 x 200)/ 100 = 60 Heaviest 60students in weight = 9 + 21 + 30 = 60 Weight = 65 kg (From table) (iii) (a) underweight students when 55.70 kg is standard = 46 (approx.) from graph (b) overweight students when 55.70 kg is standard = 200 – 55.70 = 154 (approx.) from graph

9. The distribution, given below, shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Marks obtained 5 6 7 8 9 10
No. of students 3 9 6 4 2 1

Solution:

Marks obtained(x) No. of students (f) c.f. fx
5 3 3 15
6 9 12 54
7 6 18 42
8 4 22 32
9 2 24 18
10 1 25 10
Total 25   171
Number of terms = 25 (i) Mean = 171/25 = 6.84 (ii) Median = (25 + 1)/ 2 th = 13 th term = 7 (iii) Mode = 6 since it has the maximum frequency i.e. 6

10. The mean of the following distribution is 52 and the frequency of class interval 30 – 40 is ‘f’. Find f.

Class Interval 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
Frequency 5 3 f 7 2 6 13

Solution:

C.I. Frequency(f) Mid value (x) fx
10-20 5 15 75
20-30 3 25 75
30-40 f 35 35f
40-50 7 45 315
50-60 2 55 110
60-70 6 65 390
70-80 13 75 975
Total 36 + f   1940 + 35f
Mean = ∑ fx/ ∑ f = (1940 + 35f)/ (36 + f) …… (i) But, given mean = 52 …. (ii) From (i) and (ii), we have (1940 + 35f)/ (36 + f) = 52 1940 + 35f = 1872 + 52f 17f = 68 Thus, f = 4

11. The monthly income of a group of 320 employees in a company is given below:

Monthly Income (thousands) No. of employees
6 – 7 20
7 – 8 45
8 – 9 65
9 – 10 95
10 – 11 60
11 – 12 30
12 – 13 5

Draw an ogive of the given distribution on a graph paper taking 2 cm = Rs 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine:

(i) the median wage.

(ii) number of employees whose income is below Rs 8500.

(iii) if salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.

(iv) the upper quartile.

Solution:

Monthly Income (thousands) No. of employees (f) Cumulative frequency
6-7 20 20
7-8 45 65
8-9 65 130
9-10 95 225
10-11 60 285
11-12 30 315
12-13 5 320
Total 320  
Number of employees = 320 Selina Solutions Concise Class 10 Maths Chapter 24 ex. 24(E) - 6 (i) Median = 320/2 = 160 th term Through mark 160, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B. The value of point B is the median = Rs 9.3 thousands (ii) The number of employees with income below Rs 8,500 = 95 (approx from the graph) (iii) Number of employees with income below Rs 11,500 = 305 (approx from the graph) Thus, the number of employees (senior employees) = 320 – 305 = 15 (iv) Upper quartile = Q 3 = 320 x (3/4) = 240 th term = 10.3 thousands = Rs 10,300

ICSE Class 10 Maths Selina Solutions Chapter 24 PDF

The PDF link is available below for ICSE Class 10 Maths Selina Solutions Chapter 24. These solutions are designed to help students grasp the concepts clearly and perform well in their exams. Access the PDF to enhance your understanding and improve your performance in your studies.

ICSE Class 10 Maths Selina Solutions Chapter 24 PDF

 Last Days Preparation Strategy for Chapter 24 – ICSE Class 10 Maths

As the ICSE Class 10 Board Exams have started from 17 February 2026 and the Mathematics exam is scheduled for 2 March 2026, revising Chapter 24 – Selina Solutions effectively in the last few days can boost confidence and help secure high marks. Here’s a focused strategy:

Revise Important Topics

Focus on all the key concepts and problem types in Chapter 24, including formulas, examples, and frequently asked questions. Prioritize high-weightage topics to ensure maximum scoring potential in the board exam.

Make Quick Notes

Create concise notes or flashcards of important formulas, theorems, and step-by-step methods for solving different types of problems. These will help in fast last-minute revision and quick recall during the exam.

Practice Problem Sets

Solve previous years’ questions, sample papers, and exercise problems from the chapter. Pay special attention to recurring question patterns and tricky sums to improve speed and accuracy.

Understand Concepts Clearly

Instead of memorizing steps blindly, focus on understanding the underlying concepts and logic behind each type of problem. Concept clarity ensures you can solve application-based and higher-order questions confidently.

ICSE Class 10 Maths Selina Solutions Chapter 24 FAQs

What is the mean?

The mean is the average of a set of numbers, calculated by adding all the numbers together and dividing by the total count of numbers.

How do you calculate the median?

The median is the middle value in a list of numbers ordered from smallest to largest. If the list has an odd number of values, the median is the middle one. If it has an even number of values, the median is the average of the two middle numbers.

What are quartiles?

Quartiles divide a ranked dataset into four equal parts. The first quartile (Q1) is the median of the lower half of the data, the second quartile (Q2) is the median of the dataset, and the third quartile (Q3) is the median of the upper half of the data.

How is the mode defined?

The mode is the number that appears most frequently in a dataset. A dataset may have one mode, more than one mode, or no mode at all if no number repeats.
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Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
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