Selina Solutions Concise Maths Class 10 Chapter 24 Measures of Central Tendency

The numerical expressions which represent the characteristics of a group (a large collection of numerical data) are called Measures of Central Tendency. In this chapter, students will learn and solve problems based on three types of statistical averages, namely Mean, Median and Mode. Further in the chapter, different methods of finding these central tendencies are studied and tested under the exercise problems. Students can refer to the Selina Solutions for Class 10 Mathematics for clarifying any doubts regarding solving problems. It also helps students for their exam preparations and mainly improves problem-solving skills, which is vital. The solutions of the Selina Solutions for Class 10 Mathematics Chapter 24 Measures of Central Tendency exercises are available in the links given below.

Selina Solutions Concise Maths Class 10 Chapter 24 Measures of Central Tendency Download PDF

Exercises of Concise Selina Solutions Class 10 Maths Chapter 24 Measures of Central Tendency

Exercise 24(A) Solutions

Exercise 24(B) Solutions

Exercise 24(C) Solutions

Exercise 24(D) Solutions

Exercise 24(E) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 24 Measures of Central Tendency

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:

Find the arithmetic mean.

Solution:

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.

Solution:

Mean = ∑f_i x_i/ ∑f_i = 1074/50 = 21.48

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

Calculate the mean by using:

(i) Direct Method

(ii) Short – Cut Method

Solution:

(i) Direct Method

Mean = ∑f_i x_i/ ∑f_i = 5520/80 = 69

(ii) Short – cut method

Here, A = 72.5

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

Calculate the mean by:

(i) Short – cut method

(ii) Step – deviation method

Solution:

(i) Short – cut method

Here, A = 65

(ii) Step – deviation method

Here, A = 65 and h = 10

4. Find mean by step – deviation method:

Solution:

Here, A = 87.50 and h = 7

5. The mean of the following frequency distribution is. Find the value of ‘f’.

Solution:

Given,

9324 + 148f = 8645 + 245f

245f – 148f = 9324 – 8645

f = 679/97

Thus, f = 7

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

(iii) Upper quartile =

(iv) Interquartile range = q₃ – q₁ =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 =

(iii) Interquartile range is given by,

q₁ = 16^th/4 term = 18; q₃ = 65

Interquartile range = q₃ – q₁

Thus,

q₃ – q₁ = 65 – 18 = 47

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

Find the median.

Solution:

Number of terms (n) = 37

Median =

And, the 19^th term is 14

Therefore, the median = 14

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:

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:

Solution:

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:

Solution:

Clearly,

Mode is in 30-35 because it has the maximum frequency.

Exercise 24(E) Page No: 375

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

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:

Now, let’s draw an ogive taking height of student along x-axis and cumulative frequency along y-axis.

(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 (Q₁) = (160/4) = 40^th term = 152

Upper quartile (Q₃) = (3 x 160/4) = 120^th term = 164

Inner Quartile range = Q₃ – Q₁

= 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.

Solution:

Scale:

1cm = 10 marks on X axis

1cm = 20 students on Y axis

(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:

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

Solution:

(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.

 (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:

(i) Cumulative Frequency Curve

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:

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:

(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 = Q₁ = 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.

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:

(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.

Solution:

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.

Solution:

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:

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:

Number of employees = 320

(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 = Q₃ = 320 x (3/4) = 240^th term = 10.3 thousands = Rs 10,300

The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2020-21 will be updated soon.