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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

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Selina Textbook Solutions Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode).

All Selina textbook questions of Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam. 

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Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 24(B)

Solution 1

Age in years

C.I.

xi

Number of students (fi)

xifi

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

Solution 2

(i) Direct Method

Weekly Wages

(Rs)

Mid-Value

xi

No. of Workers (fi)

fixi

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

(ii) Short - cut method

Weekly wages (Rs)

No. of workers (fi)

Mid-value

xi

A = 72.5

di=x-A

fidi

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

Solution 3

(i) Short - cut method

Marks

No. of boys (fi)

Mid-value xi

A = 65

di=x-A

fidi

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

(ii) Step - deviation method

Marks

No. of boys (fi)

Mid-value xi

A = 65

fiui

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

Solution 4

C. I.

Frequency (fi)

Mid-value xi

A = 87.50

fiui

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

Solution 5

C. I.

frequency

Mid-value (xi)

fixi

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

Solution 6

Let the assumed mean A= 72.5

C.I

fi

Mid value (xi)

di=xi -; A

fidi

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

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

Solution 7

C.I.

Frequency

Mid value x

fx

15-25

10

20

200

25-35

20

30

600

35-45

25

40

1000

45-55

15

50

750

55-65

5

60

300

Total

75

2850

Solution 8

Class

Frequency (f)

Mid Value (x)

fx

0 - 20

7

10

70

20 - 40

p

30

30p

40 - 60

10

50

500

60 - 80

9

70

630

80 - 100

13

90

1170

Total

39 + p

 

2370 + 30p

 

Here mean = 54 ..(ii)

from (i) and (ii)

Solution 9

Class

Freq (f)

Mid value

fx

0-20

5

10

50

20-40

f1

30

30f1

40-60

10

50

500

60-80

f2

70

70f2

80-100

7

90

630

100-120

8

110

880

Total

30+f1+f2

2060+30f1+70f2

Now, and

from (i)

using (i) and (ii)

Solution 10

  

Solution 11

  

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 24(C)

Solution 1

Arranging the given data in descending order:

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

The middle term is 4 which is the 5th term.

Median = 4

Solution 2

Arranging the given data in descending order:

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

The middle terms are 24 and 24, 5th and 6th terms

Solution 3

Arranging in ascending order:

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

(i) Middle term is 10th term i.e. 29

Median = 29

(ii) Lower quartile =

(iii) Upper quartile =

(iv) Interquartile range = q3 - q1 =35 - 26 = 9

Solution 4

Arrange in ascending order:

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

(i) Median is the mean of 8th and 9th term

(ii) Upper quartile =

(iii) Interquartile range =

Solution 5

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 = 37

Median =

Median = 14

Solution 6

Weight

(kg) x

no. of boys

f

cumulative frequency

37

10

10

38

14

24

39

18

42

40

12

54

41

6

60

Number of terms = 60

(i) median = the mean of the 30th and the 31st terms

(ii) lower quartile (Q1) =

(iii) upper quartile (Q3) =

(iv) Interquartile range = Q3 - Q1 = 40 - 38 = 2

Solution 7

Class

Frequency

Cumulative Frequency

0-10

4

4

10-20

9

13

20-30

15

28

30-40

14

42

40-50

8

50

Number of terms = 50

Through mark of 25.5 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.

The value of B is the median which is 28.

Solution 8

Weight (kg)

No. of boys

Cumulative Frequency

10-15

11

11

15-20

25

36

20-25

12

48

25-30

5

53

30-35

2

55

Number of terms = 55

Through mark of 28 on the y-axis, draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis, which meets x-axis at B.

The value of B is the median which is 18.4 kg

Solution 9

Marks

(less than)

Cumulative frequency

10

5

20

24

30

37

40

40

50

42

60

48

70

70

80

77

90

79

100

80

Number of terms = 80

\Median = 40th term.

(i) Median = Through 40th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 40

(ii) Lower quartile (Q1) = 20th term = 18

(iii) Upper Quartile (Q3) = 60th term = 66

Solution 10

Height

(in cm)

No. of

pupils

Cumulative

Frequency

121 - 130

12

12

131 - 140

16

28

141 - 150

30

58

151 - 160

20

78

161 - 170

14

92

171 - 180

8

100

Number of terms = 100

Through 50th term mark draw a line parallel to the x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

Value of B is the median = 148

Median height = 148cm

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 24(D)

Solution 1

(i) Mode = 7

Since 7 occurs 4 times

(ii) Mode = 11

Since it occurs 4 times

Solution 2

Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.

Solution 3

Mode is in 20-30, because in this class there are 20 frequencies.

Solution 4

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

Solution 5

which is 5.

Mode = 5 because it occurs maximum number of times.

Solution 6

Arranging the given data in ascending order:

7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19

(i) Mode = 16 as it occurs maximum number of times.

(ii)

(iii)Total marks = 7+10+12+12+14+15+16+16+16+17+19 =

154

(iv)

Solution 7

(i)

(ii) Median = mean of 8th and 9th term

(iii) Mode = 5 as it occurs maximum number of times.

Solution 8

Score

x

No. of shots

f

fx

0

0

0

1

3

3

2

6

12

3

4

12

4

7

28

5

5

25

Total

25

80

(i) Modal score = 4 as it has maximum frequency 7.

(ii)

(iii) Total score = 80

(iv)

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 24(E)

Solution 1

 

 

Taking Height of student along x-axis and cumulative frequency along y-axis we will draw an ogive.

(i)

  

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)Since, number of terms = 160

(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.

The number of students whose height is above 172 cm

= 160 - 144 = 16

Solution 2

Weekly wages

(in Rs)

No. of

workers

(f)

Cumulative frequency

Class Marks

(x)

fx

50-55

5

5

52.5

262.5

55-60

20

25

57.5

1150.0

60-65

10

35

62.5

625.0

65-70

10

45

67.5

675.0

70-75

9

54

72.5

652.5

75-80

6

60

77.5

465.0

80-85

12

72

82.5

990.0

85-90

8

80

87.5

700.0

Total

80

5520.0

(i)

(ii) Modal class = 55-60 as it has maximum frequencies.

(iii) Number of workers getting wages below Rs.80 = 60

(iv) Number of workers getting Rs.65 or more and less than Rs.85 = 72 - 35 = 37

Solution 3

Marks

No. of Students

Cumulative frequency

0.5-9.5

5

5

9.5-19.5

9

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

Total number of terms = 120

(i)

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 =

The number of students getting more than 75% marks = 120 - 111 = 9

Solution 4

Mean of 1, 7, 5, 3, 4 and 4 =

m=4

Now, mean of 3, 2, 4, 2, 3, 3 and p = m-1 = 4-1 = 3

Therefore, 17+p = 3 x n …. Where n = 7

17+p = 21

p = 4

Arranging in ascending order:

2, 2, 3, 3, 3, 3, 4, 4

Mean = 4th term = 3

Therefore, q = 3

Solution 5

 

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 = 5th July as it has maximum frequencies.

(ii) Total number of terms = 206

Upper quartile =

Lower quartile =

Solution 6

 

We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:

 

 

 

 

 

 

 

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

 

 

Solution 7

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

(i)

(ii)

(iii) Mode = 15 as it has maximum frequencies i.e. 3

Solution 8

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

 

(i)

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.

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 =

Solution 9

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

(i) Number of students weighing more than 55 kg = 200-44 = 156

Therefore, percentage of students weighing 55 kg or more

(ii) 30% of students =

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

Solution 10

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 =

(ii)

(iii) Mode = 6 as it has maximum frequencies i.e. 6

Solution 11

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

 

Solution 12

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

(i)

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 8500 = 95 (approx from the graph)

(iii) Number of employees with income below Rs 11500 = 305 (approx from the graph)

Therefore number of employees (senior employees) = 320-305 =15

(iv) Upper quartile =

Solution 13

 

(i)Draw the histogram

(ii) In the highest rectangle which represents modal class draw two lines AC and BD intersecting at P.

(iii) From P, draw a perpendicular to x-axis meeting at Q.

(iv) Value of Q is the mode = 82 (approx)

Solution 14

Marks

No. of students

 

Cumulative frequency

0-10

5

5

10-20

11

16

20-30

10

26

30-40

20

46

40-50

28

74

50-60

37

111

60-70

40

151

70-80

29

180

80-90

14

194

90-100

6

200

Number of students = 200

(i)

Through mark 100, 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 = 57 marks (approx)

(ii) The number of students who failed (if minimum marks required to pass is 40)= 46 (approx from the graph)

(iii) The number of students who secured grade one in the examination = 200 - 188 = 12 (approx from the graph)

Solution 15

 

Solution 16

Since the frequency for x = 14 is maximum.

So Mode = 14.

 

According to the table it can be observed that the value of x from the 13th term to the 17th term is 13.

So the median = 13.

Solution 17

  

Solution 18

  

Solution 19

Histogram is as follows:

 

 

 

In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E. 

From E, draw a perpendicular to x-axis meeting at L.

Value of L is the mode. Hence, mode = 21.5

Solution 20

 

Marks

Number of students

(Frequency)

Cumulative Frequency

0-10

3

3

10-20

7

10

20-30

12

22

30-40

17

39

40-50

23

62

50-60

14

76

60-70

9

85

70-80

6

91

80-90

5

96

90-100

4

100

 

The ogive is as follows:

 

 

 

 

 

 

 

Solution 21

  

Solution 22

The cumulative frequency table of the given distribution table is as follows:

 

Weight in Kg

Number of workers (f)

Cumulative frequency

50-60

4

4

60-70

7

11

70-80

11

22

80-90

14

36

90-100

6

42

100-110

5

47

110-120

3

50

 

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50) on a graph paper and join them to get an ogive.

 

  

 

 

 

 

  

Solution 23

 

  

Solution 24

 

Scores

f

c.f.

0 - 10

9

9

10 - 20

13

22

20 - 30

20

42

30 - 40

26

68

40 - 50

30

98

50 - 60

22

120

60 - 70

15

135

70 - 80

10

145

80 - 90

8

153

90 - 100

7

160

 

n = 160

 

 

The ogive is shown below:

 

  

 

  

  

 

Solution 25

i. The frequency distribution table is as follows:

 

Class interval

Frequency

0-10

2

10- 20

5

20-30

8

30-40

4

40-50

6

 

 

ii.

 

Class interval

Frequency

(f)

Mean value (x)

fx

0-10

2

5

10

10- 20

5

15

75

20-30

8

25

200

30-40

4

35

140

40-50

6

45

270

 

 Sf = 25 

 

 Sf = 695 

 

 

iii. Here the maximum frequency is 8 which is corresponding to class 20 - 30.

 Hence, the modal class is 20 - 30. 

Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Page/Excercise 24(A)

Solution 1

(i)

(ii)

Solution 2

(a) Here n = 9

(b)

 

If marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

Solution 3

Numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Here n = 10

Solution 4

(a) The mean of 7, 11, 6, 5 and 6

(b)

If we subtract 2 from each number, then the mean will be 7-2 = 5

Solution 5

No. of terms = 5

Mean = 8

Sum of numbers = 8 x 5 = 40 .(i)

But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)

From (i) and (ii)

27+a = 40

a = 13

Solution 6

No. of terms = 5 and mean = 8

Sum of numbers = 5 x 8 = 40 ..(i)

but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)

from (i) and (ii)

27 + y + x = 40

x + y = 13

y = 13 - x

Solution 7

Age in yrs

xi

Frequency

(fi)

fixi

12

2

24

13

4

52

14

6

84

15

9

135

16

8

128

17

7

119

18

4

72

Total

40

614

Solution 8

No. of terms = 10

Mean = 69.5

Sum of the numbers = 69.5 x 10 = 695 ..........(i)

But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82

= 619 + x ......(ii)

from (i) and (ii)

619 + x = 695

x = 76

Solution 9

Height (cm)

xi

No. of Plants

fi

fixi

50

2

100

55

4

220

58

10

580

60

f

60f

65

5

325

70

4

280

71

3

213

Total

28+f

1718 + 60f

Mean = 60.95

Solution 10

Wages

(Rs/day) (x)

No. of Workers

(f)

fx

50

2

100

60

4

240

70

8

560

80

12

960

90

10

900

100

6

600

Total

42

3360

(i) Mean remains the same if the number of workers in each category is doubled.

Mean = 80

(ii) Mean will be increased by 60% if the wages per day per worker is increased by 60%

New mean =

(iii) No change in the mean if the number of workers is doubled but if wages per worker is reduced by 40%, then

New mean =

Solution 11

No. of matches

(x)

No. of boxes

(f)

fx

35

6

210

36

10

360

37

18

666

38

25

950

39

21

819

40

12

480

41

8

328

Total

100

3813

(i)

(ii) In the second case,

New mean = 39 matches

Total contents = 39 x 100 = 3900

But total number of matches already given = 3813

Number of new matches to be added = 3900 - 3813 = 87

Solution 12

  

Solution 13

  

Solution 14

  

Solution 15

  

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well. 

Text Book Solutions

ICSE X - Mathematics

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