SELINA Solutions for Class 10 Maths Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 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

  

Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 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

  

Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 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

Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 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)

Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 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

 

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

  

 

(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 =   marks

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

 

Solution 3

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 4

 

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 5

 

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 6

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 7

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 8

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 9

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 10

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 11

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 12

 

(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 13

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 14

 

Solution 15

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 16

  

Solution 17

  

Solution 18

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 19

 

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 20

 

  

Solution 21

Here the number of observations i. e is 10, which is even.'

 

 

So, the given data is 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data, 46 occurs most frequently.

Mode = 46 

Solution 22

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

Wages (Rs.)

Upper limit

No. of workers

C.f.

400-450

450

2

2

450-500

500

6

8

500-550

550

12

20

550-600

600

18

38

600-650

650

24

62

650-700

700

13

75

700-750

750

5

80

The ogive is as follows:

  

 

Number of workers = n = 80

1) Median =  term = 40th term, draw a horizontal line which meets the curve at point A.

Draw vertical line parallel to y axis from A to meet x axis at B.

The value of point B is 605.

2) Lower quartile (Q1)=  term=20th term = 550

3) Through mark of point 625 on x axis draw a vertical line which meets the graph at point C Then through point C, draw a horizontal line which meets the y axis at the mark of 50.

Thus, the number of workers that earn more than Rs. 625 daily = 80 - 50 = 30

Solution 23

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.