Frank Solutions for Class 10 Maths Chapter 24 Measure of Central Tendency

Frank Solutions for Class 10 Maths Chapter 24 Measure of Central Tendency are useful for students as it helps them in scoring high marks in the examination. These solutions are prepared by subject matter experts at BYJU’S, describing the complete method of solving problems. Learners can download the pdf of Frank Solutions for Class 10 Maths Chapter 24 from the below links. It also provides the foundation for higher studies.

Chapter 24 – Measure of Central Tendency, is a single value that attempts to describe a set of data by identifying the central position within that set of data. Here we calculate the Frequency Distribution by using Frequency table, presentation of data in groups result in Grouping of Data. In the Frank Solutions, many such exercise problems are given which helps in better learning.

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Access answers to Frank Solutions for Class 10 Maths Chapter 24 Measure of Central Tendency

1. Find the mean of first 12 even numbers.

Solution:-

We know that, the first 12 even numbers are,

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 12

x̅ = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24)/12

x̅ = 156/12

x̅ = 13

Hence, mean of first 12 even numbers is 13.

2. Find the mean of first 10 prime numbers.

Solution:-

We know that, the first 10 prime numbers are,

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 10

x̅ = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29)/10

x̅ = 129/10

x̅ = 12.9

Hence, mean of first 10 prime numbers is 12.9.

3. Find the mean of all numbers from 7 to 17.

Solution:-

All numbers from 7 to 17 are,

7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 11

x̅ = (7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 +16 +17)/11

x̅ = 132/11

x̅ = 12

Hence, mean of all numbers from 7 to 17.

4. Find the mean of all odd numbers from 5 to 20. Find the new mean when each number is multiplied by 4.

Solution:-

All odd numbers from 5 to 20 are,

5, 7, 9, 11, 13, 15, 17, 19

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 8

x̅ = (5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/11

x̅ = 96/8

x̅ = 12

Hence, mean of all odd numbers from 5 to 20 is 12.

Then, all odd numbers from 5 to 20 multiplied by 4 are,

20, 28, 36, 44, 52, 60, 68, 76

n = 8

x̅ = (20 + 28 + 36 + 44 + 52 + 60 + 68 + 76)/8

x̅ = 384/8

x̅ = 48

Hence, mean all odd numbers from 5 to 20 multiplied by 4 is 48.

5. Find the mean of all natural numbers from 32 to 46. Find the new mean when each number is diminished by 5.

Solution:-

All natural numbers from 32 to 46 are,

32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46.

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 15

x̅ = (32 + 33 +34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46)/15

x̅ = 585/15

x̅ = 39

Hence, mean of all natural numbers from 32 to 46 is 39.

Then, all natural numbers from 32 to 46 diminished by 5 are,

27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41

n = 15

x̅ = (27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41)/15

x̅ = 510/15

x̅ = 34

Hence, mean all natural numbers from 32 to 46 diminished by 5 is 34.

6. If the mean of 8, 14, 20, x and 12 is 13, find x.

Solution:-

Form the question it is given that, 8, 14, 20, x, 12

Mean = 13

We have to find the value of x,

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 5

13 = (8 + 14 + 20 + x + 12)/5

13 × 5 = (54 + x)

65 = 54 + x

x = 65 – 54

x = 11

Therefore, the value of x is 11.

7. If the mean of 11, 14, p, 26, 10, 12, 18 and 6 is 15, find p.

Solution:-

Form the question it is given that, 11, 14, p, 26, 10, 12, 18 and 6.

Mean = 15

We have to find the value of p,

Then, x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 8

15 = (11 + 14 + p + 26 + 10 + 12 + 18 + 6)/8

15 × 8 = (97 + p)

120 = 97 + p

p = 120 – 97

p = 23

Therefore, the value of p is 23.

8. The mean monthly income of 10 persons is Rs 8,670. If a new member with a monthly income of Rs 9,000 jons the group, find the new monthly income.

Solution:-

From the question it is given that,

The mean monthly income of 10 persons is ₹ 8,670.

Number of persons, n = 10

We know that,

x̅ = (x1 + x2 + x3 + … + xn)/n

8,670 = ∑xn/10

∑xn = 8,670 × 10

∑xn = ₹ 86,700

Also it is given that, a new member with a monthly income of ₹ 9,000.

So, ∑xn = ₹ (86,700 + 9,000)

∑xn = ₹ 95,700

Then, n = 11

x̅ = (x1 + x2 + x3 + … + xn)/n

x̅ = ₹ 95,700/11

x̅ = ₹ 8,700

Therefore, the new mean monthly income is ₹ 8,700.

9. The height of 9 persons are 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 148 cm and 151 cm. Find the mean height.

Solution:-

From the question it is given that,

The height of 9 persons are, 142 cm, 158 cm, 152 cm, 143 cm, 139 cm, 144 cm, 148 cm and 151 cm.

We know that,

x̅ = (x1 + x2 + x3 + … + xn)/n

Where n is the total numbers,

n = 9

x̅ = (142 + 158 + 152 + 143 + 139 + 144 + 146 + 148 + 151)/9

x̅ = 1323/9

x̅ = 147 cm

Therefore, the mean height is 147 cm.

10. Find the mean of the following frequency distribution:

(i)

Class

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

Frequency

4

7

6

3

5

Solution:-

So, now we have to prepare the frequency distribution table

Class Interval

xi

fi

fixi

0 – 10

5

4

20

10 – 20

15

7

105

20 – 30

25

6

150

30 – 40

35

3

105

40 – 50

45

5

225

Total

25

605

we know that,

x̅ = ∑fixi/∑fi

x̅ = 605/25

x̅ = 24.2

Therefore, the mean is 24.2.

(ii)

Class

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

60 – 70

Frequency

4

4

7

10

12

8

5

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

fixi

0 – 10

5

4

20

10 – 20

15

4

60

20 – 30

25

7

175

30 – 40

35

10

350

40 – 50

45

12

540

50 – 60

55

8

440

60 – 70

65

5

325

Total

50

1910

we know that,

x̅ = ∑fixi/∑fi

x̅ = 1910/50

x̅ = 38.2

Therefore, the mean is 38.2.

(iii)

Class

0 – 6

6 – 12

12 – 18

18 – 24

24 – 30

Frequency

7

5

10

12

6

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

fixi

0 – 6

3

7

21

6 – 12

9

5

45

12 – 18

15

10

150

18 – 24

21

12

252

24 – 30

27

6

162

Total

40

630

we know that,

x̅ = ∑fixi/∑fi

x̅ = 630/40

x̅ = 15.75

Therefore, the mean is 15.75.

(iv)

Class

25 – 35

35 – 45

45 – 55

55 – 65

65 – 75

Frequency

6

10

8

12

4

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

fixi

25 – 35

30

6

180

35 – 45

40

10

400

45 – 55

50

8

400

55 – 65

60

12

720

65 – 75

70

4

280

Total

40

1980

we know that,

x̅ = ∑fixi/∑fi

x̅ = 1980/40

x̅ = 49.5

Therefore, the mean is 49.5.

(v)

Class

50 – 60

60 – 70

70 – 80

80 – 90

90 – 100

Frequency

8

6

12

11

13

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

fixi

50 – 60

55

8

440

60 – 70

65

6

390

70 – 80

75

12

900

80 – 90

85

11

935

90 – 100

95

13

1235

Total

50

3900

we know that,

x̅ = ∑fixi/∑fi

x̅ = 3900/50

x̅ = 78

Therefore, the mean is 78.

(vi)

Class

1 – 10

11 – 20

21 – 30

31 – 40

41 – 50

Frequency

9

12

15

10

14

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

fixi

1 – 10

5.5

9

49.5

11 – 20

15.5

12

186

21 – 30

25.5

15

382.5

31 – 40

35.5

10

355

41 – 50

45.5

14

637

Total

60

1610

we know that,

x̅ = ∑fixi/∑fi

x̅ = 1610/60

x̅ = 26.83

Therefore, the mean is 26.83.

(vi)

Class

101 – 110

111 – 120

121 – 130

131 – 140

141 – 150

151 – 160

Frequency

9

12

15

10

14

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

fixi

101 – 110

105.5

11

1160.5

111 – 120

115.5

16

1848

121 – 130

125.5

20

2510

131 – 140

135.5

30

4065

141 – 150

145.5

14

2037

151 – 160

155.5

9

1399.5

Total

100

13020

we know that,

x̅ = ∑fixi/∑fi

x̅ = 13020/100

x̅ = 130.2

Therefore, the mean is 130.2.

11. The mean of the following frequency distribution is 25.8 and the sum of all the frequencies is 50. Find x and y.

Class

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

Frequency

7

x

15

y

10

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

fixi

0 – 10

5

7

35

10 – 20

15

x

15x

20 – 30

25

15

375

30 – 40

35

y

35y

40 – 50

45

10

450

Total

50

860 + 15x + 35y

We know that,

∑fi = x1 + x2 + … + xn

50 = 7 + x + 15 + y + 10

x + y + 32 = 50

x + y = 18 … (i)

Also we know that, x̅ = ∑fixi/∑fi

So,

25.8 = (860 + 15x + 35y)/50

By cross multiplication we get,

15x + 35y + 860 = 1290

15x + 35y = 1290 – 860

15x + 35y = 430 … [divide both side by 5]

3x + 7y = 86 … (ii)

Now multiplying equation (i) by 3 we get,

3x + 3y = 54 … (iii)

Subtract equation (ii) from equation (iii) we get,

4y = 32

y = 32/4

y = 8

Substitute value of y in equation (i) to get the value of x,

x + y = 18

x + 8 = 18

x = 18 – 8

x = 10

Hence the value of x = 10 and y = 8.

11. Find the mean of the following frequency distribution by the short cut method.

Class

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

Frequency

9

12

15

10

14

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

A = 25

d = x – A

fid

0 – 10

5

9

– 20

-180

10 – 20

15

12

– 10

-120

20 – 30

A = 25

15

0

0

30 – 40

35

10

10

100

40 – 50

45

14

20

280

Total

60

80

we know that,

x̅ = A + ∑fid/∑fi

x̅ = 25 + 80/60

x̅ = 25 + 1.33

x̅ = 26.33

Therefore, the value of mean is 26.33.

13. Find the mean of the following frequency distribution by the short cut method:

Class

1 – 10

11 – 20

21 – 30

31 – 40

41 – 50

51 – 60

61 – 70

Frequency

7

10

14

17

15

11

6

Solution:-

So, now we have to prepare the frequency distribution table,

Class Interval

xi

fi

A = 25

d = x – A

fid

1 – 10

5.5

7

– 30

-210

11 – 20

15.5

10

20

-200

21 – 30

25.5

14

-10

-140

31 – 40

35.5

17

0

0

41 – 50

45.5

15

10

150

51 – 60

55.5

11

20

220

61 – 70

65.5

6

30

180

Total

80

0

we know that,

x̅ = A + ∑fid/∑fi

x̅ = 35.5 + 0/80

x̅ = 35.5 + 0

x̅ = 35.5

Therefore, the value of mean is 35.5.

14. Find the mean of the following frequency distribution by the step deviation method:

Class

1 – 10

11 – 20

21 – 30

31 – 40

41 – 50

51 – 60

61 – 70

Frequency

7

10

14

17

15

11

6

Solution:-

Class Interval

xi

fi

A = 125

u = (x – A)/hi

fiu

100 – 110

105

15

– 2

-30

110 – 120

115

18

-1

-18

120 – 130

A = 125

32

0

0

130 – 140

315

25

1

25

140 – 150

145

10

2

20

Total

100

-3

So, from the table A = 125 and hi = 10

we know that,

x̅ = A + h × ∑fiu/∑fi

x̅ = 125 + 10 × (-3/100)

x̅ = 125 – 0.3

x̅ = 124.70

Therefore, the mean is 124.70.

15. Find the mean of the following frequency distribution by the step deviation method:

Class

0 – 20

20 – 40

40 – 60

60 – 80

80 – 100

100 – 120

120 – 140

Frequency

12

24

52

88

66

42

16

Solution:-

Class Interval

xi

fi

A = 125

u = (x – A)/hi

fiu

0 – 20

10

12

– 3

-36

20 – 40

30

24

-2

-48

40 – 60

50

52

-1

-52

60 – 80

A = 70

88

0

0

80 – 100

90

66

1

66

100 – 120

110

42

2

84

120 – 140

130

16

3

48

Total

300

62

So, from the table A = 70 and hi = 20

we know that,

x̅ = A + h × ∑fiu/∑fi

x̅ = 70 + 20 × (62/300)

x̅ = 70 + 4.13

x̅ = 74.13

Therefore, the mean is 74.13.

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