RS Aggarwal Solutions Chapter 9 Mean, Mode and Median Exercise 9A Class 10 Maths

Chapter Name

RS Aggarwal Chapter 9 Mean, Mode and Median

Book Name

RS Aggarwal Mathematics for Class 10

Other Exercises

  • Exercise 9B
  • Exercise 9C
  • Exercise 9D
  • Exercise 9E
  • Exercise 9F

Related Study

NCERT Solutions for Class 10 Maths

Exercise 9A Solutions

1. If the mean of 5 observation x, x + 2, x + 4, x + 6 and x + 8, find the value of x.

Solution

Mean of given observations = (sum of given observations)/(total number of observation)

∴ 11 = [x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]/5

⇒ 55 = 5x + 20

⇒ 5x = 55 – 20

⇒ 5x = 35

⇒ x = 35/5

⇒ x = 7

Hence, the value of x is 7.


2. If the mean of 25 observations is 27 and each observation is decreased by 7, what will be new mean?

Solution

Mean of given observations = (sum of given observations)/(total number of observations)

Mean of 25 observations = 27

∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 – (25 × 7)

= 675 – 175

= 500

Then, new mean = 500/25 = 20

Thus, the new mean will be 20.


 3. Compute the mean for following data:

Class

1 - 3

3 - 5

5 - 7

7 - 9

Frequency

12

22

27

19

Solution

The given data is shown as follows:

Class

Frequency (fi)

Class mark (xi)

5 - 7

1 - 3

12

2

24

3 - 5

22

4

88

5 - 7

27

6

162

7 - 9

19

8

152

Total

Ʃfi = 80

 

Ʃ fi xi = 426

The mean of given data is given by 

Thus, the mean of the following data is 5.325.


4. Find the mean using direct method:

Class

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

Frequency

7

5

6

12

8

2

Solution


 5. Find the mean of the following data, using direct method:<

Class

25 – 35

35 – 45

45 – 55

55 – 65

65 – 75

Frequency

6

10

8

12

4

Solution

Class

Frequency (fi)

Mid values (xi)

(f× xi)

25 - 35

6

10

180

35 - 45

10

40

400

45 - 55

8

50

400

55 - 65

12

60

720

65 - 75

4

70

280

 

∑ f= 40

 

∑(fi × xi)  = 1980


 6. Find the mean of the following data, using direct method:

Class

0 – 100

100-200

200-300

300-400

400-500

Frequency

6

9

15

12

8

Solution 

Class

Frequency (fi)

Mid values (xi)

(f× xi)

0 - 100

6

50

300

100 - 200

5

150

1350

200 - 300

6

250

3750

300 - 400

12

350

4200

400 - 500

8

450

3600

 

∑ f= 40

 

∑(fi × xi) = 13200


 7. Using an appropriate method, find the mean of the following frequency distribution:

Class

84 - 90

90 - 96

96 - 102

102 - 108

108 - 114

114 - 120

Frequency

8

10

16

23

12

11

which method did you see, and why ? 

Solution 

Class

Frequency (fi)

Mid values (xi)

(f× xi)

84 - 90

8

87

696

90 - 96

10

93

930

96 - 102

16

99

1584

102 - 108

23

105

2415

108 - 114

12

111

1332

114 – 120

11

117

1287

Total

∑ f= 80

 

∑(fi × xi) = 8244

The mean of the data is given by, 

= 8244/80
= 103.05
Thus, the mean of the following data is 103.05.

8. If the mean of the following frequency distribution is 24, find the value of of p.

Class

0 - 10

10 - 20

20 – 30

30 - 40

40 - 50

Frequency

3

4

P

3

2

Solution 

The given data is shown as follows: 

Class

Frequency (fi)

Mid values (xi)

(fi xi)

0 - 10

3

5

15

10 - 20

4

15

60

20 - 30

p

25

25p

30 - 40

3

35

105

40 - 50

2

45

90

Total

∑fi = 12 + p

∑fi xi = 270 + 25p

The mean of the given data is given by, 

24 = (270 + 25p)/(12 + p)

24 (12 + p) = 270 + 25p 

288 + 24p = 270 + 25p 

25p – 24p = 288 – 270 

p = 18 

Hence, the value of p is 18. 


9. The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹ 18, find the missing frequency f.

Daily pocket allowance (in ₹)

11 - 13

13 - 15

15 - 17

17 - 19

19 - 21

21 - 23

23 - 25

Number of children

7

6

9

13

f

5

4

Solution 

The given data is shown as follows: 

Daily pocket allowance (in ₹)

Number of children (fi)

Class mark (xi)

fi xi

11 - 13

7

12

84

13 - 15

6

14

84

15 - 17

9

16

144

17 - 19

13

18

234

19 - 21

f

20

20f

21 - 23

5

22

110

23 - 25

4

24

96

Total

∑fi = 44 + f

 

∑fx= 752 + 20f

The mean of the given data is given by,

⇒ 18 = (750 + 20f)/(44 + f)

⇒ 18 (44 + f) = 752 + 20f 

⇒ 792 + 18f = 752 + 20f 

⇒ 20f – 18f = 792 – 752

⇒ 2f = 40 

⇒ f = 20 

Hence, the value of f is 20.


10. The mean of following frequency distribution is 54. Find the value of p. 

Class

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

Frequency

7

p

10

9

13

Solution 

The given data is shown as follows: 

Class

Frequency (fi)

Class mark (xi)

fi xi

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

∑f= 39 + p

∑fi xi = 2370 + 30p

The mean of the given data is given by, 

⇒ 54 = (2370 + 30p)/(39 + p)

⇒ 54 (39 + p) = 2370 + 30p 

⇒ 2106 + 54p = 2370 – 2106

⇒ 24p = 264 

⇒ p = 11 

Hence, the value of p is 11. 


11. The mean of the following frequency data is 42, Find the missing frequencies x and y if the sum of frequencies is 100. 

Class interval

0 – 10 

10 – 20 

20 – 30 

30 – 40 

40 – 50 

50 – 60 

60 - 70

70 – 80 

Frequency

10 

13 

10 

14 

9

Find x and y. 

Solution 

The given data is shown as follows: 

Class interval

Frequency (fi)

Class mark (xi)

fi xi

0 – 10 

35

10 – 20 

10 

15 

150

20 – 30 

25 

25x

30 – 40 

13 

35 

455

f40 – 50 

45 

45y

50 – 60 

10 

55 

550

60 – 70 

14 

65 

910

70 – 80 

75 

675

Total 

Ʃ f= 63 + x + y

Ʃ fi xi = 2775 + 25x + 45y

Sum of the frequencies = 100 

⇒ ∑i  fi = 100 

⇒ 63 + x + y = 100 

⇒ x + y = 100 – 63

⇒ x + y = 37 

⇒ y = 37 – x ...(1) 

Now, the mean of the given data is given by,

⇒ 42 = (2775 + 25x + 45y)/100

⇒ 4200 = 2775 + 25x + 45y 

⇒ 4200 – 2775 = 25x + 45y 

⇒ 1425 = 25x + 45(37 – x) [from (1)] 

⇒ 1425 = 25x + 1665 – 45x

⇒ 20x = 1665 – 1425

⇒ 20x = 240 

⇒ x = 12 

If x = 12, then y = 37 – 12 = 25 

Thus, the value of x is 12 and y is 25.


12. The daily expenditure of 100 families are given below. Calculate fand fif the mean daily expenditure is ₹ 188. 

Expenditure (in ₹)

140 - 160

160 - 180

180 - 200

200 - 220

220 - 240

Number of families

5

25

f1

f2

5

Solution 

The given data is shown as follows: 

Expenditure (in ₹)

Number of families (fi)

Class mark (xi)

fi xi

140 - 160

5

150

750

160 - 180

25

170

4250

180 - 200

F1

190

190f1

200 - 220

F2

210

210f2

220 – 240

5

230

1150

Total

∑f= 35 + f1 + f2.

 

∑f x= 6150 + 190f1 + 210f2

Sum of the frequencies = 100 

⇒ ∑i fi = 100 

⇒ 35 + f1 + f2 = 100 

⇒ f1 + f2 = 100 – 35 

⇒ f1 + f2 = 65

⇒ f2 = 65 – f….(1) 

Now, the mean of the given data is given by, 

⇒ 188 = (6150 + 190f1 + 210f2)/100

⇒ 1880 = 6150 + 190f1 + 210f2

⇒ 18800 - 6150 = 190f1 + 210f2

⇒ 12650 = 190f1 + 210(65 – f1) [From (1)]

⇒ 12650 = 190f1 – 210f1 + 13650

⇒ 20f1 = 13650 – 12650

⇒ 20f1 = 1000

⇒ f1 = 50

If f1 = 50, then f2 = 65 – 50 = 15

Thus, the value of f1 is 50 and f2 is 15.


13. Find the mean of the following frequency distribution is 57.6 and the total number of observation is 50. 

Class

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

100 - 120

Frequency

7

f1

12

f2

8

5

Solution 

Class

Frequency (fi)

Mid values (xi)

(fi × xi)

0 - 20

7

10

70

20 - 40

F2

30

30f1

40 – 60

12

50

600

60 – 80

18 - f1

70

1260 – 70f1

80 – 100

8

90

720

100 - 120

5

110

550

Total

∑f= 50

 

∑(fi × xi) = 3200 – 40f1

We have:

7 + f1 + 12 + f2 + 8 + 5 = 50

⇒ f1 + f2 = 18

⇒ f2 = 18 – f1

⇒ 57.6 = (3200 – 40f1)/50

⇒ 40f1 = 320

∴ f1 = 8

And f2 = 18 – 8

⇒ f2 = 10

∴ The missing frequencies are f1 = 8 and f2 = 10.


14. During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarized as follows:

Number of heartbeats

65 - 68

68 - 71

71 - 74

74 - 77

77 - 80

80 - 83

83 - 86

Number of patients

2

4

3

8

7

4

2

Find the mean heartbeats per minute for these patients, choosing a suitable method.

Solution

Using direct method, the given data is shown as follows:

Number of heartbeats per minute

Number of patients (fi)

Class mark (xi

fi xi

65 - 68

2

66.5

133

68 – 71

4

69.5

278

71 - 74

3

72.5

217.5

74 - 77

8

75.5

604

77 - 80

7

78.5

549.5

80 - 83

4

81.5

326

83 – 86

2

84.5

169

Total

∑fi = 30

 

∑fi xi = 2277

The mean of the data is given by,

= 2277/30

= 75.9

Thus, the mean heartbeats per minute for these patients is 75.9.


15. Find the mean marks per student, using assumed-mean method:

Marks

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

Number of students

12

18

27

20

17

6

Solution 

Class

Frequency (fi)

Mid values (xi)

Deviation (di) di = (xi – 25)

(fi × di)

0 - 10

12

5

-20

-240

10 - 20

18

15

-10

-180

20 - 30

27

25 = A

0

0

30 - 40

20

35

10

200

40 - 50

17

45

20

340

50 - 60

6

55

30

180

Total

∑f= 100

∑(f× di) = 300


16. Find the mean of the following frequency distribution, using the assumed-mean method:

Class

100 - 120

120 - 140

140 - 160

160 - 180

180 - 200

Frequency

10

20

30

15

5

Solution 

Class

Frequency (fi)

Mid values (xi)

Deviation (di) d= (xi – 150)

(f× di)

100 - 120

10

110

-40

-400

120 - 140

20

130

-20

-400

140 - 160

30

150 = A

0

0

160 - 180

15

170

20

300

180 - 200

5

190

40

200

 

∑fi = 80

   

∑(fi × di) = -300

Let A = 150 be the assumed mean, Then we have:


17. Find the mean of the following data, using assumed-mean method:

Class

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

Frequency

20

35

52

44

38

Solution 

Class

Frequency (fi)

Mid values (xi)

Deviation (di) di = (xi – 50)

(fi × di)

0 - 20

20

10

-40

-800

20 - 40

35

30

-20

-700

40 - 60

52

50 = A

0

0

60 - 80

44

70

20

880

80 - 100

38

90

40

1520

100 - 120

31

110

60

1860

 

∑fi = 220

   

∑(fi × di) = 2760

Let A = 50 be the assumed mean. Then we have:


18. The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.

Literacy rate (%)

45 - 55

55 - 65

65 - 75

75 - 85

85 - 95

Number of cities

4

11

12

9

4

Solution

Using Direct method, the given data is shown as follows:

Literacy rate (%)

Number of cities (fi)

Class mark (xi)

(fi xi)

45 - 65

4

50

200

55 - 65

11

60

660

65 - 75

12

70

840

75 – 85

9

80

720

85 - 95

4

90

360

Total

∑f= 40

 

∑fi xi = 2780

The mean of the data is given by,

= 2780/40

= 69.5

Thus, the mean literacy is 69.5 %.


19. Find the mean of the following frequency distribution using step-deviation method.

Class

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

Frequency

7

10

15

8

10

Solution

Let us choose a = 25, h = 10, then di = x– 25 and u= (xi – 25)/10

Using step-deviation method, the given data is shown as follows:

Class

Frequency (fi)

Class mark (xi)

d= xi – 25

u= (xi – 25)/10

(fi ui)

0 - 10

7

5

-20

-2

-14

10 - 20

10

15

-10

-1

-10

20 - 30

15

25

0

0

0

30 - 40

8

35

10

1

8

40 - 50

10

45

20

2

20

Total

∑fi = 50

 

 

 

∑fu= 4

 The mean of the data is given by,

= 25 + 4/50 × 10

= 25 + 4/5

= (125 + 4)/5

= 129/5

= 25.8

Thus, the mean is 25.8.


20. Find the mean of the following data, using step-deviation method:

Class

5 - 15

15 - 25

25 - 35

35 - 45

45 - 55

55 - 65

65 - 75

Frequency

6

10

16

15

24

8

7

Solution

Let us choose a = 40, h = 10, then di = xi – 40 and u= (x– 40)/10

Using step-deviation method, the given data is shown as follows:

Class

Frequency (fi)

Class mark (xi)

di = xi – 40

ui = (xi – 40)/10

(fi ui)

5 - 15

6

10

-30

-3

-18

15 - 25

10

20

-20

-2

-20

25 - 35

16

30

-10

-1

-16

35 – 45

15

40

0

0

0

45 – 55

24

50

10

1

24

55 – 65

8

60

20

2

16

65 – 75

7

70

30

3

21

Total

∑fi = 86

     

∑fu= 7

The mean of the data is given by,

= 40 + 7/86 × 10

= 40 + 70/86

= 40 + 0.81

= 40.81


21The weights of tea in 70 packets are shown in the following table:

Weight

200 - 201

201 - 202

202 - 203

203 - 204

204-205

205-206

Number of packets

13

27

18

10

1

1

Find the mean weight of packets using step deviation method.

Solution

Let us choose a = 202.5, h = 1, then d= x– 202.5 and ui = (x– 202.5)/1

Using step-deviation method, the given data is shown as follows:

Weight

Number of packets (fi)

Class mark (xi)

di = x– 202.5

ui = (xi – 202.5)/1

(fi ui)

200 - 201

13

200.5

-2

-2

-26

201 – 202

27

201.5

-1

-1

-27

202 – 203

18

202.5

0

0

0

203 - 204

10

203.5

1

2

10

204 – 205

1

204.5

2

3

2

205 - 206

1

205.5

3

 

3

Total

∑fi = 70

 

 

 

∑fi ui = - 38

The mean of the given data is given by,

= 202.5 + (-38/70) × 1

= 202.5 – 0.542

= 201.96

Hence, the mean is 201.96 g.


22. Find the mean of the following frequency distribution table using a suitable method:

Class

20 - 30

30 - 40

40 - 50

50 - 60

60 - 70

Frequency

25

40

42

33

10

Solution

Let us choose a = 45, h = 10, di = x– 45 and u= (xi – 45)/10

Using step-deviation method, the given data is shown as follows:

Weight

Number of packets (fi)

Class mark (xi)

di = x- 45

ui = (xi – 45)/10

(fi ui)

20 - 30

25

35

-20

-2

-50

30 - 40

40

35

-10

-1

-40

40 - 50

42

45

0

0

0

50 - 60

33

55

10

1

33

60 – 70

10

65

20

2

20

Total

∑fi = 150

 

 

 

∑fi ui = - 37

The mean of the given data is given by,

= 45 – (37/150) × 10

= 45 – 37/15

= 45 – 2.466

= 42.534

Hence, the mean is 42.534.


23. In an annual examination, marks (out of 90) obtained by students of Class X in mathematics are given below:

Marks Obtained

0 - 15

15 - 30

30 - 45

45 - 60

60 - 75

75 - 90

Number of students

2

4

5

20

9

10

Find the mean marks.

Solution

Let us choose a = 52.5, h = 15, then di = xi – 52.5 and ui = (xi – 52.5)/15

Using step-deviation method, the given data is shown as follows:

Weight

Number of students (fi)

Class mark (xi)

di = xi – 37.5

ui = (x– 52.5)/15

(fi ui)

0 - 15

2

7.5

-45

-3

-6

15 - 30

4

22.5

-30

-2

-8

30 - 45

5

37.5

-15

-1

-5

45 - 60

20

52.5

0

0

0

60 – 75

9

67.5

15

1

9

75 – 90

10

82.5

30

2

20

Total

∑fi = 50

     

∑fi ui = 10

The mean of the given data is given by,

= 52.5 + (10/50) × 15

= 52.5 + 3

= 55.5

Thus, the mean is 55.5.


24. Find the arithmetic mean of the following frequency distribution using step-deviation method:

Age (in years)

18 - 24

24 - 30

30 - 36

36 - 42

42 – 48

48 - 54

Number of workers

6

8

12

8

4

2

 

Solution

Class

Frequency (fi)

Mid values (xi)

u= (xi – A)/h = (xi – 33)/6

(fi × ui)

18 - 24

6

21

-2

-12

24 - 30

8

27

-1

-8

30 - 36

12

33 = A

0

0

36 - 42

8

39

1

8

42 – 48

4

45

2

8

48 – 54

2

51

3

6

Total

∑fi = 40

 

 

∑(f× ui) = 2

Now, A = 33, h = 6, ∑fi = 40 and ∑(fi × ui) = 2


25. Find the mean of the following data using step-deviation method:

Class

500 - 520

520 - 540

540 - 560

560 - 580

580 - 600

600 – 620

Frequency

14

9

5

4

3

5

Solution

Class

Frequency (fi)

Mid values (xi)

ui = (xi - A)/h = (xi – 550)/20

(fi × ui)

500 - 520

14

510

-2

-28

520 - 540

9

530

-1

-9

540 - 560

5

550 = A

0

0

560 - 580

4

570

1

4

580 - 600

3

590

2

6

600 - 620

5

610

3

15

 

 ∑fi = 40

 

 

∑(fi × ui) = - 12

Now, a = 550, h = 20∑fi = 40 and ∑(f× ui) = - 12


26. Find the mean age from the following frequency distribution:

Age (in years)

25 - 29

30 - 34

35 - 39

40 - 44

45 - 49

50 - 54

55 - 59

Number of persons

4

14

22

16

6

5

3

 Solution

Class

Frequency (fi)

Mid values (xi)

ui = (x– A)/h = (x– 42)/5

(fi × ui)

24.5 – 29.5

4

27

-3

-12

29.5 – 34.5

14

32

-2

-28

34.5 – 39.5

22

37

-1

-22

39.5 – 44.5

16

42 = A

0

0

44.5 – 49.5

6

47

1

6

49.5 – 54.5

5

52

2

10

54.5 – 59.5

3

57

3

9

 

∑fi = 70

 

 

∑ (fi × ui) = - 37

Now, A = 42, h = 5, ∑ fi = 70 and ∑ (fi × ui) = - 37


27. The following table shows the age distribution of patients of malaria in a village during a particular month:

Age in (years)

5 - 14

15 - 24

25 - 34

35 - 44

45 - 54

55 - 64

No. of cases

6

11

21

23

14

5

Find the average age of the patients. 

Solution 

Class

Frequency (fi)

Mid values (xi)

ui = (xi – A)/h = (xi – 29.5)/10

(fi × ui)

4.5 – 14.5

6

9.5

-2

-12

14.5 – 24.5

11

19.5

-1

-11

24.5 – 34.5

21

29.5 = A

0

0

34.5 – 44.5

23

39.5

1

23

44.5 – 54.5

14

49.5

2

28

54.5 – 64.5

5

59.5

3

15

 

∑fi = 80

 

 

∑(f× ui) = 43

Now, A = 29.5, h = 10, ∑ fi = 80 and ∑ (fi × ui) = 43


28. Weight of 60 eggs were recorded as given below:

Weight (in grams)

75 - 79

80 - 84

85 - 89

90 - 94

95 - 99

100 - 104

105 - 109

No. of eggs

4

9

13

17

12

3

2

Calculate their mean weight to the nearest gram.

Solution

Let us choose a = 92, h = 5, then d= xi – 92 and ui = (xi – 92)/5

Using step-deviation method, the given data is shown as follows

Weight (in grams)

Number of eggs (fi)

Class mark (xi)

d= xi - 92

ui = (x– 92)/5

(fi ui)

74.5 – 79.5

4

77

-15

-3

-12

79.5 – 84.5

9

82

-10

-2

-18

84.5 – 89.5

13

87

-5

-1

-13

89.5 – 94.5

17

92

0

0

0

94.5 – 99.5

12

97

5

1

12

99.5 – 104.5

3

102

10

2

6

104.5 – 109.5

2

107

15

3

6

Total

 ∑fi = 60

 

 

 

 ∑fui = - 19

The mean of the given data is given by, 

Thus, the mean weight to the nearest gram is 90 g.
 
29. The following table shows the marks scored by 80 students in an examination.

Marks

0 - 5

5 - 10

10 - 15

15 - 20

20 - 25

25 - 30

30 - 35

35 - 40

No. of students

3

10

25

49

65

73

78

80

Solution

Let us choose a = 17.5, h = 5, then di = xi – 17.5 and u= (xi – 17.5)/5

Using step-deviation method, the given data is shown as follows:

Marks

Number of students (cf)

Frequency (fi)

Class mark (xi)

di = xi – 17.5

u= (xi – 17.5)/3

(fi ui)

0 - 5

3

3

2.5

-15

-3

-9

5 - 10

10

7

7.5

-10

-2

-14

10 - 15

25

15

12.5

-5

-1

-15

15 - 20

49

24

17.5

0

0

0

20 - 25

65

16

22.5

5

1

16

25 - 30

73

8

27.5

10

2

16

30 – 35

78

5

32.5

15

3

15

35 – 40

80

2

37.5

20

4

8

Total

 

∑ f= 80

     

∑ fi u= 17

The mean of the given data is given by,

= 17.5 + (17/80) × 5

= 17.5 + 1.06

= 18.56

Thus, the mean marks correct to 2 decimal places is 18.56.

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