Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 9 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Math Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 359:

Question 1:

Find the mean, using direct method:

Class 0-10 10-20 20-30 30-40 40-50
Frequency 3 5 9 5 3

Answer:

Class

Frequency (fi)

Mid Values (xi)

(fi×xi)

0-10

3

5

15

10-20

5

15

75

20-30

9

25

225

30-40

5

35

175

40-50

3

45

135

 

fi=25

 

fi×xi=625

  Mean, x¯=(fi×xi)fi=62525=25x¯=25

Page No 359:

Question 2:

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

Answer:

Class

Frequency (fi)

Mid Values (xi)

(fi×xi)

0-10

7

5

35

10-20

5

15

75

20-30

6

25

150

30-40

12

35

420

40-50

8

45

360

50-60

2

55

110

 

fi=40

 

(fi×xi)=1150

  Mean, x¯=(fi×xi)fi115040=28.75x¯=28.75

Page No 359:

Question 3:

Find the mean, using direct method:

Class 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 11 15 20 30 14 10

Answer:

Class

Frequency (fi)

Mid Values xi

fi×xi

10-20

11

15

165

20-30

15

25

375

30-40

20

35

700

40-50

30

45

1350

50-60

14

55

770

60-70

10

65

650

 

fi=100

 

(fi×xi)=4010

  Mean, x¯=(fi×xi)fi4010100=40.1x¯=40.1



Page No 360:

Question 4:

Find the mean, using direct method:

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Number of students 6 8 13 7 3 2 1

Answer:

Class

Frequency fi

Mid Values xi

(fi×xi)

10-20

6

15

90

20-30

8

25

200

30-40

13

35

455

40-50

7

45

315

50-60

3

55

165

60-70

2

65

130

70-80

1

75

75

 

fi=40

 

(fi×xi)=1430

  Mean, x¯=(fi×xi)fi143040=35.75x¯=35.75

Page No 360:

Question 5:

Find the mean, using direct method:

Class 25-35 35-45 45-55 55-65 65-75
Frequency 6 10 8 12 4

Answer:

Class

Frequency fi

Mid values xi

fi×xi

25-35

6

30

180

35-45

10

40

400

45-55

8

50

400

55-65

12

60

720

65-75

4

70

280

 

fi=40

 

(fi×xi)=1980

  Mean, x¯=(fi×xi)fi198040=49.5x¯=49.5

Page No 360:

Question 6:

Find the mean, using direct method:

Class 0-100 100-200 200-300 300-400 400-500
Frequency 6 9 15 12 8

Answer:

Class

Frequency fi

Mid values xi

(fi×xi)

0-100

6

50

300

100-200

9

150

1350

200-300

15

250

3750

300-400

12

350

4200

400-500

8

450

3600

 

fi=50

 

(fi×xi)=13200

  Mean, x¯=(fi×xi)fi1320050=264x¯=264

Page No 360:

Question 7:

The mean of the following frequency distribution is 24. Find the value of p.

Marks 0-10 10-20 20-30 30-40 40-50
Number of students 15 20 35 p 10

Answer:

Class

Frequency fi

Mid Values xi

(fi×xi)

0-10

15

5

75

10-20

20

15

300

20-30

35

25

875

30-40

p

35

35 p

40-50

10

45

450

 

fi=80+p

 

(fi×xi)=1700+35p

  Mean, x¯=(fi×xi)fi24  = 1700+35p80+p      [Mean=24]1920+24p=1700+35p11p=220p=20p=20

Page No 360:

Question 8:

Find the missing frequencies f1 and f2 in the table in given below, it is being given that the mean of the given frequency distribution is 50.

Class 0-20 20-40 40-60 60-80 80-100 Total
Frequency 17 f1 32 f2 19 120

Answer:

Class

Frequency fi

Mid values xi

(fi×xi)

0-20

17

10

170

20-40

f1

30

30 f1

40-60

32

50

1600

60-80

52- f1

70

3640-70 f1

80-100

19

90

1710

 

fi=120

 

(fi×xi)=7120-40f1

  We have:17+f1+32+f2+19=120f1+f2=52f2=52f1 Mean, x¯=(fi×xi)fi50=712040f1120               [Mean=50]40f1=1120f1=28And f2=5228f2=24The missing frequencies are f1=28 and f2 =24.

Page No 360:

Question 9:

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50

Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 7 f1 12 f2 8 5

Answer:

Class

Frequency fi

Mid values xi

(fi×xi)

0-20

7

10

70

20-40

f1

30

30 f1

40-60

12

50

600

60-80

18- f1

70

1260-70 f1

80-100

8

90

720

100-120

5

110

550

 

fi=50

 

(fi×xi)=3200-40f1

  We have: 7+f1+12+f2+8+5=50f1+f2=18f2=18f1 Mean, x¯=(fi×xi)fi57.6=320040f150        [Mean=57.6]40f1=320f1=8And f2=188f2=10The missing frequencies are f1=8 and f2 =10.

Page No 360:

Question 10:

Find the mean, 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

Answer:

Class

Frequency fi

Mid valuesxi

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

 

fi=100

 

 

(fi×di)=300

 Let A=25 be the assumed mean. Then we have:Mean, x¯=A+(fi×di)fi= 25+300100=28x¯=28

Page No 360:

Question 11:

Find the mean, using assumed-mean method:

Class 0-40 40-80 80-120 120-160 160-200
Frequency 12 20 35 30 23

Answer:

Class

Frequency fi

Mid valuesxi

Deviation di
di=xi-100

(fi×di)

0-40

12

20

-80

-960

40-80

20

60

-40

-800

80-120

35

100=A

0

0

120-160

30

140

40

1200

160-200

23

180

80

1840

 

fi=120

 

 

(fi×di)=1280

 Let A=100 be the assumed mean. Then we have:Mean, x¯=A+(fi×di)fi= 100+1280120=100+10.67x=110.67

Page No 360:

Question 12:

Find the mean, using assumed-mean method:

Class 100-120 120-140 140-160 160-180 180-200
Frequency 10 20 30 15 5

Answer:

Class

Frequency fi

Mid valuesxi

Deviation di
di=xi-150

(fi×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:Mean, x¯=A+(fi×di)fi= 15030080=1503.75x¯=146.25

Page No 360:

Question 13:

Find the mean, using assumed-mean method:

Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 20 35 52 44 38 31

Answer:

Class

Frequency fi

Mid Valuesxi

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:Now, mean, x¯=A+(fi×di)fi=50+2760220=50+12.55x¯=62.55



Page No 361:

Question 14:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Marks 0-10 10-20 20-30 30-40 40-50 50-60
Number of students 12 18 27 20 17 6

Answer:

Class

Frequencyfi

Mid Valuesxi

ui=(xiA)h
=(xi25)10

(fi×ui)

0-10

12

5

−2

24

10-20

18

15

1

18

20-30

27

25=A

0

0

30-40

20

35

1

20

40-50

17

45

2

34

50-60

6

55

3

18

 

fi=100

 

 

(fi×ui)=30

 Now, A=25, h=10, fi=100 and (fi×ui)=30 Mean, x¯=A+h×(fi×ui)fi=25+10×30100 =25+3=28x¯=28

Page No 361:

Question 15:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class Number of students
4-8 2
8-12 12
12-16 15
16-20 25
20-24 18
24-28 12
28-32 13
32-36 3

Answer:

Class

Frequency fi

Mid valuesxi

ui=(xiA)h
=(xi18)4

(fi×ui)

4-8

2

6

-3

-6

8-12

12

10

-2

-24

12-16

15

14

-1

-15

16-20

25

18=A

0

0

20-24

18

22

1

18

24-28

12

26

2

24

28-32

13

30

3

39

32-36

3

34

4

12

 

fi=100

 

 

(fi×ui)=48

 Now, A=18, h=4, fi=100 and (fi×ui)=48Mean, x¯=A+{h×(fi×ui)fi}=18+{4×48100}=18+1.92=19.92x¯=19.92

Page No 361:

Question 16:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class 0-30 30-60 60-90 90-120 120-150 150-180
Frequency 12 21 34 52 20 11

Answer:

Class

Frequency fi

Mid valuesxi

ui=(xiA)h
=(xi75)30

(fi×ui)

0-30

12

15

2

24

30-60

21

45

1

21

60-90

34

75 = A

0

0

90-120

52

105

1

52

120-150

20

135

2

40

150-180

11

165

3

33

 

fi=150

 

 

(fi×ui)=80

 Now, A=75, h=30, fi=150 and (fi×ui)=80Mean, x¯=A+{h×(fi×ui)fi}=75+{30×80150}=75+16=91∴ x¯=91

Page No 361:

Question 17:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class 0-20 20-40 40-60 60-80 80-100 100-120 120-140
Frequency 12 18 15 25 26 15 9

Answer:

Class

Frequency fi

Mid valuesxi

ui=(xiA)h=(xi70)20

(fi×ui)

0-20

12

10

3

36

20-40

18

30

2

36

40-60

15

50

1

15

60-80

25

70 = A

0

0

80-100

26

90

1

26

100-120

15

110

2

30

120-140

9

130

3

27

 

fi=120

 

 

(fi×ui)=4

 Now, A=70, h=20, fi=120 and (fi×ui)=4Mean, x¯=A+{h×(fi×ui)fi}=70+{20×(4)120}=70-0.67=69.33x¯=69.33

Page No 361:

Question 18:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Marks 0-14 14-28 28-42 42-56 56-70
Number of students 7 21 35 11 16

Answer:

Class

Frequency fi

Mid valuesxi

ui=(xiA)h=(xi35)14

(fi×ui)

0-14

7

7

2

14

14-28

21

21

1

21

28-42

35

35 = A

0

0

42-56

11

49

1

11

56-70

16

63

2

32

 

fi=90

 

 

(fi×ui)=8

 Now, A=35, h=14, fi=90 and(fi×ui)=8 Mean, x¯=A+{h×(fi×ui)fi}=35+{14×890}=35+1.24=36.24x¯=36.24 cm

Page No 361:

Question 19:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class 10-15 15-20 20-25 25-30 30-35 35-40
Frequency 5 6 8 12 6 3

Answer:

Class

Frequency fi

Mid valuesxi

ui=(xiA)h=(xi22.5)5

(fi×ui)

10-15

5

12.5

2

10

15-20

6

17.5

1

6

20-25

8

22.5 = A

0

0

25-30

12

27.5

1

12

30-35

6

32.5

2

12

35-40

3

37.5

3

9

 

fi=40

 

 

(fi×ui)=17

 Now, A=22.5, h=5, fi=40 and (fi×ui)=17 Mean, x¯=A+{h×(fi×ui)fi}=22.5+{5×1740}=22.5+2.125=24.625x=24.625

Page No 361:

Question 20:

Find the arithmetic mean of each of the following frequency distributions 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

Answer:

Class

Frequency fi

Mid valuesxi

ui=(xiA)h=(xi33)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

 

fi=40

 

 

(fi×ui)=2

 Now, A=33, h=6, fi=40 and (fi×ui)=2 Mean, x¯=A+{h×(fi×ui)fi}=33+{6×240}=33+0.3=33.3x=33.3 years

Page No 361:

Question 21:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class 84-90 90-96 96-102 102-108 108-114 114-120
Frequency 15 22 20 18 20 25

Answer:

Class

Frequencyfi

Mid valuesxi

ui=(xiA)h=(xi99)6

(fi×ui)

84-90

15

87

2

30

90-96

22

93

1

22

96-102

20

99 = A

0

0

102-108

18

105

1

18

108-114

20

111

2

40

114-120

25

117

3

75

 

fi=120

 

 

(fi×ui)=81

 Now, A=99, h=6, fi=120 and (fi×ui)=81 Mean, x¯=A+{h×(fi×ui)fi}=99+{6×81120}=99+4.05=103.05∴ x¯=103.05



Page No 362:

Question 22:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

Class 500-520 520-540 540-560 560-580 580-600 600-620
Frequency 14 9 5 4 3 5

Answer:

Class

Frequencyfi

Mid valuesxi

ui=(xiA)h=(xi550)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