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

Chapter Name

RS Aggarwal Chapter 9 Mean, Mode and Median

Book Name

RS Aggarwal Mathematics for Class 10

Other Exercises

  • Exercise 9A
  • Exercise 9B
  • Exercise 9C
  • Exercise 9E
  • Exercise 9F

Related Study

NCERT Solutions for Class 10 Maths

Exercise 9D Solutions

1. Find the mean, median and mode of the following data:

Class

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

60 - 70

Frequency

4

4

7

10

12

8

5

Solution

To find the mean let us put the data in the table given below:

Class

Frequency (fi)

Class mark (xi)

fi xi

0 - 10

4

5

20

10 - 20

4

15

60

20 - 30

7

25

175

30 - 40

10

35

350

40 - 50

12

45

540

50 – 60

8

55

440

60 70

5

65

325

Total

Σf= 50

 

Σfi x= 1910

= 1910/50

= 38.2

Thus, the mean of the given data is 38.2.

Now, to find the median let us put the data in the table given below:

Class

Frequency (fi)

Cumulative Frequency (cf)

0 - 10

4

4

10 - 20

4

8

20 - 30

7

15

30 - 40

10

25

40 - 50

12

37

50 – 60

8

45

60 – 70

5

50

Total

N = Σfi = 50

 

Now, N = 50

⇒ N/2 = 25

The cumulative frequency just greater than 25 is 37 and the corresponding class is 40 – 50.

Thus, the median class is 40 – 50.

∴ l = 40, h = 10, N = 50, f = 12 and cf = 25.

Now,

Median = l + ((N/2 – cf)/f) × h

= 40 + (25 – 25)/12 × 10

= 40.

Thus, the median is 40.

We know that,

Mode = 3(Median) – 2(Mean)

= 3 × 40 – 2 × 38.2

= 120 – 76.4

= 43.6

Hence, Mean = 38.2, Median = 40 and Mode = 43.6


2. Find the mean, median and mode of the following data:

Class

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

100 - 120

120 - 140

Frequency

6

8

10

12

6

5

3

 Solution

To find the mean let us put the data in the table given below:

Class

Frequency (fi)

Class mark (xi)

fxi

0 - 20

6

10

60

20 - 40

8

30

240

40 - 60

10

50

500

60 - 80

12

70

840

80 - 100

6

90

540

100 - 120

5

110

550

120 - 140

3

130

390

Total

Σf= 50

 

Σfi xi = 3120

= 3120/50

= 62.4

Thus, the mean of the given data is 62.4.

Now, to find the median let us put the data in the table given below:

Class

Frequency (fi)

Cumulative Frequency (cf)

0 - 20

6

6

20 - 40

8

14

40 - 60

10

24

60 - 80

12

36

80 - 100

6

42

100 – 120

5

47

120 – 140

3

50

Total

N = Σf= 50

 

Now, N = 50

⇒ N/2 = 25

The cumulative frequency just greater than 25 is 36 and the corresponding class is 60 – 80.

Thus, the median class is 60 – 80.

∴ l = 60, h = 20, N = 50, f = 12 and cf = 24.

Now,

Median = l + ((N/2 – cf)/f) × h

= 60 + (25 – 24)/12 × 20

= 60 + 1.67

= 61.67

Thus, the median is 61.67.

We know that,

Mode = 3(Median) – 2(mean)

= 3 × 61.67 – 2 × 62.4

= 185.01 – 124.8

= 60.21

Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21


3. Find the mean, median and mode of the following data:

Class

0 - 50

5 - 100

100 - 150

150 - 200

200 - 250

250 - 300

300 - 350

Frequency

2

3

5

6

5

3

1

Solution

To find the mean let us put the data in the table given below:

Class

Frequency (fi)

Class mark (xi)

fi xi

0 - 50

2

25

50

50 - 100

3

75

225

100 - 150

5

125

625

150 - 200

6

175

1050

200 - 250

5

225

1125

250 - 300

3

275

825

300 – 350

1

325

325

Total

Σf = 25

 

Σfi xi = 4225

= 4225/25

= 169

Thus, mean of the given data is 169.

Now, to find the median let us put the data in the table given below:

Class

Frequency (fi)

Cumulative frequency (cf)

0 - 50

2

2

50 - 100

3

5

100 - 150

5

10

150 - 200

6

16

200 – 250

5

21

250 – 300

3

24

300 – 350

1

25

Total

N = Σfi = 25

 

Now, N = 25, ⇒ N/2 = 12.5

The cumulative frequency just greater than 12.5 is 16 and the corresponding class is 150 – 200.

Thus, the median class is 150 – 200.

∴ l = 150, h = 50, N = 25, f = 6 and cf = 10.

Now,

Median = l + ((N/2 – cf)/f) × h

 = 150 + (12.5 – 10)/6 × 50

= 150 + 20.83

= 170.83

Thus, the median is 170.83

We know that,

Mode = 3(Median) – 2(mean)

= 3 × 170.83 – 2 × 169

= 512.49 – 33.8

= 174.49

Hence, mean = 169, Median = 170.83 and Mode = 174.49


4. Find the mean, median and mode of the following data: 

Marks Obtained

25 - 35

35 - 45

45 - 55

55 - 65

65 - 75

75 - 85

No. of students

7

31

33

17

11

1

Solution 

To find the mean let us put the data in the table given below:

Marks Obtained

Number of students (fi)

Class mark (xi)

fxi

25 - 35

7

30

210

35 - 45

31

40

1240

45 - 55

33

50

1650

55 - 65

17

60

1020

65 - 75

11

70

770

75 - 85

1

80

80

Total

Σfi = 100

 

Σfi xi = 4970

= 4970/100

= 49.7

Thus, mean of the given data is 49.7.

Now, to find the median let us put the data in the table given below:

Class

Frequency (fi)

Cumulative frequency (cf)

25 - 35

7

7

35 - 45

31

38

45 - 55

33

71

55 - 65

17

88

65 - 75

11

99

75 - 85

1

100

Total

N = Σfi = 100

 

Now, N = 100 ⇒ N/2 = 50.

The cumulative frequency just greater than 50 is 71 and the corresponding class is 45 – 55.

Thus, the median class is 45 – 55.

∴ l = 45, h = 10, N = 100, f = 33 and cf = 38.

Now,

Median = l + ((N/2 – cf)/f) × h

= 45 + (50 – 38)/33 × 10

= 45 + 3.64

= 48.64

Thus, the median is 48.64

We know that,

Mode = 3(Median) – 2(mean)

= 3 × 48.64 – 2 × 49.70

= 145.92 – 99.4

= 46.52

Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52


5. A survey regarding the heights (in cm) of 50 girls of a class was conducted and the following data was obtained:

Height in cm

120 - 130

130 - 140

140 - 150

150 - 160

160 - 170

No. of girls

2

8

12

20

8

Find the mean, median and mode of the above data.

Solution

We have the following:

Height in cm

Mid value (xi)

Frequency (fi)

Cumulative frequency

(fi × xi)

120 - 130

125

2

2

250

130 - 140

135

8

10

1080

140 - 150

145

12

22

1740

150 - 160

155

20

42

3100

160 - 170

165

8

50

1320

 

 

Σf= 50

 

Σfi × xi = 7490


= 7490/50

= 149.8

Now, N = 50

⇒ N/2 = 25.

The cumulative frequency just greater than 25 is 42 and the corresponding class is 150 – 160.

Thus, the median class is 150 – 160.

∴ l = 150, h = 10, f = 20, c = cf of preceding class=22 and N/2 = 25

Now,

Median, Me = l + {h × ((N/2 – c)/f)}

= 150 + {10 × ((25 – 22)/20)}

= (150 + 10 × 3/20)

= 151.5

Mode = 3(median) – 2(Mean)

= 3 × 151.5 – 2 × 149.8

= 154.9


6. The following table gives the daily income of 50 workers of a factory:

Daily income (in Rs)

100 - 120

120 - 140

140 - 160

160 - 180

180 - 200

No. of workers

12

14

8

6

10

Find the mean, median and mode of the above data.

Solution

We have the following:

Daily income

Mid values (xi)

Frequency (fi)

Cumulative frequency

(fi × xi)

100 - 120

110

12

12

1320

120 - 140

130

14

26

1820

140 - 160

150

8

34

1200

160 - 180

170

6

40

1020

180 - 200

190

10

50

1900

 

 

Σfi = 50

 

Σfi × xi = 7260

= 7260/50

= 145.2

Now, N = 50

⇒ N/2 = 25

The cumulative frequency just greater than 25 is 26 and the corresponding class is 120 – 140.

Thus, the median class is 120 – 140.

∴ l = 120, h = 20, f = 14, c = cf of preceding class=12 and N/2 = 25

Now,

Median, Me = l + {h × ((N/2 – c)/f)}

= 120 + {20 × (25 – 12)/14}

= (120 + 20 × 13/14)

= 138.57

Mode = 3(median) – 2(mean)

= 3 × 138.57 – 2 × 145.2

= 125.31


7. The table shows the daily expenditure on food of 30 households in a locality:

Daily expenditure (in Rs)

Number of households

100 - 150

6

150 - 200

7

200 - 250

12

250 - 300

3

300 - 350

2

Find the mean and median daily expenditure on food.

Solution

We have the following:

Daily expenditure (in Rs)

Mid value (xi)

Frequency (fi)

Cumulative frequency

(f× xi)

100 - 150

125

6

6

750

150 - 200

175

7

13

1225

200 - 250

225

12

25

2700

250 - 300

275

3

28

825

300 - 350

325

2

30

650

 

 

Σfi = 30

 

Σfi × xi = 6150


= 6150/30

= 205

Now, N = 30

N/2 = 15

The cumulative frequency just greater than 15 is 25 and the corresponding class is 200 – 250.

Thus, the median class is 200 – 250.

l = 200, h = 50, f = 12, c = cf of preceding class = 13 and N/2 = 15

Now,

Median, Me = l + {h × (N/2 – c)/f)}

= 200 + {50 + (15 – 13)/12)}

= (200 + 50 × 2/12)

= 200 + 8.33

= 208.33

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