RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18B

RS Aggarwal Solutions for Class 9 Maths Exercise 18B PDF

The students can use RS Aggarwal Solutions as a guide to solve the exercise wise problems based on latest CBSE guidelines. The solutions are accurate and explained in a stepwise manner for a better understanding of concepts among students. This exercise mainly contains the formula which can be used in finding means of ungrouped frequency distribution. The students can download PDF and use them as a source of reference for exam preparation. RS Aggarwal Solutions for Class 9 Maths Chapter 18 Mean, Median and Mode of Ungrouped Data Exercise 18B are provided here.

RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18B Download PDF

 

rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B
rs aggarwal solution class 9 maths chapter 18B

 

Access RS Aggarwal Solutions for Class 9 Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18B

1. Obtain the mean of the following distribution:

Variable (xi)

4

6

8

10

12

Frequency (fi)

4

8

14

11

3

Solution:

Variable (xi)

Frequency (fi)

fi xi

4

4

16

6

8

48

8

14

112

10

11

110

12

3

36

 

 Σ fi = 40

 Σ fi xi = 322

We know that

Mean = Σ fi xi/ Σ fi

By substituting the values

Mean = 322/40 = 8.05

2. The following table shows the weights of 12 workers in a factory:

Weight (in kg)

60

63

66

69

72

No. of workers

4

3

2

2

1

Find the mean weight of the workers.

Solution:

Weight (in kg) (xi)

No. of workers (fi)

fi xi

60

4

240

63

3

189

66

2

132

69

2

138

72

1

72

 

Σ fi = 12

Σ fi xi =771

We know that

Mean weight of the workers = Σ fi xi/ Σ fi

By substituting the values

Mean weight of the workers = 771/12 = 64.250kg

Therefore, the mean weight of the workers is 64.250kg.

3. The measurements (in mm) of the diameters of the heads of 50 screws are given below:

Diameter (in mm) (xi)

34

37

40

43

46

Number of screws (fi)

5

10

17

12

6

Calculate the mean diameter of the heads of the screws.

Solution:

Diameter (in mm) (xi)

Number of screws (fi)

 fi xi

34

5

170

37

10

370

40

17

680

43

12

516

46

6

276

 

Σ fi = 50

Σ fi xi = 2012

We know that

Mean diameter of the heads of the screws = Σ fi xi/ Σ fi

By substituting the values

Mean diameter of the heads of the screws = 2012/50 = 40.24mm

Therefore, the mean diameter of the heads of the screws is 40.24mm.

4. The following data give the number of boys of a particular age in a class of 40 students.

Age (in years)

15

16

17

18

19

20

Frequency (fi)

3

8

9

11

6

3

Calculate the mean age of the students.

Solution:

Age (in years) (xi)

Frequency (fi)

 fi xi

15

3

45

16

8

128

17

9

153

18

11

198

19

6

114

20

3

60

 

Σ fi = 40

Σ fi xi = 698

We know that

Mean age of the students = Σ fi xi/ Σ fi

By substituting the values

Mean age of the students = 698/40 = 17.45 years

Therefore, the mean age of the students is 17.45 years.

5. Find the mean of the following frequency distribution:

Variables (xi)

10

30

50

70

89

Frequency (fi)

7

8

10

15

10

Solution:

Variables (xi)

Frequency (fi)

 fi xi

10

7

70

30

8

240

50

10

500

70

15

1050

89

10

890

 

Σ fi = 50

Σ fi xi = 2750

We know that

Mean = Σ fi xi/ Σ fi

By substituting the values

Mean = 2750/50 = 55

6. Find the mean of daily wages of 40 workers in a factory as per data given below:

Daily wages (in ₹ ) (xi)

250

300

350

400

450

Number of workers (fi)

8

11

6

10

5

Solution:

Daily wages (in ₹ ) (xi)

Number of workers (fi)

  fi xi

250

8

2000

300

11

3300

350

6

2100

400

10

4000

450

5

2250

 

Σ fi = 40

Σ fi xi = 13650

We know that

Mean of daily wages of 40 workers in a factory = Σ fi xi/ Σ fi

By substituting the values

Mean of daily wages of 40 workers in a factory = 13650/40 = ₹ 341.25

Therefore, the mean of daily wages of 40 workers in a factory is ₹ 341.25.

7. If the mean of the following data is 20.2, find the value of p.

Variable (xi)

10

15

20

25

30

Frequency (fi)

6

8

p

10

6

Solution:

Variable (xi)

Frequency (fi)

   fi xi

10

6

60

15

8

120

20

p

20p

25

10

250

30

6

180

 

Σ fi = 30 + p

Σ fi xi = 610 + 20p

It is given that

Mean = 20.2

We know that

Mean = Σ fi xi/ Σ fi

So we get

(610 + 20p)/ (30 + p) = 20.2

On further calculation

610 + 20p = 606 + 20.2p

So we get

0.2p = 4

By division

p = 20

Therefore, the value of p is 20.

8. If the mean of the following data is 8, find the value of p.

x

3

5

7

9

11

13

f

6

8

15

p

8

4

Solution:

x

f

    fi xi

3

6

18

5

8

40

7

15

105

9

p

9p

11

8

88

13

4

52

 

Σ fi = 41 + p

Σ fi xi = 303 + 9p

It is given that

Mean = 8

We know that

Mean = Σ fi xi/ Σ fi

So we get

(303 + 9p)/ (41 + p) = 8

On further calculation

303 + 9p = 8 (41 + p)

So we get

303 + 9p = 328 + 8p

It can be written as

9p – 8p = 328 – 303

We get

p = 25

Therefore, the value of p is 25.

9. Find the missing frequency p for the following frequency distribution whose mean is 28.25.

x

15

20

25

30

35

40

f

8

7

p

14

15

6

Solution:

x

f

    fi xi 

15

8

120

20

7

140

25

p

25p

30

14

420

35

15

525

40

6

240

 

Σ fi = 50 + p

Σ fi xi = 1445 + 25p

It is given that

Mean = 28.25

We know that

Mean = Σ fi xi/ Σ fi

So we get

(1445 + 25p)/ (50 + p) = 28.25

On further calculation

1445 + 25p = 1412.50 + 28.25p

So we get

-28.25p + 25p = -1445 + 1412.50

It can be written as

-3.25p = -32.5

By division

p = 10

Therefore, the value of p is 10.

10. Find the value of p for the following frequency distribution whose mean is 16.6.

x

8

12

15

p

20

25

30

f

12

16

20

24

16

8

4

Solution:

x

f

     fi xi 

8

12

96

12

16

192

15

20

300

p

24

24p

20

16

320

25

8

200

30

4

120

 

Σ fi = 100

Σ fi xi = 1228 + 24p

It is given that

Mean = 16.6

We know that

Mean = Σ fi xi/ Σ fi

So we get

(1228 + 24p)/ 100 = 16.6

On further calculation

1228 + 24p = 1660

So we get

24p = 1660 – 1228

By subtraction

24p = 432

We get

p = 18

Therefore, the value of p is 18.

11. Find the missing frequencies in the following frequency distribution whose mean is 34.

x

10

20

30

40

50

6

Total

f

4

f1

8

f2

3

4

35

Solution:

x

f

 fi xi 

10

4

40

20

f1

20f1

30

8

240

40

f2

40f2

50

3

150

6

4

240

Total

35 = Σ fi =19 + f1 + f2

Σ fi xi = 670 + 20f1 + 40f2

We know that

Σ fi =19 + f1 + f2

It can be written as

35 = 19 + f1 + f2

So we get

f1 + f2 = 16 …… (1)

It is given that

Mean = 34

We know that

Mean = Σ fi xi/ Σ fi

So we get

(670 + 20f1 + 40f2)/ 35 = 34

On further calculation

(335 + 10f1 + 20f2)/ 35 = 17

So we get

335 + 10f1 + 20f2 = 595

It can be written as

10f1 + 20f2 = 260

Dividing the equation by 10

f1 + 2f2 = 26 …….. (2)

By subtracting equation (1) from (2)

f2 = 10

Substituting in equation (1)

f1 + 10 = 16

So we get

f1 = 6

Therefore, the missing frequencies are 6 and 10.

12. Find the missing frequencies in the following frequency distribution whose mean is 50.

x

10

30

50

70

90

Total

f

17

f1

32

f2

19

120

Solution:

x

f

  fi xi 

10

17

170

30

f1

30f1

50

32

1600

70

f2

70f2

90

19

1710

Total

120 = Σ fi =68 + f1 + f2

Σ fi xi = 3480 + 30f1 + 70f2

We know that

Σ fi =68 + f1 + f2

It can be written as

120 = 68 + f1 + f2

So we get

f1 + f2 = 52

f2 = 52 – f1 ……. (1)

It is given that

Mean = 50

We know that

Mean = Σ fi xi/ Σ fi

So we get

(3480 + 30f1 + 70f2)/ 120 = 50

Substituting equation (1)

3480 + 30f1 + 70(52 – f1) / 120 = 50

On further calculation

(3480 + 30f1 + 3640 – 70f1)/ 120 = 50

So we get

(7120 – 40f1)/ 120 = 50

It can be written as

6000 = 7120 – 40f1

By subtraction

40f1 = 1120

By division

f1 = 28

By substituting it in equation (1)

f2 = 52 – 28 = 24

Therefore, the missing frequencies are 28 and 24.

13. Find the value of p, when the mean of the following distribution is 20.

x

15

17

19

20 + p

23

f

2

3

4

5p

6

Solution:

x

f

   fi xi 

15

2

30

17

3

51

19

4

76

20 + p

5p

100p + 5p2

23

6

138

 

Σ fi = 15 + 5p

Σ fi xi = 5p2 + 100p + 295

It is given that

Mean = 20

We know that

Mean = Σ fi xi/ Σ fi

So we get

(5p2 + 100p + 295)/ (15 + 5p) = 20

On further calculation

5p2 + 100p + 295 = 300 + 100p

So we get

5p2 = 5

By division

p2 = 1

By taking square root

p = 1

Therefore, the value of p is 1.

14. The mean of the following distribution is 50.

x

10

30

50

70

90

f

17

5a + 3

32

7a – 11

19

Find the value of a and hence the frequencies of 30 and 70.

Solution:

x

f

 fi xi 

10

17

170

30

5a + 3

150a + 90

50

32

1600

70

7a – 11

490a – 770

90

19

171

 

Σ fi = 60 + 12a

Σ fi xi = 2800 + 640a

It is given that

Mean = 50

We know that

Mean = Σ fi xi/ Σ fi

So we get

(2800 + 640a)/ (60 + 12a) = 50

On further calculation

2800 + 640a = 3000 + 600a

So we get

40a = 200

By division

a = 5

We know that

Frequency of 30 = 5a + 3

By substituting the values

Frequency of 30 = 5 (5) + 3

So we get

Frequency of 30 = 28

We know that

Frequency of 70 = 7a – 11

By substituting the values

Frequency of 70 = 7 (5) – 11

So we get

Frequency of 70 = 24

Therefore, the value of a is 5 and the frequencies of 30 and 70 is 28 and 24.


 

s 9 Maths Chapter 18: Mean, Median and Mode of Ungrouped Data

Exercise 18A Solutions 31 Questions

Exercise 18C Solutions 10 Questions

Exercise 18D Solutions 8 Questions

RS Aggarwal Solutions Class 9 Maths Chapter 18 – Mean, Median and Mode of Ungrouped Data Exercise 18B

RS Aggarwal Solutions Class 9 Maths Chapter 18 Mean, Median and Mode of Ungrouped Data Exercise 18B explains the method of finding means of an ungrouped frequency distribution otherwise known as the direct method.

Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 18: Mean, Median and Mode of Ungrouped Data Exercise 18B

  • The students can use PDF of solutions while solving the RS Aggarwal textbook to find the other methods of answering problems.
  • Time management abilities are improved among students, which is one of the important aspects from the exam point of view.
  • The answers are explained in simple language which match the understanding capacity of students.
  • The subject experts prepare answers based on CBSE guidelines to help students obtain better academic scores.