RS Aggarwal Solutions for Class 9 Chapter 17: Bar Graph, Histogram and Frequency Polygon Exercise 17B

RS Aggarwal Solutions for Class 9 Maths Exercise 17B PDF

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RS Aggarwal Solutions for Class 9 Chapter 17: Bar Graph, Histogram and Frequency Polygon Exercise 17B Download PDF

 

RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B
RS Aggarwal Solutions Class 9 Maths Chapter 17B

 

Access RS Aggarwal Solutions for Class 9 Chapter 17: Bar Graph, Histogram and Frequency Polygon Exercise 17B

Exercise 17(B) page: 658

1. The daily wages of 50 workers in a factory are given below:

Daily wages (in ₹)

Number of workers

340 – 380

16

380 – 420

9

420 – 460

12

460 – 500

2

500 – 540

7

540 – 580

4

Construct a histogram to represent the above frequency distribution.

Solution:

The frequency distribution is as follows:

Daily wages (in ₹)

Number of workers

340 – 380

16

380 – 420

9

420 – 460

12

460 – 500

2

500 – 540

7

540 – 580

4

We know that of the upper limit of one class is the lower limit of next class then exclusive method of classification is necessary. So the frequency distribution is in the exclusive form.

Consider the class intervals that is Daily wages (in ₹) on the X – axis and number of workers on the Y- axis and draw rectangles which provides the required histogram.

A break is indicated near the origin on the X axis since the scale starts at 340.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 1

2. The following table shows the average daily earnings of 40 general stores in a market, during a certain week:

Daily earning (in rupees)

Number of stores

700 – 750

6

750 – 800

9

800 – 850

2

850 – 900

7

900 – 950

11

950 – 1000

5

Draw a histogram to represent the above data.

Solution:

The frequency distribution is as follows:

Daily earning (in rupees)

Number of stores

700 – 750

6

750 – 800

9

800 – 850

2

850 – 900

7

900 – 950

11

950 – 1000

5

We know that of the upper limit of one class is the lower limit of next class then exclusive method of classification is necessary. So the frequency distribution is in the exclusive form.

Consider the class intervals that is Daily earning in rupees on the X – axis and number of stores on the Y- axis and draw rectangles which provides the required histogram.

A break is indicated near the origin on the X axis since the scale starts at 700.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 2

3. The heights of 75 students in a school are given below:

Height (in cm)

Number of students

130 – 136

9

136 – 142

12

142 – 148

18

148 – 154

23

154 – 160

10

160 – 166

3

Draw a histogram to represent the above data.

Solution:

The frequency distribution is as follows:

Height (in cm)

Number of students

130 – 136

9

136 – 142

12

142 – 148

18

148 – 154

23

154 – 160

10

160 – 166

3

We know that of the upper limit of one class is the lower limit of next class then exclusive method of classification is necessary. So the frequency distribution is in the exclusive form.

Consider the class intervals that is height in cm on the X – axis and number of students on the Y- axis and draw rectangles which provides the required histogram.

A break is indicated near the origin on the X axis since the scale starts at 130.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 3

4. The following table gives the lifetimes of 400 neon lamps:

Lifetime (in hr)

Number of lamps

300 – 400

14

400 – 500

56

500 – 600

60

600 – 700

86

700 – 800

74

800 – 900

62

900 – 1000

48

(i) Represent the given information with the help of a histogram.

(ii) How many lamps have a lifetime of more than 700 hours?

Solution:

(i) Histogram is given below:

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 4

(ii) Number of lamps having a lifetime of more than 700 hours = (74 + 62 + 48) = 184.

5. Draw a histogram for the frequency distribution of the following data:

Class interval

Frequency

8 – 13

320

13 – 18

780

18 – 23

160

23 – 28

540

28 – 33

260

33 – 38

100

38 – 43

80

Solution:

The frequency distribution is as follows:

Class interval

Frequency

8 – 13

320

13 – 18

780

18 – 23

160

23 – 28

540

28 – 33

260

33 – 38

100

38 – 43

80

We know that of the upper limit of one class is the lower limit of next class then exclusive method of classification is necessary. So the frequency distribution is in the exclusive form.

Consider the class intervals on the X – axis and frequency on the Y- axis and draw rectangles which provides the required histogram.

A break is indicated near the origin on the X axis since the scale starts at 8.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 5

6. Construct a histogram for the following frequency distribution:

Class interval

Frequency

5 – 12

6

13 – 20

15

21 – 28

24

29 – 36

18

37 – 44

4

45 – 52

9

Solution:

Convert the given inclusive form into an exclusive form by taking intervals as given below:

Class interval

Frequency

4.5 – 12.5

6

12.5 – 20.5

15

20.5 – 28.5

24

28.5 – 36.5

18

36.5 – 44.5

4

454.5 – 52.5

9

Consider the class intervals on the X – axis and frequency on the Y- axis and draw rectangles which provides the required histogram.

A break is indicated near the origin on the X axis since the scale starts at 4.5.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 6

7. The following table shows the number of illiterate persons in the age group (10 – 58 years) in a town:

Age group (in years)

Number of illiterate persons

10 – 16

175

17 – 23

325

24 – 30

100

31 – 27

150

38 – 44

250

45 – 51

400

52 – 58

525

Draw a histogram to represent the above data.

Solution:

Convert the given inclusive form into an exclusive form by taking intervals as given below:

Age group (in years)

Number of illiterate persons

9.5 – 16.5

175

16.5 – 23.5

325

23.5 – 30.5

100

30.5 – 37.5

150

37.5 – 44.5

250

44.5 – 51.5

400

51.5 – 58.5

525

Consider the class intervals that is age group in years on the X – axis and number of illiterate persons on the Y- axis and draw rectangles which provides the required histogram.

A break is indicated near the origin on the X axis since the scale starts at 9.5.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 7

8. Draw a histogram to represent the following data:

Class interval

Frequency

10 – 14

5

14 – 20

6

20 – 32

9

32 – 52

25

52 – 80

21

Solution:

The frequency distribution is as follows:

Class interval

Frequency

10 – 14

5

14 – 20

6

20 – 32

9

32 – 52

25

52 – 80

21

The class sizes are 4, 6, 12, 20 and 28

So the minimum class size is 4

We know that

Adjusted frequency = (minimum class size × its frequency)/ class size of this class

On further calculation we get

4/4 × 5 = 5

4/6 × 6 = 4

4/12 × 9 = 3

4/20 × 25 = 5

4/28 × 21 = 3

So the adjusted frequency table is given below:

Class interval

Frequency

10 – 14

5

14 – 20

4

20 – 32

3

32 – 52

5

52 – 80

3

Consider the class intervals on the X – axis and frequency on the Y- axis and draw rectangles which provides the required histogram.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 8

9. 100 surnames were randomly picked up from a local telephone directory and frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of letters

Number of surnames

1 – 4

6

4 – 6

30

6 – 8

44

8 – 12

16

12 – 20

4

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.

Solution:

(i) The frequency distribution is as follows:

Number of letters

Number of surnames

1 – 4

6

4 – 6

30

6 – 8

44

8 – 12

16

12 – 20

4

The class sizes are 3, 2, 2, 4 and 8

So the minimum class size is 2

We know that

Adjusted frequency = (minimum class size × its frequency)/ class size of this class

On further calculation we get

2/3 × 6 = 4

2/2 × 30 = 30

2/2 × 44 = 44

2/4 × 16 = 8

2/8 × 4 = 1

So the adjusted frequency table is given below:

Number of letters

Number of surnames

1 – 4

4

4 – 6

30

6 – 8

44

8 – 12

8

12 – 20

1

Consider the class intervals that is number of letters on the X – axis and number of surnames on the Y- axis and draw rectangles which provides the required histogram.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 9

(ii) Maximum surnames lie in class interval 6 – 8.

10. Draw a histogram to represent the following information:

Class interval

Frequency

5 – 10

6

10 – 15

12

15 – 25

10

25 – 45

8

45 – 75

18

Solution:

The frequency distribution is as follows:

Class interval

Frequency

5 – 10

6

10 – 15

12

15 – 25

10

25 – 45

8

45 – 75

18

The class sizes are 5, 5, 10, 2 and 30

So the minimum class size is 5

We know that

Adjusted frequency = (minimum class size × its frequency)/ class size of this class

On further calculation we get

5/5 × 6 = 6

5/5 × 12 = 12

5/10 × 10 = 5

5/20 × 8 = 2

5/30 × 18 = 3

So the adjusted frequency table is given below:

Class interval

Frequency

5 – 10

6

10 – 15

12

15 – 25

5

25 – 45

2

45 – 75

3

Consider the class intervals on the X – axis and frequency on the Y- axis and draw rectangles which provides the required histogram.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 10

11. Draw a histogram to represent the following information:

Marks

Number of students

0 – 10

8

10 – 30

32

30 – 45

18

45 – 50

10

50 – 60

6

Solution:

The frequency distribution is as follows:

Marks

Number of students

0 – 10

8

10 – 30

32

30 – 45

18

45 – 50

10

50 – 60

6

The class sizes are 10, 20, 15, 5 and 10.

So the minimum class size is 5

We know that

Adjusted frequency = (minimum class size × its frequency)/ class size of this class

On further calculation we get

5/10 × 8 = 4

5/20 × 32 = 8

5/15 × 18 = 6

5/5 × 10 = 10

5/10 × 6 = 3

So the adjusted frequency table is given below:

Marks

Number of students

0 – 10

4

10 – 30

8

30 – 45

6

45 – 50

10

50 – 60

3

Consider the class intervals that is marks on the X – axis and number of students on the Y- axis and draw rectangles which provides the required histogram.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 11

12. In a study of diabetic patients in a village, the following observations were noted.

Age in years

Number of patients

10 – 20

2

20 – 30

5

30 – 40

12

40 – 50

19

50 – 60

9

60 – 70

4

Represent the above data by a frequency polygon.

Solution:

The frequency distribution is as follows:

Age in years

Number of patients

10 – 20

2

20 – 30

5

30 – 40

12

40 – 50

19

50 – 60

9

60 – 70

4

We know that

Class mark of a class interval = (lower limit + upper limit)/ 2

Age in years

Class marks

Frequency

0 – 10

5

0

10 – 2

15

2

20 – 30

25

5

30 – 40

35

12

40 – 50

45

19

50 – 60

55

9

60 – 70

65

4

70 – 80

75

0

By taking imaginary class intervals 0 – 10 and 70 – 80 with frequency 0 plot the points by taking class marks on X-axis and frequencies on the Y-axis.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 12

13. Draw a frequency polygon for the following frequency distribution:

Class interval

Frequency

1 – 10

8

11 – 20

3

21 – 30

6

31 – 40

12

41 – 50

2

51 – 60

7

Solution:

The frequency distribution is as follows:

Class interval

Frequency

1 – 10

8

11 – 20

3

21 – 30

6

31 – 40

12

41 – 50

2

51 – 60

7

Convert the given inclusive form into an exclusive form by taking intervals as 0.5 – 1.5, 10.5 – 20.5, 20.5 – 30.5, 40.5 – 40.5, 40.5 – 50.5 and 50.5 – 60.5

We know that

Class mark of a class interval = (lower limit + upper limit)/ 2

Consider – 9.5 – 0.5 and 60.5 – 70.5 with frequency zero the table is as follows:

Class interval

True class intervals

Class marks

Frequency

(-9) – 0

(-9.5) – 0.5

-4.5

0

1 – 10

0.5 – 10.5

5.5

8

11 – 20

10.5 – 20.5

15.5

3

21 – 30

20.5 – 30.5

25.5

6

31 – 40

30.5 – 40.5

35.5

12

41 – 50

40.5 – 50.5

45.4

2

51 – 60

50.5 – 60.5

55.5

7

61 – 70

60.5 – 70.5

65.5

0

Consider the class intervals that is marks on the X – axis and frequency on the Y- axis.

Plot the points and join them to obtain the frequency polygon.

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 13

14. The age (in years) of 360 patients treated in a hospital on a particular day are given below:

Age in years

Number of patients

10 – 20

90

20 – 30

40

30 – 40

60

40 – 50

20

50 – 60

120

60 – 70

30

Draw a histogram and frequency polygon on the same graph to represent the above data.

Solution:

The frequency distribution is as follows:

Age in years

Number of patients

10 – 20

90

20 – 30

40

30 – 40

60

40 – 50

20

50 – 60

120

60 – 70

30

Consider the class intervals that is age in years on the X – axis and number of patients on the Y- axis and draw rectangles which provides the required histogram.

To construct a frequency polygon, take 0 – 10 and 70 – 80 as imaginary class intervals having class marks 5 and 75 with frequency zero and join the midpoints of rectangles i.e. A (5, 0) and B (75, 0).

So the frequency polygon is as given below:

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 14

15. Draw a histogram and the frequency polygon from the following data:

Class interval

Frequency

20 – 25

30

25 – 30

24

30 – 35

52

35 – 40

28

40 – 45

46

45 – 50

10

Solution:

The frequency distribution is as follows:

Class interval

Frequency

20 – 25

30

25 – 30

24

30 – 35

52

35 – 40

28

40 – 45

46

45 – 50

10

Consider the class intervals on the X – axis and frequency on the Y- axis and draw rectangles which provides the required histogram.

To construct a frequency polygon, take 15 – 20 and 50 – 55 as imaginary class intervals with frequency zero and join the midpoints of rectangles.

So the frequency polygon is as given below:

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 15

16. Draw a histogram for the following data:

Class interval

Frequency

600 – 640

18

640 – 680

45

680 – 720

153

720 – 760

288

760 – 800

171

800 – 840

63

Using this histogram, draw the frequency polygon on the same graph.

Solution:

The frequency distribution is as follows:

Class interval

Frequency

600 – 640

18

640 – 680

45

680 – 720

153

720 – 760

288

760 – 800

171

800 – 840

63

Consider the class intervals on the X – axis and frequency on the Y- axis and draw rectangles which provides the required histogram.

To construct a frequency polygon, take 560 – 600 and 840 – 880 as imaginary class intervals with frequency zero and join the midpoints of rectangles.

So the frequency polygon is as given below:

RS Aggarwal Solutions for Class 9 Maths Chapter 17 Ex 17B Image 16


Access other exercise solutions of Class 9 Maths Chapter 17: Bar Graph, Histogram and Frequency Polygon

Exercise 17A Solutions 12 Questions

RS Aggarwal Solutions Class 9 Maths Chapter 17 – Bar Graph, Histogram and Frequency Polygon Exercise 17B

RS Aggarwal Solutions Class 9 Maths Chapter 17 Bar Graph, Histogram and Frequency Polygon Exercise 17B explains the methods used in drawing a histogram and frequency polygon for the exercise problems.

Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 17: Bar Graph, Histogram and Frequency Polygon Exercise 17B

  • Solving the examples and exercise wise problems improve the speed in answering questions in the board exam.
  • The problems in RS Aggarwal textbook are based on the latest CBSE syllabus and guidelines.
  • The students by solving huge numbers of problems can understand which concepts require more importance based on weightage.
  • This exercise helps students with the various methods which can be used in drawing a frequency polygon and histogram.