# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 17 – Bar Graph, Histogram and Frequency Polygon

## Chapter 17 – Bar Graph, Histogram and Frequency Polygon Exercise Ex. 17A

Question 1

The following table shows the number of students participating in various games in a school.

 Game Cricket Football Basketball Tennis Number of students 27 36 18 12

Draw a bar graph to represent the above data.

Hint: Along the y-axis, take 1 small square=3 units.

Solution 1

Take the various types of games along the x-axis and the number of students along the y-axis.

Along the y-axis, take 1 small square=3 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below: Question 2

Given below are the seats won by
different political parties in the polling outcome of a state assembly
elections:

 Political party A B C D E F Seats won 65 52 34 28 10 31

Draw a bar graph to represent the
polling results.

Solution 2 Question 3

Various modes of transport used by 1850 students of a school are given below.

 School bus Private bus Bicycle Rickshaw By foot 640 360 490 210 150

Draw a bar graph to represent the above data.

Solution 3

Take themode of transport along the x-axis and the number of students along the y-axis.

Along the y-axis, take 1 big division = 100 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below: Question 4

Look at the bar graph given below. (i) What information does the bar graph give?

(ii) In which subject does the student very good?

(iii) In which subject is he poor?

(iv) What is the average of the marks?

Solution 4

(i) The bar graph shows the marks obtained by a student in various subject in an examination.

(ii) The student is very good in mathematics.

(iii) He is poor in Hindi

(iv)  Average marks = Question 5

On a certain day, the tempreture in a city was recorded as under.

 Times 5 a.m 8 a.m 11a.m 3p.m 6p.m Tempreture (in 0C) 20 24 26 22 18

Illustrate the data by a bar graph.

Solution 5

Take the timings along the x-axis and the temperatures along the y-axis.

Along the y-axis, take 1 small square=5 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below: Question 6

The approximate velocities of some
vehicles are given below:

 Name of vehicle Bicycle Scooter Car Bus Train Velocity (in km/hr) 27 45 90 72 63

Draw bar graph to represent the
above data.

Solution 6 Question 7

The following table shows the favorite sports of 250 students of a school. Represent the data by a bar graph.

 Sports Cricket Football Tennis Badminton Swimming No. of Students 75 35 50 25 65
Solution 7

Take the various types of sports along the x-axis and the number of students along the y-axis.

Along the y-axis, take 1 small square=10 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below: Question 8

Given below is a table which shows the year wise strength of a school. Represent this data by a bar graph.

 Year 2012-13 2013-14 2014-15 2015-16 2016-17 No of students 800 975 1100 1400 1625
Solution 8

Take the academic year along the x-axis and the number of students along the y-axis.

Along the y-axis, take 1 big division =200 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below: Question 9

The following table shows the
number of scooters sold by a dealer during six consecutive years. Draw a bar
graph to represent this data.

 Year 2011 2012 2013 2014 2015 2016 Number of scooters sold (in thousand) 16 20 32 36 40 48

Solution 9 Question 10

The air distances of four cities from Delhi (in km) are given below :

 City Kolkata Mumbai Chennai Hyderabad Distance from Delhi(in km) 1340 1100 1700 1220

Draw a bar graph to represent the above data.

Solution 10

Take city along the x-axis and distance from Delhi (in Km) along the y-axis.

Along the y-axis, take 1 big division =200 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below: Question 11

The birth rate per thousand in five countries over a period of time shown below:

 Country China India Germany UK Sweden Birth ratePer thousand 42 35 14 28 21

Represent the above data by a bar graph.

Solution 11

Take the countries along the x-axis and the birth rate (per thousand) along the y-axis.

Along the y-axis, take 1 big division = 5 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below: Question 12

The following table shows the life
expectancy (average age to which people live) in various countries in a
particular year. Represent the data by a bar graph.

 Country Japan India Britain Ethiopia Cambodia UK Life expectancy(in years) 84 68 80 64 62 73

Solution 12 ## Chapter 17 – Bar Graph, Histogram and Frequency Polygon Exercise Ex. 17B

Question 1

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

 Daily wages in rupees 340-380 380-420 420-460 460-500 500-540 540-580 Number ofworkers 16 9 12 2 7 4

Construct a histogram to represent the above frequency distribution.

Solution 1

Given frequency distribution is as below :

 Daily wages (in Rs) 340-380 380-420 420-460 460-500 500-540 540-580 No. of workers 16 9 12 2 7 4

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

To draw the required histogram , take class intervals , i.e. daily wages (in Rs. ) along x-axis and frequencies i.e.no.of workers alongy-axisand draw rectangles . So , we get the requiredhistogram .

Since the scale on X-axis starts at 340, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 340. Question 2

Draw a histogram to represent the
following information:

 Class interval 5-10 10-15 15-25 25-45 45-75 Frequency 6 12 10 8 18

Solution 2

Minimum class size = 10 – 5 = 5  Question 3

Draw a histogram to represent the
following information:

 Marks 0-10 10-30 30-45 45-50 50-60 Number of students 8 32 18 10 6

Solution 3

Minimum
class size = 50 – 45 = 5  Question 4

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

 Age in years 10-20 20-30 30-40 40-50 50-60 60-70 Number of patients 2 5 12 19 9 4

Represent the above data by a frequency polygon.

Solution 4

The given frequency distribution is as below:

 Age in years 10-20 20-30 30-40 40-50 50-60 60-70 No of patients 2 5 12 19 9 4

In order to draw, frequency polygon, we require class marks.

The class mark of a class interval is: The frequency distribution table with class marks is given below:

 Class- intervals Class marks Frequency 0-1010-2020-3030-4040-5050-6060-7070-80 515253545556575 0 251219940

In the above table, we have taken imaginary class intervals 0-10 at beginning and 70-80 at the end, each with frequency zero . Now take class marks along x-axis and the corresponding frequencies along y-axis.

Plot points (5,0), (15,2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4) and (75, 0) and draw line segments. Question 5

Draw a frequency polygon for the following frequency distribution

 Class-interval 1-10 11-20 21-30 31-40 41-50 51-60 Frequency 8 3 6 12 2 7
Solution 5

The given frequency distribution table is as below:

 Class intervals 1-10 11-20 21-30 31-40 41-50 51-60 Frequency 8 3 6 12 2 7

This table has inclusive class intervals and so these are to be converted into exclusive class intervals (i.e true class limits).

These are (0.5-10.5), (10.5-20.5), (20.5-30.5), (30.5-40.5),
(40.5-50.5), and (50.5-60.5)

In order to draw a frequency polygon, we need to determine the class marks. Class marks of a class interval = Take imaginary class interval ( -9.5-0.5) at the beginning and (60.5-70.5) at the end , each with frequency zero. So we have the following table

 Class intervals True class intervals Class marks Frequency (-9)-01-1011-2021-3031-4041-5051-6061-70 (-9.5)-0.50.5-10.510.5-20.520.5-30.530.5-40.540.5-50.550.5-60.560.5-70.5 -4.55.515.525.535.545.555.565.5 083612270

Now, take class marks along x-axis and their corresponding frequencies along y-axis.

Mark the points and join them.

Thus, we obtain a complete frequency polygon as shown below: Question 6

The ages (in years) of 360 patients treated in a hospital on a particular dayare given below.

 Age in years 10-20 20-30 30-40 40-50 50-60 60-70 Number of patients 90 40 60 20 120 30

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

Solution 6

The given frequency distribution is as under

 Age in years 10-20 20-30 30-40 40-50 50-60 60-70 Numbers of patients 90 40 60 20 120 30

Take class intervals i.e age in years along x-axis and number of patients of width equal to the size of the class intervals and height equal to the corresponding frequencies.

Thus we get the required histogram.

In order to draw frequency polygon,we take imaginary intervals 0-10 at the beginning and 70-80 at the end each with frequency zero and join the mid-points of top of the rectangles.

Thus, we obtain a complete frequency polygon, shown below: Question 7

Draw a histogram and frequency polygon from the following data.

 Class interval 20-25 25-30 30-35 35-40 40-45 45-50 Frequency 30 24 52 28 46 10
Solution 7

The given frequency distribution is as below :

 Class intervals 20-25 25-30 30-35 35-40 40-45 45-50 Frequency 30 24 52 28 46 10

Take class intervals along x-axis and frequencies along y-axis and draw rectangle s of width equal to the size of the class intervals and heights equal to the corresponding frequencies.

Thus we get required histogram.

Now take imaginary class intervals 15-20 at the beginning and 50-55 at the end , each with frequency zero and join the mid points of top of the rectangles to get the required frequency polygon. Question 8

Draw a histogram for the following data:

 Class interval 600-640 640-680 680-720 720-760 760-800 800-840 Frequency 18 45 153 288 171 63

Usingthis histogram, draw the frequencypolygon on the same graph.

Solution 8

The given frequency distribution table is given below :

 Class interval 600-640 640-680 680-720 720-760 760-800 800-840 Frequency 18 45 153 288 171 63

Take class intervals along x-axis and frequencies along y-axis and draw rectangles of width equal to to size of class intervals and height equal to their corresponding frequencies.

Thus we get the requiredhistogram.

Take imaginary class intervals 560-600 at the beginning and 840-880 at the end, each with frequency zero.

Now join the mid points of the top of the rectangles to get the required frequency polygon. Question 9

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

 Daily earning (in rupees) 700-750 750-800 800-850 850-900 900-950 950-1000 Number ofStores 6 9 2 7 11 5

Draw a histogram to represent the above data.

Solution 9

Given frequency distribution is as below :

 Daily earnings (in Rs) 700-750 750-800 800-850 850-900 900-950 950-1000 No of stores 6 9 2 7 11 5

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

We take class intervals, i.e. daily earnings (in Rs .) along x-axis and frequencies i.e. number of stores along y-axis. So , we get the required histogram .

Since the scale on X-axis starts at 700, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 700. Question 10

the heights of 75 students in a school are given below :

 Height(in cm) 130-136 136-142 142-148 148-154 154-160 160-166 Number of students 9 12 18 23 10 3

Draw a histogram to represent the above data.

Solution 10

 Height(in cm) 130-136 136-142 142-148 148-154 154-160 160-166 No. of students 9 12 18 23 10 3

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

We take class intervals, i.e. height (in cm ) along x-axis and frequencies i.e. number of student s along y-axis . So we get the required histogram.

Since the scale on X-axis starts at 130, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 130. Question 11

The following table gives the

 Lifetime(in hr ) 300-400 400-500 500-600 600-700 700-800 800-900 900-1000 Number of lamps 14 56 60 86 74 62 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 11

(i) Histogram is as follows: (ii) Number of lamps having lifetime
more than 700 hours = 74 + 62 + 48 = 184

Question 12

Draw a histogram for frequency distribution of the following data.

 Class -Interval 8-13 13-18 18-23 23-28 28-33 33-38 38-43 Frequency 320 780 160 540 260 100 80

Solution 12

Give frequency distribution is as below :

 Class interval 8-13 13-18 18-23 23-28 28-33 33-38 38-43 Frequency 320 780 160 540 260 100 80

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

We take class intervals along x-axis and frequency along y-axis . So , we get the required histogram.

Since the scale on X-axis starts at 8, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 8. Question 13

Construct a histogram for the following frequency distribution.

 Class interval 5-12 13-20 21-28 29-36 37-44 45-52 Frequency 6 15 24 18 4 9

Solution 13

Histogram is the graphical representation of a frequency distribution in the form of rectangles, such that there is no gap between any two successive rectangles.

Clearly the given frequency distribution is in inclusive form, that is there is a gap between the upper limit of a class and the lower limit of the next class.

Therefore, we need to convert the given frequency distribution into exclusive form, as shown below:

 Class interval 4.5-12.5 12.5-20.5 20.5-28.5 28.5-36.5 36.5-44.5 44.5-52.5 Frequency 6 15 24 18 4 9

To draw the required histogram , take class intervals, along x-axis and frequencies along y-axis and draw rectangles . So, we get the required histogram .

Since the scale on X-axis starts at 4.5, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 4.5. Question 14

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

 Age group(in years) 10-16 17-23 24-30 31-37 38-44 45-51 52-58 Number of illiterate persons 175 325 100 150 250 400 525

Draw a histogram to represent the above data.

Solution 14

Given frequency distribution is as below :

 Age group (in years ) 10-16 17-23 24-30 31-37 38-44 45-51 52-58 No. of illiterate persons 175 325 100 150 250 400 525

Histogram is the graphical representation of a frequency distribution in the form of rectangles, such that there is no gap between any two successive rectangles.

Clearly the given frequency distribution is in inclusive form, that is there is a gap between the upper limit of a class and the lower limit of the next class.

Therefore, we need to convert the frequency distribution in exclusive form, as shown below:

 Age group(in years) 9.5-16.5 16.5-23.5 23.5-30.5 30.5-37.5 37.5-44.4 44.5-51.5 51.5-58.5 No of illiterate persons 175 325 100 150 250 400 525

To draw the required histogram , take class intervals, that is age group, along x-axis and frequencies, that is number of illiterate persons along y-axis and draw rectangles . So , we get the required histogram.

Since the scale on X-axis starts at 9.5, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 9.5. Question 15

Draw a histogram to represent the following data.

 Class -Interval 10-14 14-20 20-32 32-52 52-80 Frequency 5 6 9 25 21
Solution 15

given frequency distribution is as below :

 Class interval 10-14 14-20 20-32 32-52 52-80 Frequency 5 6 9 25 21

In the above table , class intervals are of unequal size, so we calculate the adjusted frequency by using the following formula : Thus , the adjusted frequency table is

 Class intervals frequency Adjusted Frequency 10-1414-2020-3232-5252-80 5692521  Now take class intervals along x-axis and adjusted frequency along y-axis and constant rectangles having their bases as class size and heights as the corresponding adjusted frequencies.

Thus, we obtain the histogram as shown below: Question 16

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 1-4 4-6 6-8 8-12 12-20 Number of surnames 6 30 44 16 4

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

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

Solution 16

(i) Minimum class size = 6 – 4 = 2  (ii) Maximum number of surnames lies in
the class interval 6 – 8.

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