RS Agarwal Solution | Class 9th | Chapter-17 |  Bar Graph, Histogram and Frequency Polygon| Edugrown

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

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 2

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 3

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 3

Question 4

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 4

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 5

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 5

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 6

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 6

Question 7

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 7

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 8

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 8

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 9

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 9

Question 10

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 10

Question 11

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 11

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 12

Look at the bar graph given below.

Read it carefully and answer the following questions.

(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 12

(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 =

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

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 2

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 3

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 3

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 4

The following table gives the lifetimes of 400 neon lamps:

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 4

(i) Histogram is as follows:

(ii) Number of lamps having lifetime more than 700 hours = 74 + 62 + 48 = 184Question 5

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 5

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 6

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 6

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 7

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 7

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 8

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 8

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

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 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	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 9

(i) Minimum class size = 6 – 4 = 2

(ii) Maximum number of surnames lies in the class interval 6 – 8.Question 10

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 10

Minimum class size = 10 – 5 = 5

Question 11

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 11

Minimum class size = 50 – 45 = 5

Question 12

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 12

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	0251219940

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 13

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 13

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 14

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 14

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 15

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 15

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 16

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 16

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.