R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 17  Bar Graph, Histogram and Frequency Polygon
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Chapter 17  Bar Graph, Histogram and Frequency Polygon Exercise Ex. 17A
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 yaxis, take 1 small square=3 units.
Take the various types of games along the xaxis and the number of students along the yaxis.
Along the yaxis, 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:
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.
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.
Take themode of transport along the xaxis and the number of students along the yaxis.
Along the yaxis, 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:
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?
(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 =
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 ^{0}C) 
20 
24 
26 
22 
18 
Illustrate the data by a bar graph.
Take the timings along the xaxis and the temperatures along the yaxis.
Along the yaxis, 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:
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.
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 
Take the various types of sports along the xaxis and the number of students along the yaxis.
Along the yaxis, 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:
Given below is a table which shows the year wise strength of a school. Represent this data by a bar graph.
Year 
201213 
201314 
201415 
201516 
201617 
No of students 
800 
975 
1100 
1400 
1625

Take the academic year along the xaxis and the number of students along the yaxis.
Along the yaxis, 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:
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 
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.
Take city along the xaxis and distance from Delhi (in Km) along the yaxis.
Along the yaxis, 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:
The birth rate per thousand in five countries over a period of time shown below:
Country 
China 
India 
Germany 
UK 
Sweden 
Birth rate Per thousand 
42 
35 
14 
28 
21 
Represent the above data by a bar graph.
Take the countries along the xaxis and the birth rate (per thousand) along the yaxis.
Along the yaxis, 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:
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 
Chapter 17  Bar Graph, Histogram and Frequency Polygon Exercise Ex. 17B
The daily wages of 50 workers in a factory are given below :
Daily wages in rupees 
340380 
380420 
420460 
460500 
500540 
540580 
Number of workers 
16 
9 
12 
2 
7 
4 
Construct a histogram to represent the above frequency distribution.
Given frequency distribution is as below :
Daily wages (in Rs) 
340380 
380420 
420460 
460500 
500540 
540580 
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 xaxis and frequencies i.e.no.of workers alongyaxisand draw rectangles . So , we get the requiredhistogram .
Since the scale on Xaxis starts at 340, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 340.
Draw a histogram to represent the following information:
Class interval 
510 
1015 
1525 
2545 
4575 
Frequency 
6 
12 
10 
8 
18 
Minimum class size = 10  5 = 5
Draw a histogram to represent the following information:
Marks 
010 
1030 
3045 
4550 
5060 
Number of students 
8 
32 
18 
10 
6 
Minimum class size = 50  45 = 5
In a study of diabetic patients in a village , the following observations were noted.
Age in years 
1020 
2030 
3040 
4050 
5060 
6070 
Number of patients 
2 
5 
12 
19 
9 
4 
Represent the above data by a frequency polygon.
The given frequency distribution is as below:
Age in years 
1020 
2030 
3040 
4050 
5060 
6070 
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 
010 1020 2030 3040 4050 5060 6070 7080 
5 15 25 35 45 55 65 75 
0 2 5 12 19 9 4 0 
In the above table, we have taken imaginary class intervals 010 at beginning and 7080 at the end, each with frequency zero . Now take class marks along xaxis and the corresponding frequencies along yaxis.
Plot points (5,0), (15,2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4) and (75, 0) and draw line segments.
Draw a frequency polygon for the following frequency distribution
Classinterval 
110 
1120 
2130 
3140 
4150 
5160 
Frequency 
8 
3 
6 
12 
2 
7 
The given frequency distribution table is as below:
Class intervals 
110 
1120 
2130 
3140 
4150 
5160 
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 (0.510.5), (10.520.5), (20.530.5), (30.540.5),
(40.550.5), and (50.560.5)
In order to draw a frequency polygon, we need to determine the class marks. Class marks of a class interval =_{}
Take imaginary class interval ( 9.50.5) at the beginning and (60.570.5) at the end , each with frequency zero. So we have the following table
Class intervals 
True class intervals 
Class marks 
Frequency 
(9)0 110 1120 2130 3140 4150 5160 6170

(9.5)0.5 0.510.5 10.520.5 20.530.5 30.540.5 40.550.5 50.560.5 60.570.5 
4.5 5.5 15.5 25.5 35.5 45.5 55.5 65.5 
0 8 3 6 12 2 7 0 
Now, take class marks along xaxis and their corresponding frequencies along yaxis.
Mark the points and join them.
Thus, we obtain a complete frequency polygon as shown below:
The ages (in years) of 360 patients treated in a hospital on a particular dayare given below.
Age in years 
1020 
2030 
3040 
4050 
5060 
6070 
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.
The given frequency distribution is as under
Age in years 
1020 
2030 
3040 
4050 
5060 
6070 
Numbers of patients 
90 
40 
60 
20 
120 
30 
Take class intervals i.e age in years along xaxis 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 010 at the beginning and 7080 at the end each with frequency zero and join the midpoints of top of the rectangles.
Thus, we obtain a complete frequency polygon, shown below:
Draw a histogram and frequency polygon from the following data.
Class interval 
2025 
2530 
3035 
3540 
4045 
4550 
Frequency 
30 
24 
52 
28 
46 
10 
The given frequency distribution is as below :
Class intervals 
2025 
2530 
3035 
3540 
4045 
4550 
Frequency 
30 
24 
52 
28 
46 
10 
Take class intervals along xaxis and frequencies along yaxis 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 1520 at the beginning and 5055 at the end , each with frequency zero and join the mid points of top of the rectangles to get the required frequency polygon.
Draw a histogram for the following data:
Class interval 
600640 
640680 
680720 
720760 
760800 
800840 
Frequency 
18 
45 
153 
288 
171 
63 
Usingthis histogram, draw the frequencypolygon on the same graph.
The given frequency distribution table is given below :
Class interval 
600640 
640680 
680720 
720760 
760800 
800840 
Frequency 
18 
45 
153 
288 
171 
63 
Take class intervals along xaxis and frequencies along yaxis 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 560600 at the beginning and 840880 at the end, each with frequency zero.
Now join the mid points of the top of the rectangles to get the required frequency polygon.
The following table shows the average daily earnings of 40 general stores in a market, during a certain week.
Daily earning (in rupees) 
700750 
750800 
800850 
850900 
900950 
9501000 
Number of Stores 
6 
9 
2 
7 
11 
5 
Draw a histogram to represent the above data.
Given frequency distribution is as below :
Daily earnings (in Rs) 
700750 
750800 
800850 
850900 
900950 
9501000 
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 xaxis and frequencies i.e. number of stores along yaxis. So , we get the required histogram .
Since the scale on Xaxis starts at 700, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 700.
the heights of 75 students in a school are given below :
Height (in cm) 
130136 
136142 
142148 
148154 
154160 
160166 
Number of students 
9 
12 
18 
23 
10 
3 
Draw a histogram to represent the above data.
Height (in cm) 
130136 
136142 
142148 
148154 
154160 
160166 
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 xaxis and frequencies i.e. number of student s along yaxis . So we get the required histogram.
Since the scale on Xaxis starts at 130, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 130.
The following table gives the lifetimes of 400 neon lamps:
Lifetime (in hr ) 
300400 
400500 
500600 
600700 
700800 
800900 
9001000 
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?
(i) Histogram is as follows:
(ii) Number of lamps having lifetime more than 700 hours = 74 + 62 + 48 = 184
Draw a histogram for frequency distribution of the following data.
Class Interval 
813 
1318 
1823 
2328 
2833 
3338 
3843 
Frequency 
320 
780 
160 
540 
260 
100 
80 
Give frequency distribution is as below :
Class interval 
813 
1318 
1823 
2328 
2833 
3338 
3843 
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 xaxis and frequency along yaxis . So , we get the required histogram.
Since the scale on Xaxis starts at 8, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 8.
Construct a histogram for the following frequency distribution.
Class interval 
512 
1320 
2128 
2936 
3744 
4552 
Frequency 
6 
15 
24 
18 
4 
9 
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 
4.512.5 
12.520.5 
20.528.5 
28.536.5 
36.544.5 
44.552.5 
Frequency 
6 
15 
24 
18 
4 
9 
To draw the required histogram , take class intervals, along xaxis and frequencies along yaxis and draw rectangles . So, we get the required histogram .
Since the scale on Xaxis 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.
The following table shows the number of illiterate persons in the age group (1058 years) in a town:
Age group (in years) 
1016 
1723 
2430 
3137 
3844 
4551 
5258 
Number of illiterate persons 
175 
325 
100 
150 
250 
400 
525 
Draw a histogram to represent the above data.
Given frequency distribution is as below :
Age group (in years ) 
1016 
1723 
2430 
3137 
3844 
4551 
5258 
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) 
9.516.5 
16.523.5 
23.530.5 
30.537.5 
37.544.4 
44.551.5 
51.558.5 
No of illiterate persons 
175 
325 
100 
150 
250 
400 
525 
To draw the required histogram , take class intervals, that is age group, along xaxis and frequencies, that is number of illiterate persons along yaxis and draw rectangles . So , we get the required histogram.
Since the scale on Xaxis 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.
Draw a histogram to represent the following data.
Class Interval 
1014 
1420 
2032 
3252 
5280 
Frequency 
5 
6 
9 
25 
21 
given frequency distribution is as below :
Class interval 
1014 
1420 
2032 
3252 
5280 
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 
1014
1420
2032
3252
5280 
5
6
9
25
21 
_{} 
Now take class intervals along xaxis and adjusted frequency along yaxis and constant rectangles having their bases as class size and heights as the corresponding adjusted frequencies.
Thus, we obtain the histogram as shown below:
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 
14 
46 
68 
812 
1220 
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.
(i) Minimum class size = 6  4 = 2
(ii) Maximum number of surnames lies in the class interval 6  8.
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