R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 17  Bar Graph, Histogram and Frequency Polygon
Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 17  Bar Graph, Histogram and Frequency Polygon.
All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.
R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 17  Bar Graph, Histogram and Frequency Polygon Page/Excercise 17B
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.
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.
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.
(i) Histogram is as follows:
(ii) Number of lamps having lifetime more than 700 hours = 74 + 62 + 48 = 184
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.
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.
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.
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:
(i) Minimum class size = 6  4 = 2
(ii) Maximum number of surnames lies in the class interval 6  8.
Minimum class size = 10  5 = 5
Minimum class size = 50  45 = 5
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.
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 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:
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.
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.
R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 17  Bar Graph, Histogram and Frequency Polygon Page/Excercise 17A
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:
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:
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:
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:
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:
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:
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:
(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 =
R. S. Aggarwal and V. Aggarwal  Mathematics  IX Class 9 Chapter Solutions
 Chapter 1  Number Systems
 Chapter 2  Polynomials
 Chapter 3  Factorisation of Polynomials
 Chapter 4  Linear Equations in Two Variables
 Chapter 5  Coordinate Geometry
 Chapter 6  Introduction to Euclid's Geometry
 Chapter 7  Lines And Angles
 Chapter 8  Triangles
 Chapter 9  Congruence of Triangles and Inequalities in a Triangle
 Chapter 10  Quadrilaterals
 Chapter 11  Areas of Parallelograms and Triangles
 Chapter 12  Circles
 Chapter 13  Geometrical Constructions
 Chapter 14  Areas of Triangles and Quadrilaterals
 Chapter 15  Volume and Surface Area of Solids
 Chapter 16  Presentation of Data in Tabular Form
 Chapter 17  Bar Graph, Histogram and Frequency Polygon
 Chapter 18  Mean, Median and Mode of Ungrouped Data
 Chapter 19  Probability
CBSE Class 9 Textbook Solutions
 NCERT Biology Class 9 Textbook Solution
 NCERT Chemistry Class 9 Textbook Solution
 NCERT Mathematics Class 9 Textbook Solution
 NCERT Physics Class 9 Textbook Solution
 NCERT Beehive Class 9 Textbook Solution
 NCERT क्षितिज भाग  १ Class 9 Textbook Solution
 NCERT India and the Contemp. World  I Class 9 Textbook Solution
 NCERT Democratic Politics  I Class 9 Textbook Solution
 NCERT Contemporary India  I Class 9 Textbook Solution
 NCERT Economics Class 9 Textbook Solution
Browse Study Material
Browse questions & answers
TopperLearning provides stepbystep solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 17  Bar Graph, Histogram and Frequency Polygon for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 17  Bar Graph, Histogram and Frequency Polygon now.