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
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.
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:
Given below are the seats won by
different political parties in the polling outcome of a state assembly
elections:
Political | A | B | C | D | E | F |
Seats | 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 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:
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 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:
The approximate velocities of some
vehicles are given below:
Name | Bicycle | Scooter | Car | Bus | Train |
Velocity | 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 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:
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 |
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:
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 | 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 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:
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 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:
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 (in | 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 | 340-380 | 380-420 | 420-460 | 460-500 | 500-540 | 540-580 |
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) | 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.
Draw a histogram to represent the
following information:
Class | 5-10 | 10-15 | 15-25 | 25-45 | 45-75 |
Frequency | 6 | 12 | 10 | 8 | 18 |
Minimum class size = 10 – 5 = 5
Draw a histogram to represent the
following information:
Marks | 0-10 | 10-30 | 30-45 | 45-50 | 50-60 |
Number | 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 | 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.
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 |
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 | 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 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.
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 |
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)-0 1-10 11-20 21-30 31-40 41-50 51-60 61-70 | (-9.5)-0.5 0.5-10.5 10.5-20.5 20.5-30.5 30.5-40.5 40.5-50.5 50.5-60.5 60.5-70.5 | -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 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:
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.
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:
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 |
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.
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.
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.
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 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) | 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.
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.
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.
The following table gives the
lifetimes of 400 neon lamps:
Lifetime (in | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 | 800-900 | 900-1000 |
Number | 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 | 8-13 | 13-18 | 18-23 | 23-28 | 28-33 | 33-38 | 38-43 |
Frequency | 320 | 780 | 160 | 540 | 260 | 100 | 80 |
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.
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 |
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.
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.
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.
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 |
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:
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 | 1-4 | 4-6 | 6-8 | 8-12 | 12-20 |
Number | 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.