Rs Aggarwal 2019 Solutions for Class 9 Math Chapter 17 Bar Graph, Histogram And Frequency Polygon are provided here with simple step-by-step explanations. These solutions for Bar Graph, Histogram And Frequency Polygon are extremely popular among Class 9 students for Math Bar Graph, Histogram And Frequency Polygon Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 9 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 642:

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.

Answer:

Page No 642:

Question 2:

On a certain day, the temperature in a city was recorded as under:

Time 5 a. m. 8 a. m. 11 a. m. 3 p. m. 6 p. m.
Temperature (in °C) 20 24 26 22 18
Illustrate the data by a bar graph.

Answer:

Page No 642:

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 a bar graph to represent the above data.

Answer:

Take the name of vehicle along the x-axis and the velocity along the y-axis.
 

Page No 642:

Question 4:

The following table shows the favourite 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

Answer:

Page No 642:

Question 5:

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

Year 201213 2013−14 2014−15 2015−16 2016−17
No. of students 800 975 1100 1400 1625

Answer:

Page No 642:

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 thousands)
16 20 32 36 40 48

Answer:



Page No 643:

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.

Answer:

Page No 643:

Question 8:

The birth rate per thousand in five countries over a period of time is 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.

Answer:

Page No 643:

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

Answer:

Page No 643:

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 result.

Answer:

Page No 643:

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.

Answer:



Page No 644:

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 is the student very good?
(iii) In which subject is he poor?
(iv) What is the average of his marks?

Answer:

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

(ii) The student scores very good in mathematics, as the height of the corresponding bar is the highest.

(iii) The student scores bad in Hindi, as the height of the corresponding bar is the lowest.

(iv) Average marks = 60+35+75+50+605 = 2805= 56



Page No 658:

Question 1:

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

Daily wages (in ₹) 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.

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals [daily wages (in rupees)] along the x-axis & the corresponding frequencies [number of workers] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 40 rupees
On y-axis: 1 big division = 2 workers
Because the scale on the x-axis starts at 340, a kink, i.e., a  break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 340
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Page No 658:

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 of stores 6 9 2 7 11 5
Draw a histogram to represent the above data.

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals [daily earnings (in rupees)] along the x-axis & the corresponding frequencies [number of stores] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 50 rupees
On y-axis: 1 big division = 1 store
Because the scale on the x-axis starts at 700, a kink, i.e., a break is indicated near the origin to signify that the graph is drawn with a scale beginning at 700 and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:



Page No 659:

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.

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals [heights (in cm)] along the x-axis & the corresponding frequencies [number of students ] along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 6 cm
On y-axis: 1 big division = 2 students
Because the scale on the x-axis starts at 130, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 130
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Page No 659:

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?

Answer:

(i)
(ii) Lamps with lifetime more than 700 hours = 74 + 62 + 48 = 184 

Page No 659:

Question 5:

Draw a histogram for the 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

Answer:

The given frequency distribution is in exclusive form.
We will represent the class intervals along the x-axis & the corresponding frequencies along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 5 units
On y-axis: 1 big division = 50 units
Because the scale on the x-axis starts at 8, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 8
and not at the origin.
We will construct rectangles with the class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the following histogram:

Page No 659:

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

Answer:

The given frequency distribution is in inclusive form.
So, we will convert it into exclusive form, as shown below:
 

 Class Interval Frequency
4.5–12.5 6
12.5–20.5 15
20.5–28.5 24
28.5–36.5 18
36.5–44.5 4
44.5–52.5 9

We will mark class intervals along the x-axis and frequencies along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 8 units
On y-axis: 1 big division = 2 units
Because the scale on the x-axis starts at 4.5, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 4.5
and not at the origin.
We will construct rectangles with class intervals as bases and the corresponding frequencies as heights.
Thus, we will obtain the histogram as shown below:

Page No 659:

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.

Answer:

The given frequency distribution is inclusive form.
So, we will convert it into exclusive form, as shown below:
 

Age (in years) Number of Illiterate Persons
9.5-16.5 175
16.5-23.5 325
23.5-30.5 100
30.5-37.5 150
37.5-44.5 250
44.5-51.5 400
51.5-58.5 525

We will mark the age groups (in years) along the x-axis & frequencies (number of illiterate persons) along the y-axis.
The scale is as follows:
On x-axis: 1 big division = 7 years
On y-axis: 1 big division = 50 persons
Because the scale on the x-axis starts at 9.5, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 9.5
and not at the origin.
We will construct rectangles with class intervals (age) as bases and the corresponding frequencies (number of illiterate persons) as
heights.
Thus, we obtain the histogram, as shown below:



Page No 660:

Question 8:

Draw a histogram to represent the following data.

Clas interval 10−14 14−20 20−32 32−52 52−80
Frequency 5 6 9 25 21

Answer:

In the given frequency distribution, class sizes are different.
So, we calculate the adjusted frequency for each class.
The minimum class size is 4.
Adjusted frequency of a class =Minimum class sizeClass size of the class ×Its frequency 

We have the following table:
 

 Class Interval Frequency Adjusted Frequency
10-14 5 44×5=5
14-20 6 46×6=4
20-32 9 412×9=3
32-52 25 420×25=5
52-80 21 428×21=3


We mark the class intervals along the x-axis and the corresponding adjusted frequencies along the y-axis.
We have chosen the scale as follows:
On the x- axis,
1 big division = 5 units
On the y-axis,
1 big division = 1 unit
We draw rectangles with class intervals as the bases and the corresponding adjusted frequencies as the heights.
Thus, we obtain the following histogram:

Page No 660:

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.

Answer:

(i) Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula, 
Adjusted freq=min class sizeclass size of this class×freq
Class sizes are 
- 1 = 3
- 4 = 2
- 6 = 2
12 - 8 = 4
20 - 12 = 8
Minimum class size = 2

Number of letters Frequency Width of class Height of rectangle
1-4 6 3 23×6=4
4-6 30 2 23×30=30
6-8 44 2 22×44=44
8-12 16 4 24×16=8
12-20 4 8 28×4=1


(ii) Maximum number of surnames lie in the interval 6-8. 

Page No 660:

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

Answer:

Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula, 
Adjusted freq=min class sizeclass size of this class×freq
Class sizes are 
10 - 5 = 5
15 - 10 = 5
25 - 15 = 10
45 - 25 = 20
75 - 45 = 30
Minimum class size = 5

Class interval Frequency Width of class Height of rectangle
5-10 6 5 55×6=6
10-15 12 5 55×12=12
15-25 10 10 510×10=5
25-45 8 20 520×8=2
45-75 18 30 530×18=3
​

Page No 660:

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

Answer:

Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula, 
Adjusted freq=min class sizeclass size of this class×freq
Minimum class size = 5

Marks Frequency Width of class Height of rectangle
0-10 8 10 510×8=4
10-30 32 20 520×32=8
30-45 18 15 515×18=6
45-50 10 5 55×10=10
50-60 6 10 ​510×6=3
​

Page No 660:

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.

Answer:

We take two imagined classes—one at the beginning (0–10) and other at the end (70–80)—each with frequency zero.
With these two classes, we have the following frequency table:

Age in Years  Class Mark Frequency
(Number of Patients)
0–10 5 0
10–20 15 2
20–30 25 5
30–40 35 12
40–50 45 19
50–60 55 9
60–70 65 4
70–80 75 0

Now, we plot the following points on a graph paper:
A(5, 0), B(15, 2), C(25, 5), D(35, 12), E(45, 19), F(55, 9), G(65, 4) and H(75, 0)
Join these points with line segments AB, BC, CD, DE, EF, FG, GH ,HI and IJ to obtain the required frequency polygon.

Page No 660:

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

Answer:

Though the given frequency table is in inclusive form, class marks in case of inclusive and exclusive forms are the same.
We take the imagined classes (-9)–0 at the beginning and 61–70 at the end, each with frequency zero.
Thus, we have:
 

 Class Interval Class Mark Frequency
 -9–0  –4.5 0
1–10 5.5 8
11–20 15.5 3
21–30 25.5 6
31–40 35.5 12
41–50 45.5 2
51–60 55.5 7
61–70 65.5 0

Along the x-axis, we mark –4.5, 5.5, 15.5, 25.5, 35.5, 45.5, 55.5 and 65.5.
Along the y-axis, we mark 0, 8, 3, 6, 12, 2, 7 and 0.
We have chosen the scale as follows :
On the x-axis, 1 big division = 10 units.
On the y-axis, 1 big division = 1 unit.
We plot the points A(–4.5,0), B(5.5, 8), C(15.5, 3), D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0).
We draw line segments AB, BC, CD, DE, EF, FG, GH to obtain the required frequency polygon, as shown below.

 

Page No 660:

Question 14:

The ages (in years) of 360 patients treated in a hospital on a particular day are 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.

Answer:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis:
1 big division = 10 years
On the y-axis:
1 big division = 2 patients
Thus, we obtain the histogram, as shown below.
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 0–10 and 70–80, each with frequency 0. The class marks of these classes are 5 and 75, respectively.
So, we plot the points A(5, 0) and B(75, 0). We join A with the midpoint of the top of the first rectangle and B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:



Page No 661:

Question 15:

Draw a histogram and the 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

Answer:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis, 1 big division = 5 units.
On the y-axis, 1 big division = 5 units.
Because the scale on the x-axis starts at 15, a kink, i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 15
and not at the origin.
Thus, we obtain the histogram, as shown below.
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 15–20 and 50–55, each with frequency 0. The class marks of these classes are 17.5 and 52.5, respectively.
So, we plot the points A( 17.5, 0) and B(52.5, 0). We join A with the midpoint of the top of the first rectangle and B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:

Page No 661:

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
Using this histogram, draw the frequency polygon on the same graph.

Answer:

We represent the class intervals along the x-axis and the corresponding frequencies along the y-axis.
We construct rectangles with class intervals as bases and respective frequencies as heights.
We have the scale as follows:
On the x-axis:
1 big division = 40 units
On the y-axis:
1 big division = 20 units
Thus, we obtain the histogram, as shown below:
We join the midpoints of the tops of adjacent rectangles by line segments.
Also, we take the imagined classes 560–600 and 840–880, each with frequency 0.
The class marks of these classes are 580 and 860, respectively.
Because the scale on the x-axis starts at 560, a kink; i.e., a break, is indicated near the origin to signify that the graph is drawn with a scale beginning at 560
and not at the origin.
So, we plot the points A( 580, 0) and B(860, 0). We join A with the midpoint of the top of the first rectangle and join B with the midpoint of the top of the last rectangle.
Thus, we obtain a complete frequency polygon, as shown below:

 



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