Rs Aggarwal 2021 2022 Solutions for Class 9 Maths 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 Maths 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 2021 2022 Book of Class 9 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 642:

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

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#### Page No 644:

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(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 = $\frac{60+35+75+50+60}{5}=\frac{280}{5}=56$

#### Page No 658:

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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:

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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:

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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:

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(i)

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

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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:

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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:

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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:

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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 =$\frac{\mathrm{Minimum}\mathrm{class}\mathrm{size}}{\mathrm{Class}\mathrm{size}\mathrm{of}\mathrm{the}\mathrm{class}}\times \mathrm{Its}\mathrm{frequency}$

We have the following table:

Class Interval | Frequency | Adjusted Frequency |

10-14 | 5 | $\frac{4}{4}\times 5=5$ |

14-20 | 6 | $\frac{4}{6}\times 6=4$ |

20-32 | 9 | $\frac{4}{12}\times 9=3$ |

32-52 | 25 | $\frac{4}{20}\times 25=5$ |

52-80 | 21 | $\frac{4}{28}\times 21=3$ |

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:

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(i) Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula,

$\mathrm{Adjusted}\mathrm{freq}=\frac{\mathrm{min}\mathrm{class}\mathrm{size}}{\mathrm{class}\mathrm{size}\mathrm{of}\mathrm{this}\mathrm{class}}\times \mathrm{freq}$

Class sizes are

4 $-$ 1 = 3

6 $-$ 4 = 2

8 $-$ 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 | $\frac{2}{3}\times 6=4$ |

4-6 | 30 | 2 | $\frac{2}{3}\times 30=30$ |

6-8 | 44 | 2 | $\frac{2}{2}\times 44=44$ |

8-12 | 16 | 4 | $\frac{2}{4}\times 16=8$ |

12-20 | 4 | 8 | $\frac{2}{8}\times 4=1$ |

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

#### Page No 660:

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Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula,

$\mathrm{Adjusted}\mathrm{freq}=\frac{\mathrm{min}\mathrm{class}\mathrm{size}}{\mathrm{class}\mathrm{size}\mathrm{of}\mathrm{this}\mathrm{class}}\times \mathrm{freq}$

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 | $\frac{5}{5}\times 6=6$ |

10-15 | 12 | 5 | $\frac{5}{5}\times 12=12$ |

15-25 | 10 | 10 | $\frac{5}{10}\times 10=5$ |

25-45 | 8 | 20 | $\frac{5}{20}\times 8=2$ |

45-75 | 18 | 30 | $\frac{5}{30}\times 18=3$ |

#### Page No 660:

#### Answer:

Here the class intervals are of unequal size. So, we calculate the adjusted frequency using the formula,

$\mathrm{Adjusted}\mathrm{freq}=\frac{\mathrm{min}\mathrm{class}\mathrm{size}}{\mathrm{class}\mathrm{size}\mathrm{of}\mathrm{this}\mathrm{class}}\times \mathrm{freq}$

Minimum class size = 5

Marks | Frequency | Width of class | Height of rectangle |

0-10 | 8 | 10 | $\frac{5}{10}\times 8=4$ |

10-30 | 32 | 20 | $\frac{5}{20}\times 32=8$ |

30-45 | 18 | 15 | $\frac{5}{15}\times 18=6$ |

45-50 | 10 | 5 | $\frac{5}{5}\times 10=10$ |

50-60 | 6 | 10 | $\frac{5}{10}\times 6=3$ |

#### Page No 660:

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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:

#### 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:

#### 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:

#### 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:

#### 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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