To construct the bar graph representing the given data, we follow the following steps:
(a) We take a graph paper and draw two mutually perpendicular lines OX and OY.
(b) Along the horizontal line OX, we mark 'cities' and along the vertical line, we mark the 'population'.
(c) Along the axis OX, we choose equal suitable width of each bar. The gap between the bars is chosen same.
(d) Choose a suitable scale to determine the heights of the bars, according to the availability of space.
Here, we choose 1 big division to represent 20 lakh population.
(e) Calculate the height of various bars as follows:
The height of the bar for Mumbai \(=\frac{120}{20}=\) 6 big divisions
The height of the bar for Kolkata \(=\frac{130}{20}=\) 6.5 big divisions
The height of the bar for Delhi \(=\frac{150}{20}=\) 7.5 big divisions
The height of the bar for Chennai \(=\frac{80}{20}=\) 4 big divisions​​​​​​​
(f) Now, we draw the bars as shown in Fig. below and at the top of each bar we write the population of the corresponding city.
​​​​​​​

Mode = 48 kg, Mean = 51 kg

61

i) \(\frac { 21 }{ 50 } \)
ii) \(\frac { 29 }{ 50 } \)

A double bar graph is shown below

We infer from the bar graph that maximum number of students were absent in Class VI on 15.04.2009 whereas minimum number of students were absent in Class X
(b) Clearly, class VIII has maximum number of students, i.e. 95.
(c) The difference of total number of students and number of students present is minimum for Class X, i.e. 63 - 62 = 1.
(d) Number of students present in Class IX = 65 Total number of students = 70
Hence, required radio = \(\frac { 65 }{ 70 } =\frac { 13 }{ 14 } \)or 13:14
(e) Total number of students in Class VI = 90
Number of students present in Class VI = 81
Number of students absent = 90 - 81 = 9
= \(\frac { number\quad of\quad absent\quad students }{ total\quad number\quad of\quad students } \times 100\)
= \(\frac { 9 }{ 90 } \times 100=10%\) %

(a) The above double bar graph compares the sale of brands A and 8during the months of January to June.
(b) We can clearly see from the double bar graph that sales for brand A reduced in the month of March compared to that of February.
(c) Sales of brand A in June = Rs. 57 lakh
and sales of brand 8 in June = Rs. 54 lakh
Difference in sales = 57 - 54 = Rs. 3 lakh
(d) Average sales of brand 8
=\(\frac { total\ sales\ of\ brand\ b\ in\ sin\ months\ from\ january\ to\ june }{ 6 } \)
= \(\frac { 36+38+43+35+45+54 }{ 6 } \)
= \(\frac { 251 }{ 6 } \) Rs.41.83 lakh
(e) We can clearly see from the double bar graph that sales of brand 8is less than sales of brand A in the month of April and June
(f) Sales of brand A in January = 31
and sales of brand 8 in January = 36
Required ratio = \(\frac { 31 }{ 36 } \)or 31:36

After studying the bar graph, we have
Production of rice in 2005 = 50 million tonne
Production of rice in 2006 = 40 million tonne
Production of rice in 2007 = 70 million tonne
Production of rice in 2008 = 50 million tonne
Production of rice in 2009 = 60 million tonne
(a) From above calculation, the bar graph shows
the production of rice in million tonne by a
country during years 2005 to 2009.
(b) The production of rice in 2006 was the least
40 million tonne.
(c) The maximum production of rice was in 2007.
.The production rose after 2006.
(d) For average production Average = \(\frac { sum\ of\ observations }{ number\ of\ observations } \)
Sum of productions = 50 + 40 + 70 + 50 + 60 = 270
Average = \(\frac { 270 }{ 5 } \)  = 54.4 million tonne
(e) Production in 2006 = 40 million tonne
Production in 2008 = 50 million tonne.
Difference = 50 - 40 = 10 million tonne.

(a) For average number of children,
The sum of given number of children
= 125 + 110 + 100 + 95 + 90 + 70 = 590
average/mean  \(=\frac { sum\ of\ osbervations }{ number\ of\ observations } \)
= \(\frac { 590 }{ 6 } =98.33=98\)
Hence, average number of children in each class is 98.
(b) Ratio = \(\frac { 110 }{ 95 } =\frac { 22 }{ 19 } \)
Hence, ratio is 22 : 19

As per the given information, mean of 10 observations is 40.
\(\therefore\)   Mean​​​​​​ ​\(=\frac { sum\ of\ all\ observations }{ number\ of\ observations } \)
\(\Rightarrow\)   40 = \(\frac { sum\ of\ all\ observations }{ 10 } \)
Sum of all observations = 40 x 10 = 400
But this is incorrect sum. Since, one observation was copied wrongly.
Correct sum = Incorrect sum - Incorrect observation + Correct observation
= 400 - 15 + 45 = 430
correct mean ​​​​​​​\(=\frac { Correct\ sum }{ number\ of\ observations } =\frac { 430 }{ 10 } =43\)

On the basis of given data, we can draw the following double bar graph:

The difference in the sales of the two subjects books
In the year 1999, 490 - 450 = 40
In the year 2000, 480 - 400 = 80
In the year 2001, 540 - 410 = 130
In the year 2002,610 - 540 = 70
The difference in the sale of the two subjects books is least in the year 1999