Five more situations are as follows:
1. In a school in class 6, the number of students coming in working days is shown below:

2. The marks obtained by Meenu in different subjects in annual examination are shown below:

3. The weights of five students are shown below:

4. The heights of five girls are shown below:

5. The monthly pocket expenditure of Manish during an year is shown below:

1. Bar graphs for solution 1 to 5 are as follows:

2 

3.

4.

5.
 

Watching TV is preferred by most of the studentsother than playing.

(i) April (ii) March (iii) 225 (iv) Rs. 4375000

(a) Tree is maximum in number in the garden.
(b) Creeper is minimum in number in the garden.

(a) Class X has minimum number of scouts i.e. 10.
(b) Class VIII has maximum number of scouts i.e. 60.
(c) Number of scouts in class VI = 40
(d) Number of scouts in class X =10
Hence. class VI has exactly four times the scouts as that of class X.
(e) Total number of scouts in classes VI to X
                = 40 + 20 + 60 + 30 + 10 = 160.

To draw a bar graph, we will use the following steps:
(i) Firstly, draw two perpendicular lines, one is horizontal and one is vertical. Along the horizontal line, mark 'days' and along the vertical line mark 'number of books sold'.
(ii) Now, take scale of 1 unit length = 5 books, along the vertical line and then mark the corresponding values.
Also, the heights of the bars for various days are as follows:

(iii) Drawbars of equal width and of height calculated in Step (ii) on the horizontal line with equal spacing (or gap) between them.
Thus, we get the following bar graph.

(A) According to the question, 100 students can be represented by = 1symbol 
\(\therefore\) 1 student can be represented by \(\frac{1}{100}\) symbols
Now, for year 1996,
400 students can be represented by \(\frac{400\times 1}{100}\)
                        = 4 symbols
For year 1998,
535 students can be represented by \(\frac{535\times 1}{100}\)
                       = 5 complete symbols + 35 students
                       = 5 complete symbols and 1 incomplete symbol
For year 2000,
472 students can be represented by \(\frac{472\times 1}{100}\)
                      = 4 complete symbols + 72 students
                      = 4 complete symbols and 1 incomplete symbol
For year 2002,
600 students can be represented by \(\frac{600\times 1}{100}\)
                     = 6 complete symbols
For year 2004,
623 students can be represented by \(\frac{623\times 1}{600}\)
6 complete symbol + 23 students = = 6 complete symbols and 1 incomplete symbol
Hence, the required pictograph of given data is shown below

(a) 6 symbols represent total number of students in the year 2002.
(b) 5 complete symbols and 1 incomplete symbol represent total number of students for the year 1998.
(B) According to the question, 50 students can be represented by 1symbol
\(\therefore\) 1 student can be represented by \(\frac{1}{50}\) symbols
Now,  for year 1996,
400 students can be represented by \(\frac{400\times 1}{50}\) symbols
                          = \(\frac{400}{50}\)
                          = 8 complete symbols.
For year 1998,
535 students can be represented by \(\frac{535\times 1}{50}\) 
                         = \(\frac{535}{50}\) symbols
                         = 10 complete symbols + 1 incomplete symbol
For year 2000,
472 students can be represented by \(\frac{472\times 1}{50}\) symbols
                         = \(\frac{472}{50}\) 
                         = 9 complete symbols + 1 incomplete symbol
For year 2002,
600 students can be represented by \(\frac{600\times 1}{50}\) symbols
                         = \(\frac{600}{50}\)
                         = 12 complete symbol
For year 2004,
623 students can be represented by \(\frac{623\times 1}{50}\) symbols
                         = \(\frac{623}{50}\)
                         = 12 complete symbols + 1 incomplete symbol
Hence, the required pictograph of given data is shown below

Therefore, we observe that pictograph B is more informative because it gives better approximation.

The table with tally marks is shown below:

Marks	Tally marks	Number of students
1	II	2
2	III	3
3	III	3
4		7
5		6
6		7
7		5
8	IIII	4
9	III	3
Total		40

(a) Here, we have to find out the students, who obtained marks equal to 7 or more than 7. So, we have to add the number of students who obtained marks equal to 7 or more than 7 i.e. 7, 8 and 9.
\(\therefore\) Number of those students = 5 + 4 + 3 = 12
Hence, 12 students obtained marks equal to 7 or more than 7.
(b) Here, we have to find out the students who obtained marks below 4.
So, we have to add the number of students who obtained marks below 4 i.e. 1, 2 and 3.
\(\therefore\) Number of those students = 3 + 3 + 2 = 8
Hence, 8 students obtained marks below 4.

In the given pictograph, 1 picture = 2 bulbs
Now, number of bulbs sold on Monday = 6 pictures
                                                              = 6 x 2 = 12bulbs
Number of bulbs sold on Tuesday = 8 x 2 = 16 bulbs
Number of bulbs sold on Wednesday = 4 x 2 = 8 bulbs
Number of bulbs sold on Thursday = 5 x 2 = 10 bulbs
Number of bulbs sold on Friday = 7 x 2 = 14 bulbs
Number of bulbs sold on Saturday = 4 x 2 = 8 bulbs
Number of bulbs sold on Sunday = 9 x 2 = 18 bulbs
(a) Number of bulbs sold on Friday = 7 x 2 = 14 bulbs
(b) Maximum number of bulbs were sold on Sunday i.e. 18 bulbs.
(c) The same number of bulbs were sold on Wednesday and Saturday i.e. 8 bulbs.
(d) The minimum number of bulbs were sold on Wednesday and Saturday i.e. 8 bulbs.
(e) Total number of bulbs sold in a week
                      = 12 + 16 + 8 + 10 + 14 + 8 + 18 = 86
Now, number of cartons which can hold 9 bulbs = 1 and number of carton which can hold 1 bulb = \(\frac{1}{9}\)
\(\therefore\) Number of cartons which can hold 86 bulbs = \(\frac{1\times 86}{9}\)
                     = \(\frac{86}{9}=9\frac{5}{9}=10\)
Hence, 10 cartons were needed in the given week.

(i) Observing the pictograph, it is clear that village D has the minimum number of tractors i.e. only 3 tractors.
(ii) Observing the pictograph, it is clear that village C has the maximum number of tractors i.e. 8 tractors.
(iii) Observing the pictograph, it is clear that village C has 8 tractors and village B has 5 tractors. So, village C has 8 - 5 = 3 more tractors as compared to village B.
(iv) Total number of tractors in all the five villages = Sum of all tractors in villages A, B, C, D and E = 6 + 5 + 8 + 3 + 6 = 28. Hence, there are 28 tractors in all the five villages.