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.
 

(i) 510 (ii) Tamil (iii) 284 (iv) 72 (v) 334

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

(a) 290 (b) Delhi (c) Chennai (d) Patna, Jaipur, Guwahati (e) 50

1000 newspapers can be represented by \(\boxplus\) and 500 newspaper represented by \(\Box\)
For English language,
5000 newspapers can be represented by 5 complete symbols.
For Hindi language,
8500 newspapers can be represented by 8 complete and 1 incomplete symbol.
For Tamil language,
500 newspapers can be represented by 1 incomplete symbol.
For Punjabi language,
2500 newspapers can be represented by 2 complete symbols and 1 incomplete symbol.
For Gujarati language,
1000 newspapers can be represented by 1 complete symbol.
Hence, the required pictograph of given data is shown below:
Take \(\boxplus\) = 1000 newspapers and \(\Box\) = 500 newspaper.

(a) Let 1 unit length = Rs 100

(b) In May, her saving is maximum.
(c) Economy saving for future planning.

To draw the 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 'age-group' and along vertical line mark 'number of persons'.
(ii) Now, take scale of 1 unit length = 20000 along the vertical line and then mark the corresponding values.
Also, the heights of bars for various groups are as follows:

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

(a) From bar graph, we see that the lengths of bars for age group 30-44 and 45-59 are same, so age group (30 - 44) and (45 - 59) have the same population.
(b) Persons having age 60 or above are called senior citizens.
\(\therefore\) Number of senior citizens in the town
          = Number of persons of age group (60 - 74) + Number of persons of age 75 and above
          = 80000 + 40000 = 120000.

(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 = 4 girls, half picture = 2 girls
(a) Observing the pictograph, it is clear that the minimum number of girl students are in class VIII i.e. 4 + 2 = 6
(b) Observing the pictograph, it is clear that Number of girls in class VI = 4 x 4 = 16 and number of girls in class V = 2 x 4 + 1 x 2 = 8 + 2 = 10
\(\therefore\) 10 < 16
So, it is clear that number of girls in class VI is not Iess than the number of girls in class V.
(c) Observing the pictograph, number of girls in class VII = 3 x 4 = 12.