RD Sharma solution class 8 chapter 25 Data Handling III(Pictorial Representation of Data as Pie Charts or Circle Graphs) Exercise 25.2

Exercise 25.2

Page-25.21

Question 1:

The pie-chart given in Fig. 25.17 represents the expenditure on different items in constructing a flat in Delhi. If the expenditure incurred on cement is Rs 112500, find the following:


(i) Total cost of the flat.
(ii) Expenditure incurred on labour.

Answer 1:

(i) Expenditure incurred on cement  = $\frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{cost}}{360°}$
Total cost of the flat = $\frac{360° \times 112500}{75°} = \mathrm{Rs} 540000$
                                        
(ii) Expenditure incurred on labour = $\frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{labour} \mathrm{sector} \times \mathrm{Total} \mathrm{cost}}{360°}$
$= \frac{100° \times 540000}{360°}= \mathrm{Rs} 150000$
                                                               

Question 2:

The pie-chart given in Fig. 25.18 shows the annual agricultural production of an Indian state. If the total production of all the commodities is 81000 tonnes, find the production (in tonnes) of
(i) Wheat
(ii) Sugar
(iii) Rice
(iv) Maize
(v) Gram

Answer 2:

(i)
$$\begin{gathered} \mathrm{Production} \mathrm{of} \mathrm{wheat} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{for} \mathrm{wheat} \times \mathrm{Total} \mathrm{production}}{360°} \\ = \frac{120° \times 81000}{360°} = 27000 \mathrm{tonnes} \end{gathered}$$
               
(ii)
$$\begin{gathered} \mathrm{Production} \mathrm{of} \mathrm{sugar} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{for} \mathrm{sugar} \times \mathrm{Total} \mathrm{production}}{360°} \\ = \frac{10° \times 81000}{360°} = 22500 \mathrm{tonnes} \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{Production} \mathrm{of} \mathrm{rice} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{for} \mathrm{Rice} \times \mathrm{Total} \mathrm{production}}{360°} \\ = \frac{60° \times 81000}{360°} = 13500 \mathrm{tonnes} \end{gathered}$$

(iv)
$$\begin{gathered} \mathrm{Production} \mathrm{of} \mathrm{maize} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{for} \mathrm{maize} \times \mathrm{Total} \mathrm{production}}{360°} \\ = \frac{30° \times 81000}{360°} = 6750 \mathrm{tonnes} \end{gathered}$$

(v)
$$\begin{gathered} \mathrm{Production} \mathrm{of} \mathrm{gram} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{for} \mathrm{Gram} \times \mathrm{Total} \mathrm{production}}{360°} \\ = \frac{120° \times 81000}{360°} = 11250 \mathrm{tonnes} \end{gathered}$$

Page-25.22

Question 3:

The following pie-chart shows the number of students admitted in different faculties of a college. If 1000 students are admitted in Science answer the following:


(i) What is the total number of students?
(ii) What is the ratio of students in science and arts?

Answer 3:

(i)
$$\begin{gathered} \mathrm{Students} \mathrm{in} \mathrm{science}= \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{students}}{360°} \\ 1000 = \frac{100° \times \mathrm{Total} \mathrm{students}}{360°} \\ \therefore \mathrm{Total} \mathrm{students} = 3600 \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Students} \mathrm{in} \mathrm{arts}= \frac{\mathrm{Central} \mathrm{angle} \mathrm{for} \mathrm{arts} \times \mathrm{Total} \mathrm{students}}{360°} \\ = \frac{120° \times 3600}{360°} = 1200 \\ \therefore \mathrm{Ratio} \mathrm{of} \mathrm{students} \mathrm{in} \mathrm{science} \mathrm{and} \mathrm{arts} = 1000:1200 = 5:6 \end{gathered}$$

Question 4:

In Fig. 25.20, the pie-chart shows the marks obtained by a student in an examination. If the student secures 440 marks in all, calculate his marks in each of the given subjects.

Answer 4:

Marks secured in mathematics = (108 x 440)/360 marks =  132 marks
Marks secured in science = (81 x 440)/360 marks = 99 marks
Marks secured in English = (72 x 440)/360  marks = 88 marks
Marks secured in Hindi = (54 x 440)/360 marks = 66 marks
Marks secured in social science = (45 x 440)/360 marks = 55 marks

Question 5:

In Fig. 25.21, the pie-chart shows the marks obtained by a student in various subjects. If the student scored 135 marks in mathematics, find the total marks in all the subjects. Also, find his score in individual subjects.

Answer 5:

$$\begin{gathered} \mathrm{Marks} \mathrm{scored} \mathrm{in} \mathrm{mathematics} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Marks}}{360°} \\ 135 = \frac{90° \times \mathrm{Total}}{360°} \\ \mathrm{Total} \mathrm{Marks} = 540 \end{gathered}$$

Marks scored in Hindi = (Central angle of Hindi x Total)/360
          = (60 x 540)/360 marks = 90 marks
Similarly, marks scored in science = (76 x 540) /360 marks = 114 marks
Marks scored in social science = (72 x 540) /360 marks = 108 marks
Marks scored in English = (62 x 540)/360 marks = 93 marks

Page-25.23

Question 6:

The following pie-chart shows the monthly expenditure of Shikha on various items. If she spends Rs 16000 per month, answer the following questions:


(i) How much does she spend on rent?
(ii) How much does she spend on education?
(iii) What is the ratio of expenses on food and rent?

Answer 6:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Money} \mathrm{spent} \mathrm{on} \mathrm{rent} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Money} \mathrm{spent}}{360°} \\ = \frac{81° \times 16000}{360°} = \mathrm{Rs} 3,600 \\ \left(\mathrm{ii}\right) \mathrm{Money} \mathrm{spent} \mathrm{on} \mathrm{education} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Money} \mathrm{spent}}{360°} \\ = \frac{36° \times 16000}{360°} = \mathrm{Rs} 1,600 \\ \left(\mathrm{iii}\right) \mathrm{Money} \mathrm{spent} \mathrm{on} \mathrm{food} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Money} \mathrm{spent}}{360°} \\ = \frac{135° \times 16000}{360°} = 6,000 \\ \mathrm{Ratio} \mathrm{of} \mathrm{expenses} \mathrm{on} \mathrm{food} \mathrm{and} \mathrm{rent} = \frac{6000}{3600} = \frac{5}{3} \end{gathered}$$

Question 7:

The pie chart (as shown in the figure 25.23) represents the amount spent on different sports by a sports club in a year. If the total money spent by the club on sports is Rs 1,08,000, find the amount spent on each sport.

Answer 7:

$$\begin{gathered} \mathrm{Amount} \mathrm{spent} \mathrm{on} \mathrm{cricket} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Money} \mathrm{spent}}{360°} \\ = \frac{150° \times 108000}{360°} = \mathrm{Rs} 45,000 \\ \mathrm{Amount} \mathrm{spent} \mathrm{on} \mathrm{hockey} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Money} \mathrm{spent}}{360°} \\ = \frac{100° \times 108000}{360°} = \mathrm{Rs} 30,000 \\ \mathrm{Amount} \mathrm{spent} \mathrm{on} \mathrm{football} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Money} \mathrm{spent}}{360°} \\ = \frac{60° \times 108000}{360°} = \mathrm{Rs} 18,000 \\ \mathrm{Amount} \mathrm{spent} \mathrm{on} \mathrm{tennis} = \frac{\mathrm{Central} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{corresponding} \mathrm{sector} \times \mathrm{Total} \mathrm{Money} \mathrm{spent}}{360°} \\ = \frac{50° \times 108000}{360°} = \mathrm{Rs} 15,000 \end{gathered}$$