Class 8 Exam > Class 8 Notes > Mathematics (Maths) Class 8 > Data Handling-III (Pictorial Representation of Data as Pie Charts or Circle Graphs) - Exercise 25.2
Data Handling-III (Pictorial Representation of Data as Pie Charts or Circle Graphs) - Exercise 25.2 | Mathematics (Maths) Class 8 PDF Download
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Page 1
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.
Solution:
(i) By using the formula,
Expenditure incurred on cement = (central angle × Total cost) / 360°
Total cost of the flat = (360° × 112500) / 75
o
= Rs 540000
(ii) By using the formula,
Expenditure incurred on labour = (central angle × Total cost) / 360°
= (100° × 540000) / 360
o
= Rs 150000
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
Page 2
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.
Solution:
(i) By using the formula,
Expenditure incurred on cement = (central angle × Total cost) / 360°
Total cost of the flat = (360° × 112500) / 75
o
= Rs 540000
(ii) By using the formula,
Expenditure incurred on labour = (central angle × Total cost) / 360°
= (100° × 540000) / 360
o
= Rs 150000
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
Solution:
We know that,
Total Production = 81000 Tonnes.
So,
(i) Production of wheat = (central angle of wheat × Total production) / 360°
= (120
o
× 81000) / 360
o
= 27000 tonnes
(ii) Production of sugar = (central angle of sugar × Total production) / 360°
= (100
o
× 81000) / 360
o
= 22500 tonnes
(iii) Production of rice = (central angle of rice × Total production) / 360°
= (60
o
× 81000) / 360
o
= 13500 tonnes
(iv) Production of maize = (central angle of maize × Total production) / 360°
= (30
o
× 81000) / 360
o
= 6750 tonnes
(v) Production of gram = (central angle of gram × Total production) / 360°
= (50
o
× 81000) / 360
o
= 11250 tonnes
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 :
Page 3
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.
Solution:
(i) By using the formula,
Expenditure incurred on cement = (central angle × Total cost) / 360°
Total cost of the flat = (360° × 112500) / 75
o
= Rs 540000
(ii) By using the formula,
Expenditure incurred on labour = (central angle × Total cost) / 360°
= (100° × 540000) / 360
o
= Rs 150000
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
Solution:
We know that,
Total Production = 81000 Tonnes.
So,
(i) Production of wheat = (central angle of wheat × Total production) / 360°
= (120
o
× 81000) / 360
o
= 27000 tonnes
(ii) Production of sugar = (central angle of sugar × Total production) / 360°
= (100
o
× 81000) / 360
o
= 22500 tonnes
(iii) Production of rice = (central angle of rice × Total production) / 360°
= (60
o
× 81000) / 360
o
= 13500 tonnes
(iv) Production of maize = (central angle of maize × Total production) / 360°
= (30
o
× 81000) / 360
o
= 6750 tonnes
(v) Production of gram = (central angle of gram × Total production) / 360°
= (50
o
× 81000) / 360
o
= 11250 tonnes
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?
Solution:
(i)
Students in science = (central angle × Total students) / 360°
1000 = (100
o
× Total students) / 360
o
Total students = (1000 × 360
o
)/100
o
= 3600 students
? Total number of students are 3600.
(ii) Students in arts = (central angle of arts × Total students) / 360°
= (120
o
× 3600) / 360
o
= 1200 students
? Ratio of students in science and arts is 1000:1200 = 5:6
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.
Page 4
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.
Solution:
(i) By using the formula,
Expenditure incurred on cement = (central angle × Total cost) / 360°
Total cost of the flat = (360° × 112500) / 75
o
= Rs 540000
(ii) By using the formula,
Expenditure incurred on labour = (central angle × Total cost) / 360°
= (100° × 540000) / 360
o
= Rs 150000
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
Solution:
We know that,
Total Production = 81000 Tonnes.
So,
(i) Production of wheat = (central angle of wheat × Total production) / 360°
= (120
o
× 81000) / 360
o
= 27000 tonnes
(ii) Production of sugar = (central angle of sugar × Total production) / 360°
= (100
o
× 81000) / 360
o
= 22500 tonnes
(iii) Production of rice = (central angle of rice × Total production) / 360°
= (60
o
× 81000) / 360
o
= 13500 tonnes
(iv) Production of maize = (central angle of maize × Total production) / 360°
= (30
o
× 81000) / 360
o
= 6750 tonnes
(v) Production of gram = (central angle of gram × Total production) / 360°
= (50
o
× 81000) / 360
o
= 11250 tonnes
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?
Solution:
(i)
Students in science = (central angle × Total students) / 360°
1000 = (100
o
× Total students) / 360
o
Total students = (1000 × 360
o
)/100
o
= 3600 students
? Total number of students are 3600.
(ii) Students in arts = (central angle of arts × Total students) / 360°
= (120
o
× 3600) / 360
o
= 1200 students
? Ratio of students in science and arts is 1000:1200 = 5:6
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.
Solution:
Marks secured in mathematics = (central angle of maths × Total score secured) / 360°
= (108 × 440) / 360
o
= 132 marks
Marks secured in science = (central angle of science × Total score secured) / 360°
= (81 × 440) / 360
o
= 99 marks
Marks secured in English = (central angle of English × Total score secured) / 360°
= (72 × 440) / 360
o
= 88 marks
Marks secured in Hindi = (central angle of Hindi × Total score secured) / 360°
= (54 × 440) / 360
o
= 66 marks
Marks secured in social science = (central angle of social science × Total score secured) /
360°
= (45 × 440) / 360
o
= 55 marks
Subject Mathematics Science English Hindi Social
Science
Marks
secured
132 99 88 66 55
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.
Page 5
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.
Solution:
(i) By using the formula,
Expenditure incurred on cement = (central angle × Total cost) / 360°
Total cost of the flat = (360° × 112500) / 75
o
= Rs 540000
(ii) By using the formula,
Expenditure incurred on labour = (central angle × Total cost) / 360°
= (100° × 540000) / 360
o
= Rs 150000
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
Solution:
We know that,
Total Production = 81000 Tonnes.
So,
(i) Production of wheat = (central angle of wheat × Total production) / 360°
= (120
o
× 81000) / 360
o
= 27000 tonnes
(ii) Production of sugar = (central angle of sugar × Total production) / 360°
= (100
o
× 81000) / 360
o
= 22500 tonnes
(iii) Production of rice = (central angle of rice × Total production) / 360°
= (60
o
× 81000) / 360
o
= 13500 tonnes
(iv) Production of maize = (central angle of maize × Total production) / 360°
= (30
o
× 81000) / 360
o
= 6750 tonnes
(v) Production of gram = (central angle of gram × Total production) / 360°
= (50
o
× 81000) / 360
o
= 11250 tonnes
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?
Solution:
(i)
Students in science = (central angle × Total students) / 360°
1000 = (100
o
× Total students) / 360
o
Total students = (1000 × 360
o
)/100
o
= 3600 students
? Total number of students are 3600.
(ii) Students in arts = (central angle of arts × Total students) / 360°
= (120
o
× 3600) / 360
o
= 1200 students
? Ratio of students in science and arts is 1000:1200 = 5:6
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.
Solution:
Marks secured in mathematics = (central angle of maths × Total score secured) / 360°
= (108 × 440) / 360
o
= 132 marks
Marks secured in science = (central angle of science × Total score secured) / 360°
= (81 × 440) / 360
o
= 99 marks
Marks secured in English = (central angle of English × Total score secured) / 360°
= (72 × 440) / 360
o
= 88 marks
Marks secured in Hindi = (central angle of Hindi × Total score secured) / 360°
= (54 × 440) / 360
o
= 66 marks
Marks secured in social science = (central angle of social science × Total score secured) /
360°
= (45 × 440) / 360
o
= 55 marks
Subject Mathematics Science English Hindi Social
Science
Marks
secured
132 99 88 66 55
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.
Solution:
Let us calculate the total marks.
So,
Marks scored in mathematics = (central angle of maths × Total marks) / 360°
135 = (90 × Total marks) / 360
o
Total marks = (135 × 360)/90
= 540 marks
Now,
Marks scored in Hindi = (central angle of Hindi × Total marks) / 360°
= (60 × 540) / 360
o
= 90 marks
Marks scored in Science = (central angle of Science × Total marks) / 360°
= (76 × 540) / 360
o
= 114 marks
Marks scored in Social science = (central angle of Social science × Total marks) / 360°
= (72 × 540) / 360
o
= 108 marks
Marks scored in English = (central angle of English × Total marks) / 360°
= (62 × 540) / 360
o
= 93 marks
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