RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

Class 8 : RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

The document RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

PAGE NO 25.13:

Question 12:

Represent the following data by a pie-diagram:

Items of expenditureExpenditure
Family AFamily B
Food40006400
Clothing2500480
Rent15003200
Education4001000
Miscellaneous1600600
Total1000016000

Ans.

We know:
Central angle of a component = (component value/sum of component values × 360)
Here the total expenditure of family A = 10000 and family B = 11680

Thus the central angle for each component can be calculated as follows:
 

Item Expenditure (Family A)Sector angle (Family A)Expenditure
(Family B)
Sector angle
(Family B)
Food40004000/10000 × 360 = 144 64006400/11680 × 360 = 197.3
Clothing25002500/10000 × 360 = 90480480/11680 × 360 = 14.8
Rent15001500/10000 × 360 = 5432003200/11680 × 360 = 98.6
Education400400/10000 × 360 = 14.410001000/11680 × 360 = 30.8
Miscellaneous16001600/10000 × 360 = 57.6600600/11680 × 360 = 18.5

Total expenditure of family A: 10000
Total expenditure of family B: 11680  (not 16000)

Now, the pie chat representing the given data can be constructed by following the steps below:
Step 1 : Draw circle of an appropriate radius.
Step 2 : Draw a vertical radius of the circle drawn in step 1.
Step 3 : Choose the largest central angle. Here the largest central angle is 144o. Draw a sector with the central angle 144o in such a way that one of its radii coincides with the radius drawn in step 2 and another radius is in its counter clockwise direction.
Step 4 : Construct other sectors representing the other items in the clockwise sense in descending order of magnitudes of their central angles.
Step 5 : Shade the sectors with different colours and label them, as shown as in figure below.

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev


Question 13:

Following data gives the break up of the cost of production of a book:

PrintingPaperBinding chargesAdvertisementRoyaltyMiscellaneous
30%15%15%20%10%15%

Draw a pie- diagram depicting the above information.

Ans.

We know:
Central angle of a component = (component value/sum of component values × 360)
Here, total expenditures = 105%
Thus, the central angle for each component can be calculated as follows:

Item Expenditure
(in %)
Sector angle
Printing3030/105 × 360 = 102.9
Paper1515/105 × 360 = 51.4
Binding charges1515/105 × 360 = 51.4
Advertisement2020/105 × 360 = 68.6
Royalty1010/105 × 360 = 34.3
Miscellaneous1515/105 × 360 = 51.4

Total : 105%

Now, the pie chat representing the given data can be constructed by following the steps below:
Step 1 : Draw circle of an appropriate radius.
Step 2 : Draw a vertical radius of the circle drawn in step 1.
Step 3 : Choose the largest central angle. Here the largest central angle is 102.9o. Draw a sector with the central angle 102.9o in such a way that one of its radii coincides with the radius drawn in step 2 and another radius is in its counter clockwise direction.
Step 4 : Construct other sectors representing the other items in the clockwise sense in descending order of magnitudes of their central angles.
Step 5 : Shade the sectors with different colours and label them, as shown as in figure below.
RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev


Question 14:

Represent the following data with the help of a pie-diagram:

ItemsWheatRiceTea
Production (in metric tons)32601840900

Ans.

We know:
Central angle of a component = (component value/sum of component values x 360)
Here, total production = 6000 (in metric tons)
Thus, the central angle for each component can be calculated as follows:

ItemProduction
(in metric tons)
Sector angle
Wheat32603260/6000 x 360 = 195.6
Rice18401840/6000 x 360 = 1 10.4
Tea900900/6000 x 360 = 54

Total = 6000 (in metric tons)

Now, the pie chat representing the given data can be constructed by following the steps below:
Step 1 : Draw circle of an appropriate radius.
Step 2 : Draw a vertical radius of the circle drawn in step 1.
Step 3 : Choose the largest central angle. Here, the largest central angle is 195.6o. Draw a sector with the central angle 195.6 o in such a way that one of its radii coincides with the radius drawn in step 2 and another radius is in its counter clockwise direction.
Step 4 : Construct the other sectors representing the other items in the clockwise direction  in descending order of magnitudes of their central angles.
Step 5 : Shade the sectors with different colours and label them as shown in the figure below.
RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

PAGE NO 25.14:

Question 15:

Draw a pie-diagram representing the relative frequencies (expressed as percentage) of the eight classes as given below:
  12.6, 18.2, 17.5, 20.3, 2.8, 4.2, 9.8, 14.7

Ans.

We know:
Central angle of a component = (component value/sum of component values × 360)
Here, total amount = 100.1%
Thus, central angle for each component can be calculated as follows:

ItemAmount (in %)Sector angle
Class I12.612.6/100.1 × 360 = 45.3
Class II18.218.2/100.1 × 360 = 65.5
Class III17.517.5/100.1 × 360 = 62.9
Class IV20.320.3/100.1 × 360 = 73
Class V2.82.8/100.1 × 360 = 10.1
Class VI4.24.2/100.1 × 360 = 15.1
Class VII9.89.8/100.1 × 360 = 35.2
Class VIII14.714.7/100.1 × 360 = 52.9

Total = 100.1%

Now, the pie chat representing the given data can be constructed by following the steps below:
Step 1 : Draw circle of an appropriate radius.
Step 2 :  Draw a vertical radius of the circle drawn in step 1
Step 3 : Choose the largest central angle. Here the largest central angle is 73o. Draw a sector with the central angle 73in such a way that one of its radii coincides with the radius drawn in step 2 and another radius is in its counter clockwise direction.
Step 4 : Construct other sectors representing the other items in the clockwise sense in descending order of magnitudes of their central angles.
Step 5 : Shade the sectors with different colours and label them, as shown as in the figure below.

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev


Question 16:

Following is the break up of the expenditure of a family on different items of consumption:

ItemsFoodClothingRentEducationFuel etc.MedicineMiscellaneous
Expenditure (in Rs)160020060015010080270

Draw a pie-diagram to represent the above data.

Ans.

We know:
Central angle of a component = (component value/sum of component values × 360)
Here, total expenditure = Rs 3000
Thus, central angle for each component can be calculated as follows:

ItemExpenditure (in Rs)Sector angle
Food16001600/3000 × 360 = 192
Clothing200200/3000 × 360 = 24
Rent600600/3000 × 360 = 72
Education150150/3000 × 360 = 18
Fuel etc100100/3000 × 360 = 12
Medicine8080/3000 × 360 = 9.6
Miscellaneous270270/3000 × 360 = 32.4

Total : 3000 (in Rs)

Now, the pie chat representing the given data can be constructed by following the steps below:
Step 1 : Draw a circle of an appropriate radius.
Step 2 : Draw a vertical radius of the circle drawn in step 1.
Step 3 : Choose the largest central angle. Here, the largest central angle is 192o. Draw a sector with the central angle 192o in such a way that one of its radii coincides with the radius drawn in step 2 and another radius is in its counter clockwise direction.
Step 4 : Construct other sectors representing the other items in the clockwise sense in descending order of magnitudes of their central angles.
Step 5 : Shade the sectors with different colours and label them as shown in the figure below.

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

 


Question 17:

Draw a pie-diagram for the following data of the investment pattern in a five year plan:

AgricultureIrrigation and PowerSmall IndustriesTransportSocial serviceMiscellaneous
14%16%29%17%16%8%

Ans.

We know:
Central angle of a component = (component value/sum of component values x 360)
Here the total percentage = 100%
Thus, the central angle for each component can be calculated as follows:
 

ItemAmount
(in %)
Sector angle
Agriculture1414/100 x 360 = 50.4
Irrigation and Power1616/100 x 360 = 57.6
Small Industries2929/100 x 360 = 104.4
Transport1717/100 x 360 = 61.2
Social Service1616/100 x 360  = 57.6
Miscellaneous88/100 x 360 = 28.8


Now, the pie chat representing the given data can be constructed by following the steps below:
Step 1 : Draw circle of an appropriate radius.
Step 2 : Draw a vertical radius of the circle drawn in step 1.
Step 3 : Choose the largest central angle. Here the largest central angle is 104.4o. Draw a sector with the central angle 104.4o in such a way that one of its radii coincides with the radius drawn in step 2 and another radius is in its counter clockwise direction.
Step 4 : Construct the other sectors representing the other items in the clockwise sense in descending order of magnitudes of their central angles.
Step 5 : Shade the sectors with different colours and label them as shown in the figure below.

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev
 

PAGE NO 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:

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev
(i) Total cost of the flat.
  (ii) Expenditure incurred on labour.

Ans.

(i) Expenditure incurred on cement  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev
Total cost of the flat =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev =  Rs 540000
                                       
(ii) Expenditure incurred on labour =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev =  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
RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

Ans.

(i)
Production of wheat  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  27000 tonnes
               
(ii)
Production of sugar  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  22500 tonnes

(iii)
Production of rice  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev =  13500 tonnes

(iv)
Production of maize  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  6750 tonnes

(v)
Production of gram  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  11250 tonnes

 

PAGE NO 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:

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev
  (i) What is the total number of students?
  (ii) What is the ratio of students in science and arts?

Ans.


(i) Students in science =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev 1000 =RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

 ∴ Total students  =  3600

(ii) Students in arts =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

∴ Ratio of students in science and arts  =  1000:1200  =  5:6 


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.

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

Ans.

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.
RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

Ans.

Marks scored in mathematics   =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev 135  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

Total Marks  =  540

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 NO 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:

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

(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?

Ans.

(i) Money spent on rent  = RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  Rs 3,600

(ii)  Money spent on education  = RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev =  Rs 1,600

(iii) Money spent on food  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev       RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev =  6,000
Ratio of expenses on food and rent  =  6000/3600  =  5/3


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.

RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev

Ans.

Amount spent on cricket   = RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =   RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  Rs 45,000

Amount spent on hockey   =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =   RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  Rs 30,000

Amount spent on football   =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  Rs 18,000

Amount spent on tennis  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  RD Sharma Solutions - Chapter 25 - Data Handling-III (Part - 2), Class 8, Maths Class 8 Notes | EduRev  =  Rs 15,000

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