31 Look Again Figure 83 on Page 258 and Answer the Following Questions

NCERT Exemplar Solutions Class eight Maths Affiliate 2 – Complimentary PDF Download

NCERT Exemplar for Course 8 Maths Chapter 2 Data Treatment, is provided here for students to fix for exams. These exemplars problems and solutions are designed by experts in accordance with CBSE syllabus(2021-2022) for 8th standard, which covers all the topics of Maths Chapter 2. Solving the questions from the exemplars will help the students to larn the concepts of data handling in a amend and like shooting fish in a barrel way. In this affiliate, the students will learn about different types of data and how to represent them. Solving the Exemplar questions for Form 8 volition help the students to articulate the fundamental concepts present in each chapter. Below is the list of topics covered in Chapter 2 of Grade eight Maths field of study.

• Represent the data in Pictorial form, using bar graphs and double bar graph
• Learn, organizing and group of data
• Representation of information via histogram and pie chart
• Probability basic concepts

Solve these Exemplars problems and score good marks in the master exams. Also, it is recommended to solve sample papers and previous twelvemonth question papers which gives an idea of question types asked in the board exam from Chapter 2 Data handling. BYJU'S also provide notes, exemplar books, Maths NCERT Solutions for 8th standard and question papers to help students practice well for their exams.

NCERT Exemplar Solutions for Grade eight Maths Chapter 2 Information Handling:-Download PDF Here

Access Answers to NCERT Exemplar SOluti Grade 8 Maths Chapter two

Exercise Page: 42

In questions 1 to 35 at that place are four options given, out of which 1 is correct. Choose the correct reply.

one. The peak of a rectangle in a histogram shows the

(a) Width of the grade (b) Upper limit of the class

(c) Lower limit of the class (d) Frequency of the class

Solution:-

(d) Frequency of the grade

2. A geometric representation showing the relationship between a whole and its parts is a

(a) Pie nautical chart (b) Histogram (c) Bar graph (d) Pictograph

Solution:-

(a) Pie chart

A geometric representation showing the human relationship between a whole and its parts is pie chart.

three. In a pie chart, the total angle at the centre of the circle is

(a) 180^o (b) 360^o (c) 270^o (d) ninety^o

Solution:-

(b) 360^o

4. The range of the data 30, 61, 55, 56, 60, 20, 26, 46, 28, 56 is

(a) 26 (b) thirty (c) 41 (d) 61

Solution:-

(c) 41

The difference between the lowest and the highest observation in a given information is called its Range.

And then, Range = Highest observation – Lowest observation

= 61 – 20

= 41

5. Which of the following is non a random experiment?

(a) Tossing a coin (b) Rolling a dice

(c) Choosing a card from a deck of 52 cards

(d) Throwing a stone from a roof of a building

Solution:-

(d) Throwing a stone from a roof of a edifice.

There is only one output that is the stone will autumn down therefore it is non a random experiment.

6. What is the probability of choosing a vowel from the alphabets?

(a) 21/26 (b) five/26 (c) 1/26 (d) three/26

Solution:-

(b) 5/26

Probability = Number of Vowels/ Total number of alphabets

= 5/26

7. In a schoolhouse merely, iii out of 5 students can participate in a contest. What is the probability of the students who do non make information technology to the competition?

(a) 0.65 (b) 0.4 (c) 0.45 (d) 0.6

Solution:-

(b) 0.4

Probability = Number of students who practise not make it to the contest / Total number

of students.

= 2/5

= 0.four

Students of a class voted for their favorite colour and a pie chart was prepared based on the information collected.

Discover the pie chart given below and reply questions 8 –ten based on it.

8. Which color received i/5 of the votes?

(a) Red (b) Blue (c) Green (d) Xanthous

Solution:-

(c) Green

= 1/five × 100

= 0.two × 100

= xx%

9. If 400 students voted in all, then how many did vote 'Others' color as their favorite?

(a) 6 (b) 20 (c) 24 (d) twoscore

Solution:-

(c) 24

From the pie chart, 6% out of 400 students voted for others,

So,

= (6%/100%) × 400

= 0.06 × 400

= 24

10. Which of the following is a reasonable conclusion for the given data?

(a) (one/twenty)^th student voted for bluish colour

(b) Green is the to the lowest degree popular colour

(c) The number of students who voted for ruby-red color is two times the number of students who voted for yellow colour

(d) Number of students liking together xanthous and light-green colour is approximately the same as those for red colour.

Solution:-

(d) Number of students liking together xanthous and green color is approximately the same as those for cerise colour.

xi. Listed below are the temperature in ^oC for ten days.

–half dozen, –8, 0, three, 2, 0, 1, 5, 4, 4

What is the range of the data?

Solution:-

(a) viii (b) 13^oC (c) 10^oC (d) 12^oC

Solution:-

(b) 13^oC

The difference betwixt the lowest and the highest ascertainment in a given data is called its Range.

Then, Range = Highest ascertainment – Lowest observation

= 5 – (-8)

= 5 + 8

= thirteen^oC

12. Ram put some buttons on the tabular array. There were 4 blue, 7 cerise, 3 black and vi white buttons in all. All all of a sudden, a cat jumped on the table and knocked out one button on the flooring. What is the probability that the push button on the flooring is blueish?

(a) 7/20 (b) iii/five (c) 1/5 (d) ¼

Solution:-

(c) 1/5

Probability = Number of blue buttons on the table / Total number of buttons on the

table

= 4/xx … [divide both numerator and denominator by 4]

= i/five

13. Rahul, Varun and Yash are playing a game of spinning a coloured wheel. Rahul wins if spinner lands on red. Varun wins if spinner lands on blue and Yash wins if it lands on green. Which of the following spinner should be used to brand the game fair?

(a) (i) (b) (ii) (c) iii (d) (iv)

Solution:-

(d) (iv) this figure contains equal share for three colours.

fourteen. In a frequency distribution with classes 0 –10, x –twenty etc., the size of the course intervals is 10. The lower limit of fourth class is

(a) 40 (b) 50 (c) 20 (d) 30

Solution:-

(d) xxx

The lower value of the class interval is called its Lower Class Limit.

First class = 0 -10

Second grade = ten – xx

Third class = xx – 30

Quaternary class = 30 – twoscore

5th grade = 40 – l

and Tenth class = 90 – 100

15. A money is tossed 200 times and head appeared 120 times. The probability of getting a head in this experiment is

(a) two/v (b) 3/5 (c) 1/v (d) 4/five

Solution:-

(b) 3/5

Probability = Number of times caput appeared / Full number of times coin is tossed

= 120/200

= 12/twenty … [divide both numerator and denominator by 4]

= 3/5

xvi. Data collected in a survey shows that twoscore% of the buyers are interested in ownership a particular brand of toothpaste. The central angle of the sector of the pie chart representing this information is

(a) 120^o (b) 150^o (c) 144^o (d) 40°

Solution:-

(c) 144^o

= (40/100) × 360^o

= 0.iv × 360^o

= 144^o

17. Monthly salary of a person is Rs. 15000. The central angle of the sector representing his expenses on food and house hire on a pie nautical chart is threescore°. The amount he spends on food and house rent is

(a) Rs. 5000 (b) Rs. 2500 (c) Rs. 6000 (d) Rs. 9000

Solution:-

(b) Rs. 2500

From the question,

The part of monthly salary spent on food and house rent = 60^o/360^o

The amount he spends on food and house rent is = (threescore^o/360^o) × 15000

= Rs. 2500

18. The following pie nautical chart gives the distribution of constituents

in the human body. The central angle of the sector showing the

distribution of protein and other constituents is

(a) 108^o (b) 54^o (c) 30^o (d) 216^o

Solution:-

(a) 108^o

The following pie chart gives the distribution of constituents

in the homo body = sixteen% + fourteen% = 30%

= (30/100) × 360^o

= 108^o

19. Rohan and Shalu are playing with v cards as shown

in the figure. What is the probability of Rohan picking

a card without seeing, that has the number 2 on it?

(a) 2/5 (b) 1/five (c) 3/5 (d) iv/5

Solution:-

(a) 2/five

Probability = Number of cards having ii/ Total number of cards

= ii/5

20. The following pie chart represents the distribution of

proteins in parts of a human torso. What is the ratio of

distribution of proteins in the muscles to that of proteins

in the basic?

(a) 3 : 1 (b) 1 : 2 (c) 1 : iii (d) two : one

Solution:-

(d) two : 1

From the chart,

Distribution of protein in the muscles = one/3

Distribution of poly peptide in the bones = 1/6

So,

Ratio of distribution of proteins in the muscles to that of proteins in the bones,

= i/3 : 1/6

= (1/iii) × (6/i)

= (1/1) × (2/1)

= 2/1

= two : 1

21. What is the central angle of the sector (in the above pie chart) representing pare and bones together?

(a) 36^o (b) 60^o (c) 90^o (d) 96^o

Solution:-

(d) 96^o

Distribution of protein in the peel = 1/10

Distribution of protein in the bones = 1/6

Central angle of the sector representing skin and basic,

= 1/10 + 1/half-dozen

= (three + 5)/30

= viii/30 × 360^o

= 96^o

22. What is the central bending of the sector (in the in a higher place pie chart) representing hormones enzymes and other proteins.

(a) 120^o (b) 144^o (c) 156^o (d) 176^o

Solution:-

(b) 144^o

Distribution of protein in the skin = 1/ten

Distribution of protein in the basic = i/6

Distribution of protein in the Muscles = 1/iii

Central bending of the sector representing peel, muscles and bones,

= 1/10 + ane/6 + i/3

= (iii + five + ten)/30

= 18/30 × 360^o

= 216^o

Then,

Central angle of the sector representing hormones enzymes and other proteins,

= 360^o – 216^o

= 144^o

23. A coin is tossed 12 times and the outcomes are observed as shown below:

The risk of occurrence of Head is

(a) ½ (b) 5/12 (c) 7/12 (d) 5/7

Solution:-

(b) 5/12 … [from the effigy]

24. Total number of outcomes, when a ball is drawn from a handbag which contains iii blood-red, 5 blackness and 4 blue balls is

(a) 8 (b) seven (c) 9 (d) 12

Solution:-

(d) 12

25. A graph showing two sets of data simultaneously is known as

(a) Pictograph (b) Histogram (c) Pie chart (d) Double bar graph

Solution:-

(d) Double bar graph

26. Size of the form 150 –175 is

(a) 150 (b) 175 (c) 25 (d) –25

Solution:-

(c) 25

The deviation betwixt the upper class limit and lower course limit of a course is called the Size of the class.

= Upper limit – lower limit

= 175 – 150

= 25

27. In a throw of a dice, the probability of getting the number 7 is

(a) ½ (b) 1/half dozen (c) 1 (d) 0

Solution:-

(d) 0

Only 1, 2, 3, four, 5 and half dozen are in the dice. So, no take chances of getting number 7.

28. Information represented using circles is known as

(a)Bar graph (b) Histogram (c) Pictograph (d) Pie nautical chart

Solution:-

(d) Pie chart.

29. Tally marks are used to discover

(a) Grade intervals (b) Range (c) Frequency (d) Upper limit

Solution:-

(c) Frequency

30. Upper limit of form interval 75 –85 is

(a) 10 (b) –x (c) 75 (d) 85

Solution:-

(d) 85

31. Numbers 1 to 5 are written on separate slips, i.e. one number on i skid and put in a box. Wahida pick a slip from the box without looking at it. What is the probability that the sideslip bears an odd number?

(a) i/5 (b) 2/5 (c) 3/five (d) 4/five

Solution:-

(c) three/5

Number 1 to 5 = i, 2, three, four, 5

Odd number i to 5 = 1, 3, 5

Probability = full number of odd number slips/ Full number of slips

= 3/5

32. A glass jar contains six blood-red, 5 light-green, 4 blue and v xanthous marbles

of aforementioned size. Hari takes out a marble from the jar at random.

What is the probability that the chosen marble is of ruby colour?

(a) vii/x (b) 3/10 (c) 4/5 (d) 2/5

Solution:-

(b) three/ten

Probability = total number of red marbles/ Total number of marbles

= 6/20 … [divide numerator and denominator by 2]

= 3/10

33. A coin is tossed ii times. The number of possible outcomes is

(a) 1 (b) 2 (c) 3 (d) 4

Solution:-

(d) 4

Head Head, Tail Tail, Tail Head, Head Tail.

34. A coin is tossed 3 times. The number of possible outcomes is

(a) 3 (b) iv (c) 6 (d) 8

Solution:-

(d) 8

HHH, TTT, THH, HTH, HHT, HTT, THT, TTH

35. A dice is tossed two times. The number of possible outcomes is

(a) 12 (b) 24 (c) 36 (d) 30

Solution:-

(c) 36

(ane, 1), (ane, 2), (one, 3), (1, 4), (i, 5), (one, half-dozen) = 6

(2, one), (2, 2), (2, 3), (2, 4), (2, 5), (2, half dozen) = 6

(iii, 1), (3, 2), (three, iii), (3, 4), (iii, 5), (three, half-dozen) = 6

(4, 1), (4, ii), (4, 3), (4, 4), (4, five), (4, 6) = 6

(5, 1), (5, 2), (v, 3), (five, 4), (5, v), (5, 6) = 6

(6, 1), (6, two), (6, 3), (six, four), (6, 5), (6, 6) = vi

= 36

In questions 36 to 58, fill in the blanks to brand the statements true.

36. Data available in an unorganised grade is chosen data.

Solution:-

Data available in an unorganised form is called raw information.

37. In the class interval xx – 30, the lower course limit is .

Solution:-

In the class interval twenty – 30, the lower course limit is xx.

38. In the class interval 26 – 33, 33 is known as .

Solution:-

In the class interval 26 – 33, 33 is known as upper form limit.

39. The range of the data 6, eight, 16, 22, viii, 20, 7, 25 is .

Solution:-

The difference between the lowest and the highest ascertainment in a given information is called its Range.

So, Range = Highest observation – Everyman ascertainment

= 25 – 6

= 19

The range of the data half-dozen, viii, 16, 22, 8, 20, vii, 25 is 19.

40. A pie chart is used to compare to a whole.

Solution:-

A pie chart is used to compare parts to a whole.

41. In the experiment of tossing a money one time, the outcome is either or

.

Solution:-

In the experiment of tossing a coin one time, the result is either Head or Tail.

42. When a dice is rolled, the six possible outcomes are .

Solution:-

When a dice is rolled, the vi possible outcomes are i, 2, iii, four, 5 and 6.

43. Each outcome or a collection of outcomes in an experiment makes an .

Solution:-

Each outcome or a collection of outcomes in an experiment makes an Issue.

44. An experiment whose outcomes cannot be predicted exactly in accelerate is called a

experiment.

Solution:-

An experiment whose outcomes cannot be predicted exactly in advance is chosen a random experiment.

45. The difference betwixt the upper and lower limit of a class interval is chosen the

of the class interval.

Solution:-

The difference between the upper and lower limit of a class interval is chosen the

Size/width of the class interval.

46. The sixth course interval for a grouped data whose first two class intervals are x – 15 and 15 – 20 is .

Solution:-

The 6th form interval for a grouped data whose first two course intervals are 10 – 15 and 15 – 20 is 35 – 40.

Histogram given on the right shows the number of

people owning the different number of books.

Reply 47 to 50 based on it

47. The total number of people surveyed is .

Solution:-

Course the Histogram,

The total number of people surveyed is 35.

48. The number of people owning books more than

60 is .

Solution:-

The number of people owning books more than than

60 is viii.

49. The number of people owning books less than

40 is .

Solution:-

The number of people owning books less than 40 is 22.

fifty. The number of people having books more than xx and less than 40 is .

Solution:-

The number of people having books more than than 20 and less than 40 is 14.

51. The number of times a particular observation occurs in a given data is called its

.

Solution:-

The number of times a detail ascertainment occurs in a given information is chosen its Frequency.

52. When the number of observations is large, the observations are usually organised in groups of equal width called .

Solution:-

When the number of observations is large, the observations are usually organised in groups of equal width called class intervals.

53. The total number of outcomes when a coin is tossed is .

Solution:-

The total number of outcomes when a coin is tossed is 2.

54. The course size of the interval eighty – 85 is .

Solution:-

= 85 – eighty

= v

The class size of the interval 80 – 85 is five.

55. In a histogram __________ are drawn with width equal to a course interval without leaving whatsoever gap in betwixt.

Solution:-

In a histogram confined are drawn with width equal to a class interval without leaving any gap in betwixt.

56. When a dice is thrown, outcomes 1, 2, three, 4, 5, vi are equally .

Solution:-

When a dice is thrown, outcomes 1, two, three, 4, 5, 6 are equally likely.

57. In a histogram, class intervals and frequencies are taken along __________ axis and __________ centrality.

Solution:-

In a histogram, course intervals and frequencies are taken along Ten axis and Y axis.

58. In the class intervals 10 –20, xx –thirty, etc., respectively, 20 lies in the class .

Solution:-

In the class intervals 10 –20, twenty –30, etc., respectively, 20 lies in the class 20 – 30.

In questions 59 to 81, land whether the statements are true (T) or imitation (F).

59. In a pie nautical chart a whole circumvolve is divided into sectors.

Solution:-

True.

Pie nautical chart shows the human relationship between a whole and its parts.

60. The central angle of a sector in a pie chart cannot exist more than 180^o.

Solution:-

False.

The cardinal angle of a sector in a pie chart is more than 180^o. But, cannot be more than 360^o.

61. Sum of all the central angles in a pie chart is 360°.

Solution:-

Truthful.

62. In a pie chart 2 cardinal angles tin be of 180^o.

Solution:-

True.

We know that sum of all the central angles in a pie chart is 360^o. So, if 2 angles have 180^o each and then the sum of two angles is 360^o.

63. In a pie nautical chart two or more than central angles can exist equal.

Solution:-

True.

Yes, in a pie chart two or more cardinal angles tin be equal.

64. Getting a prime number on throwing a die is an event.

Solution:-

Truthful.

Each outcome or a collection of outcomes in an experiment makes an Effect.

Using the following frequency table, reply question 65-68

Marks (obtained out of ten)	iv	5	7	viii	9	10
Frequency	5	10	8	half dozen	12	9

65. 9 students got full marks.

Solution:-

Truthful.

From the tabular array, given that nine students got ten out of x marks.

66. The frequency of less than 8 marks is 29.

Solution:-

False.

The frequency of less than 8 marks is viii + 10 + five = 23 students.

67. The frequency of more than than 8 marks is 21.

Solution:-

True.

The frequency of more than 8 marks is 12 + 9 = 21 students.

68. 10 marks the highest frequency.

Solution:-

False.

9 marks has highest frequency, i.due east. 12. But, 10 has 9 frequency.

69. If the fifth class interval is 60 – 65, 4th class interval is 55 – lx, so the first form interval is 45 –fifty.

Solution:-

False.

The first course interval is 40 – 45, second class interval is 45 – 50, tertiary class interval is

l – 55, fourth class interval 55 – 60, fifth class interval is 60 – 65.

70. From the histogram given on the right, nosotros

can say that 1500 males above the age of 20 are

literate.

Solution:-

False.

From the histogram given on the correct, we

can say that 1900 (600 + 800 + 500) males above the historic period of 20 are

literate.

71. The class size of the class interval sixty – 68 is eight.

Solution:-

Truthful.

The difference between the upper and lower limit of a course interval is called the

Size of the class interval. i.e. 68 – 60 = 8

72. If a pair of coins is tossed, then the number of outcomes are ii.

Solution:-

False.

If a pair of coins is tossed, then the number of outcomes are 4, i.east. HH, TT, TH and HT.

73. On throwing a die once, the probability of occurence of an even number is ½.

Solution:-

Truthful.

The even numbers in a dice are ii, 4 and six.

Probability = full number of even numbers/ Total numbers

= 3/6 … [divide numerator and denominator by 3]

= ½

74. On throwing a dice in one case, the probability of occurence of a composite number is ½.

Solution:-

Imitation.

The blended numbers in a die are 4 and 6.

Probability = total number of blended numbers/ Full numbers

= 2/6 … [carve up numerator and denominator by 2]

= i/3

75. From the given pie chart, nosotros can infer that production of Manganese is least in state B.

Solution:-

False.

It is not possible to declare, from the given pie chart that the production of Manganese is least in state B. Because, we exercise not know the central angle of the sectors.

76. 1 or more than outcomes of an experiment make an event.

Solution:-

True.

Each outcome or a drove of outcomes in an experiment makes an Event.

77. The probability of getting number half-dozen in a throw of a dice is one/6. Similarly the probability of getting a number 5 is 1/5.

Solution:-

Fake.

The probability of getting a number v is 1/6.

78. The probability of getting a prime is the same as that of a composite number in a throw of a dice.

Solution:-

False.

The composite numbers in a dice are four and 6.

Probability = total number of composite numbers/ Full numbers

= 2/6 … [split numerator and denominator past 2]

= 1/3

The prime numbers in a die are 2, iii and 5.

Probability = total number of prime numbers/ Total numbers

= 3/6 … [divide numerator and denominator by 3]

= ½

79. In a throw of a die, the probability of getting an even number is the aforementioned equally that of getting an odd number.

Solution:-

True.

The even numbers in a dice are 2, iv and 6.

Probability = total number of even numbers/ Total numbers

= 3/6 … [dissever numerator and denominator by 3]

= ½

The odd numbers in a die are 1, 3 and five.

Probability = total number of odd numbers/ Total numbers

= 3/6 … [split up numerator and denominator by three]

= ½

80. To verify Pythagoras theorem is a random experiment.

Solution:-

False.

We know the result i.e. only one effect before going to verify the Pythagoras theorem. So, it is non a random experiment.

81. The following pictorial representation of data is a histogram.

Solution:-

Truthful.

Histogram is a type of bar diagram, where the class intervals are shown on the horizontal centrality and the heights of the confined (rectangles) show the frequency of the class interval, just there is no gap between the bars as there is no gap between the class intervals.

82. Given below is a frequency distribution table. Read information technology and answer the question that follow.

Class Interval	Frequency
10 – xx	5
20 – thirty	ten
xxx – 40	4
40 – fifty	fifteen
50 – 60	12

(a) What is the lower limit of the second grade interval?

Solution:-

The lower limit of the second grade interval xx – 30 is xx.

(b) What is the upper limit of the last class interval?

Solution:-

The upper limit of the last form interval l – lx is 60.

(c) What is the frequency of the third class?

Solution:-

The frequency of the third form 30 – 40 is 4.

(d) Which interval has a frequency of 10?

Solution:-

Second form interval xx – 30 has a frequency of x.

(due east) Which interval has the lowest frequency?

Solution:-

Third interval thirty – 40 has the lowest frequency i.east. 4.

(f) What is the form size?

Solution:-

The departure between the upper and lower limit of a grade interval is called the Size of the form interval.

Size = Upper limit – Lower limit

= xx – x

= 10

83. The top speeds of thirty different land animals have been organised into a frequency table. Draw a histogram for the given data.

Maximum Speed (Km/h)	Frequency
ten – 20	5
20 – 30	5
30 – xl	10
40 – 50	viii
50 – lx	0
60 – 70	2

Solution:-

Histogram is a type of bar diagram, where the class intervals are shown on the horizontal centrality and the heights of the bars (rectangles) testify the frequency of the form interval, only at that place is no gap between the bars every bit there is no gap between the class intervals.

84. Given below is a pie nautical chart showing the time spend by a grouping of 350 children in dissimilar games. Observe it and respond the questions that follow.

(a) How many children spend at least one 60 minutes in playing games?

Solution:-

From the given pie nautical chart,

Percentage of children who at least spend one hour in playing = 16 + 30 + 34 + x + 4

= 94 %

Number of children out of 350 who at to the lowest degree spend one 60 minutes in playing = (94/100) × 350

= 0.94 × 350

= 329 children

(b) How many children spend more than 2 hours in playing games?

Solution:-

From the given pie chart,

Per centum of children who spend ii hours in playing = 34 + ten + iv

= 48 %

Number of children out of 350 who at to the lowest degree spend one hr in playing = (48/100) × 350

= 0.48 × 350

= 168 children

(c) How many children spend 3 or lesser hours in playing games?

Solution:-

From the given pie chart,

Percentage of children who at least spend ane hour in playing = 6 + 16 + 30 + 34

= 86 %

Number of children out of 350 who at least spend ane 60 minutes in playing = (86/100) × 350

= 0.86 × 350

= 301 children

(d) Which is greater — number of children who spend 2 hours or more than per mean solar day or number of children who play for less than ane hour?

Solution:-

Number of children who spend 2 hours or more per twenty-four hour period = 30 + 34 + 10 + 4 = 78%

= (78/100) × 350

= 0.78 × 350

= 273 children

Number of children who play for less than i hour = vi%

= (6/100) × 350

= 0.06 × 350

= 21 children

So, by comparison both it is clear that number of children who spend 2 hours or more per twenty-four hour period is greater

85. The pie chart on the right shows the result of a survey

Carried out to detect the modes of travel used past the children to become

to school. Written report the pie chart and respond the questions that

follow.

(a) What is the well-nigh common style of ship?

Solution:-

By observing the given pie chart the well-nigh common way of transport is Passenger vehicle i.e. its fundamental angle is 120^o.

(b) What fraction of children travel by car?

Solution:-

By observing the given pie nautical chart, central angle of the children travel past auto is 90^o.

Then,

The fraction of children travel past car,

= (90^o/360^o)

= ix/36 … [separate both numerator and denominator by nine]

= ¼

(c) If 18 children travel by car, how many children took office in the survey?

Solution:-

Let us assume total number of children took part in the survey be y

Number of children travel by machine = (¼) of total number of children

18 = ¼ × y

Y = 18 × 4

Y = 72

And so, 72 children took part in the survey.

(d) How many children use taxi to travel to school?

Solution:-

Primal angle of children employ taxi to travel to school = (360^o – (120^o + 60^o + 90^o + 60^o)

= (360^o – 330^o)

= 30^o

And so,

Out of 72 children number of children employ taxi to travel to school = (thirty^o/360^o) × 72

= 6

(eastward) Past which two modes of send are equal number of children travelling?

Solution:-

By observing the given pie chart, cycle and walk are the two modes of transport are equal number of children travelling. Because, central bending of the two modes are same i.due east. lx^o.

86. A die is rolled one time. What is the probability that the number on top volition be

(a) Odd

(b) Greater than 5

(c) A multiple of 3

(d) Less than 1

(e) A gene of 36

(f) A factor of 6

Solution:-

A dice is rolled once, the possible outcomes are 1, ii, 3, four, five and 6.

Then,

The probability that the number on top will be

(a) Odd

Odd numbers in the dice are ane, 3, five

Then,

Probability = total number of odd numbers/ Total number of outcomes

= 3/vi … [divide numerator and denominator past 3]

= ½

(b) Greater than 5

Greater than five is 6

Then,

Probability = total number Greater than 5/ Total number of outcomes

= i/6

(c) A multiple of 3

Multiple of 3 = 3 and 6

Then,

Probability = total number of multiple of iii/ Total number of outcomes

= 2/6 … [dissever numerator and denominator past ii]

= 1/3

(d) Less than 1

There is no number is less than 1.

So, probability of less than 1 = 0

(e) A gene of 36

Factors of 36 are 1, 2, three, 4 and 6

Probability = total number of factor of 36/ Total number of outcomes

= five/half-dozen

(f) A gene of half-dozen

Factors of half dozen are 1, 2, 3 and 6

Probability = total number of factor of vi/ Full number of outcomes

= 4/6 … [separate numerator and denominator by two]

= two/3

87. Classify the following statements under appropriate headings.

(a) Getting the sum of angles of a triangle as 180°.

(b) India winning a cricket match confronting Pakistan.

(c) Sun setting in the evening.

(d) Getting seven when a dice is thrown.

(e) Sun rising from the west.

(f) Winning a racing competition by you.

Certain to happen	Impossible to happen	May or may not happen

Solution:-

(a) Information technology is certain to happen. Considering, the sum of angles of a triangle as 180^o.

(b) It may or may non happen. Because, the issue of lucifer is unpredictable.

(c) Information technology is certain to happen. Because, the sun always sets in the evening.

(d) It is impossible to happen. Because, vii is non an outcome when a dice is thrown.

(eastward) It is impossible to happen. Because, Sun e'er rises from the east.

(f) It may or may not happen. Considering, the consequence is unpredictable.

89. Ritwik draws a ball from a bag that contains white and yellowish balls. The probability of choosing a white ball is ii/nine. If the total number of balls in the purse is 36, discover the number of xanthous balls.

Solution:-

From the question is given that,

Probability of choosing a white ball is = 2/nine

The total number of assurance in the bag is = 36

Number of white ball chosen = (2/9) × 36

= 8 white balls

Then, the number of yellow balls = Total assurance in bag – number of white balls

= 36 – eight

= 28 yellowish balls

ninety. Look at the histogram below and answer the questions that follow.

(a) How many students accept height more than or equal to 135 cm simply less than 150 cm?

Solution:-

By observing the given histogram,

Students have height more than or equal to 135 cm but less than 150 = 14 + 18 + ten

= 42

(b) Which grade interval has the least number of students?

Solution:-

150 – 155 has the least number of students i.e. iv.

(c) What is the course size?

Solution:-

The divergence between the upper and lower limit of a course interval is called the

Size of the form interval.

Size = Upper limit – Lower limit

= 130 – 125

= 5

(d) How many students have elevation less than 140 cm?

Solution:-

By observing the given histogram,

Students have height less than 140 cm = half-dozen + 8 + 14

= 28

91. Following are the number of members in 25 families of a village:

6, eight, 7, vii, 6, five, 3, 2, 5, half-dozen, 8, seven, 7, iv, 3, half-dozen, half dozen, 6, vii, v, iv, 3, iii, 2, 5. Prepare a frequency distribution table for the information using course intervals 0 –2, 2 –4, etc.

Solution:-

Commencement, we take to adjust the number of members in 25 families of a hamlet ascending order.

= 2, 2, three, iii, 3, 3, 4, iv, v, five, 5, 5, vi, vi, 6, 6, half-dozen, 6, 7, 7, 7, vii, 7, 8, viii.

Now, we will draw the frequency tabular array of the given data.

92. Describe a histogram to represent the frequency distribution in question 91.

Solution:-

Scale:

On Ten – centrality, one large partition = 2 fellow member

On Y – axis, 1 large partition = ii family

93. The marks obtained (out of twenty) past xxx students of a class in a exam are as follows:

14, 16, 15, xi, fifteen, 14, xiii, xvi, 8, 10, 7, eleven, 18, 15, fourteen, 19, 20, 7, 10, xiii, 12, xiv, 15, 13, 16, 17, 14, 11, 10, 20.

Prepare a frequency distribution table for the above information using form intervals of equal width in which one class interval is 4 –8 (excluding 8 and including 4).

Solution:-

Offset, we have to arrange marks obtained (out of 20) by xxx students in an ascending order.

= 7, vii, eight, 10, 10, 10, xi, 11, 11, 12, thirteen, 13, 13, 14, 14, 14, 14, 14, 15, xv, 15, xv, xvi, xvi, 16, 17, eighteen, 19, 20, xx.

Now, nosotros will draw the frequency table of the given data.

94. Gear up a histogram from the frequency distribution table obtained in question 93.

Solution:-

Scale:

On Ten – centrality, 1 big division = 4 marks

On Y – axis, 1 big division = 2 students

95. The weights (in kg) of 30 students of a class are:

39, 38, 36, 38, 40, 42, 43, 44, 33, 33, 31, 45, 46, 38, 37, 31, 30, 39, 41, 41, 46, 36, 35, 34, 39, 43, 32, 37, 29, 26.

Set up a frequency distribution table using ane grade interval equally (xxx – 35), 35 not included.

(i) Which class has the least frequency?

(ii) Which form has the maximum frequency?

Solution:-

First, we have to arrange the weights (in kg) of xxx students of a class in an ascending order.

= 26, 29, 30, 31, 31, 32, 33, 33, 34, 35, 36, 36, 37, 37, 38, 38, 38, 39, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 46, 46

Now, nosotros will draw the frequency table of the given information.

(i) Grade first has the least frequency.

(2) Class third has the maximum frequency.

96. Shoes of the following brands are sold in Nov. 2007 at a shoe store. Construct a pie nautical chart for the information.

Make	Number of pair of shoes sold
A	130
B	120
C	90
D	40
E	twenty

Solution:-

From the table,

Full number of pairs of shoes sold = 130 + 120 + 90 + twoscore +20

= 400

Primal bending of pie nautical chart for each brands,

A = (130/400) × 360^o

= 117^o

B = (120/400) × 360^o

= 108^o

C = (ninety/400) × 360^o

= 81^o

D = (40/400) × 360^o

= 36^o

East = (xx/400) × 360^o

= eighteen^o

97. The following pie nautical chart depicts the expenditure of a country government under unlike heads.

(i) If the total spending is 10 crores, how much money was spent on roads?

Solution:-

Given, full amount spent by state authorities under different heads = 10 crores

From the given pie nautical chart,

Percent of coin spent on Roads = 10 %

Total money spent on Roads = (x/100) × 100000000

= 1,00,00,000

(ii) How many times is the amount of money spent on teaching compared to the amount spent on roads?

Solution:-

Given, total corporeality spent past state authorities nether dissimilar heads = 10 crores

From the given pie chart,

Pct of money spent on Roads = 10 %

Full money spent on Roads = (10/100) × 100000000

= ₹ 1,00,00,000

Percentage of money spent on Education = 25 %

Total money spent on Teaching = (25/100) × 100000000

= ₹ 2,50,00,000

Then,

= Total coin spent on Education/ Total coin spent on Roads

= ₹ ii,50,00,000/₹ 1,00,00,000

= ₹ 25/10

∴Money spent on instruction is 2.5 times of money spent on roads.

(iii) What fraction of the total expenditure is spent on both roads and public welfare together?

Solution:-

Given, total amount spent by state government nether different heads = 10 crores

From the given pie nautical chart,

Percentage of money spent on Roads = x %

Percentage of money spent on Public Welfare = 20 %

So,

Fraction of the total expenditure is spent on both roads and public welfare together,

= 10% + twenty%

= (10/100) + (20/100)

= (1/10) + (two/ten)

= (3/x)

98. The post-obit data represents the dissimilar number of animals in a zoo. Prepare a pie chart for the given information.

Animals	Number of animals
Deer	42
Elephant	15
Giraffe	26
Reptiles	24
Tiger	13

Solution:-

From the table,

Total number of animals in a zoo = 42 + 15 + 26 + 24 +xiii

= 120

Central angle of pie chart for each animals,

Deer= (42/120) × 360^o

= 126^o

Elephant = (15/120) × 360^o

= 45^o

Giraffe = (26/120) × 360^o

= 78^o

Reptiles = (24/120) × 360^o

= 72^o

Tiger = (13/120) × 360^o

= 39^o

99. Playing cards

(a) From a pack of cards the post-obit cards are kept confront down:

[Queen, King and Jack cards are called face cards.]

Suhail wins if he picks upwardly a face card. Find the probability of Suhail winning?

Solution:-

From the above figure,

Number of face cards = ane

Total number of cards = 7

Number of outcomes = 7

Probability = Number of confront cards / Total number cards

= one/seven

(b) Now the following cards are added to the higher up cards:

What is the probability of Suhail winning at present? Reshma wins if she picks up a 4. What is the probability of Reshma winning?

[Queen, King and Jack cards are called face cards.]

Solution:-

Number of face cards = 4

Full number of cards = xv

Number of outcomes = 15

Probability = Number of confront cards / Total number cards

= 4/fifteen

100. Construct a frequency distribution table for the following weights (in grams) of 35 mangoes, using the equal course intervals, one of them is forty – 45 (45 not included).

30, xl, 45, 32, 43, fifty, 55, 62, seventy, 70, 61, 62, 53, 52, 50, 42, 35, 37, 53, 55, 65, 70, 73, 74, 45, 46, 58, 59, 60, 62, 74, 34, 35, 70, 68.

(a) How many classes are there in the frequency distribution table?

(b) Which weight group has the highest frequency?

Solution:-

Get-go, nosotros have to adapt weights (in grams) of 35 mangoes in an ascending gild.

= thirty, 32, 34, 35, 35, 37, 40, 42, 43, 45, 45, 46, fifty, fifty, 52, 53, 53, 55, 55, 58, 59, lx, 61, 62, 62, 62, 65, 68, lxx, 70, 70, 70, 73, 74, 74.

Now, nosotros will describe the frequency table of the given data.

(a) There are 9 classes in the frequency distribution table.

(b) 70 – 75 weight has the highest frequency.

101. Complete the following tabular array:

Find the total number of persons whose weights are given in the to a higher place table.

Solution:-

102. Draw a histogram for the following data.

Class interval	10 – 15	15 – 20	twenty – 25	25 – 30	xxx – 35	35 – 40
Frequency	30	98	80	58	29	fifty

Solution:-

Scale:

On X – axis, one big division = 5 marks

On Y – axis, 1 big sectionalization = 10 students

103. In a hypothetical sample of 20 people, the corporeality of coin (in thousands of rupees) with each was constitute to be as follows:

114, 108, 100, 98, 101, 109, 117, 119, 126, 131, 136, 143, 156, 169, 182, 195, 207, 219, 235, 118.

Draw a histogram of the frequency distribution, taking i of the class intervals as

l –100.

Solution:-

From the data given in the question, get-go we have to gear up frequency distribution table,

Scale:

On X – centrality, 1 large segmentation = 50 ₹ marks

On Y – axis, 1 big segmentation = ii persons

104. The below histogram shows the number of literate females in the age grouping of x to 40 years in a town.

(a) Write the classes bold all the classes are of equal width.

(b) What is the classes width?

(c) In which age group are literate females the least?

(d) In which age group is the number of literate females the highest?

Solution:-

(a) The classes are, 10-15, 15-20, 20-25, 25-xxx, 30-35, and 35-40

b) The class width is 15 – 10 = 5

c) In the class outset grouping i.due east. ten-15 have the least female literate.

d) In the class 2nd group i.e. 15-20 have the highest female literate.

105. The following histogram shows the frequency distribution of teaching experiences of 30 teachers in various schools:

(a) What is the class width?

Solution:-

The width of the class is 5.

(b) How many teachers are having the maximum teaching experience and how many have the least didactics feel?

Solution:-

Teachers having maximum educational activity experience is two and Teachers having least pedagogy feel is 5.

(c) How many teachers accept teaching experience of ten to twenty years?

Solution:-

Teachers accept instruction experience of 10 to twenty years is 7 + 2 = 9.

106. In a district, the number of branches of different banks is given below:

Bank	State Bank of India	Bank of Baroda	Punjab National Bank	Canara Bank
Number of Branches	30	17	15	10

Draw a pie chart for this data.

Solution:-

From the table,

Total number of branches of different bank in a district = 30 + 17 + 15 + ten

= 72

Central angle of pie chart for each banking company,

State Bank of Republic of india = (30/72) × 360^o

= 150^o

Banking company of Baroda = (17/72) × 360^o

= 85^o

Punjab National depository financial institution = (15/72) × 360^o

= 75^o

Canara Banking concern = (10/72) × 360^o

= 50^o

107. For the evolution of basic infrastructure in a district, a project of Rs 108 crore approved by Evolution Bank is every bit follows:

Item Caput	Road	Electricity	Drinking water	Sewerage
Amount in crore (Rs)	43.ii	sixteen.2	27.00	21.6

Describe a pie chart for this data.

Solution:-

From the question,

Full number approved = 108 crore

Central bending of pie chart for each infrastructure,

Road = (43.2/108) × 360^o

= 144^o

Electricity = (sixteen.two/108) × 360^o

= 54^o

Drinking water = (27/108) × 360^o

= 90^o

Sewerage = (21.6/108) × 360^o

= 72^o

108. In the time table of a school, periods allotted per week to different teaching subjects are given beneath:

Subject field	Hindi	English	Maths	Science	Social Science	Computer	Sanskrit
Periods Allotted	seven	8	8	8	7	four	3

Draw a pie nautical chart for this data.

Solution:-

From the question,

Full number periods allotted = 7 + 8 + 8 + 8 + 7 + 4 + 3 = 45 periods

Central angle of pie chart for each subject,

Hindi = (7/45) × 360^o

= 56^o

English = (8/45) × 360^o

= 64^o

Maths = (8/45) × 360^o

= 64^o

Science = (8/45) × 360^o

= 64^o

Social Science = (vii/45) × 360^o

= 56^o

Computer = (4/45) × 360^o

= 32^o

Sanskrit = (3/45) × 360^o

= 24^o

109. A survey was carried out to find the favorite beverage preferred by a certain group of young people. The post-obit pie chart shows the findings of this survey. From this pie chart answer the post-obit:

(i) Which blazon of beverage is liked by the maximum number of people.

Solution:-

Cold drink is liked past the maximum number of people.

(2) If 45 people like tea, how many people were surveyed?

Solution:-

From the question it is given that,

45 people like tea

So,

Let u.s. assume full number of people surveyed be x,

= 45 = (15/100) × (ten)

= 45 = 0.15x

= x = 45/0.xv

= ten = 300

So, the total number of people surveyed is 300.

110. The following data represents the approximate pct of water in various oceans. Gear up a pie chart for the given information.

Pacific 40%

Atlantic 30%

Indian xx%

Others x%

Solution:-

The central angle of the given approximate pct of water in various oceans,

Pacific = (40/100) × 360^o

= 144^o

Atlantic = (30/100) × 360^o

= 108^o

Indian = (twenty/100) × 360^o

= 72^o

Others = (10/100) × 360^o

= 36^o

111. At a Birthday Political party, the children spin a wheel to get a gift. Find the probability of

(a) getting a ball

Solution:-

Number of balls in wheel = two

Full number of gifts = 8

Probability of getting a ball = Number of balls in wheel/ total number of gifts

= 2/eight … [divide both by ii]

= 1/iv

(b) getting a toy machine

Solution:-

Number of cars in wheel = 3

Full number of gifts = 8

Probability of getting a ball = Number of balls in wheel/ total number of gifts

= 3/viii

(c) any toy except a chocolate

Solution:-

Number of toys except a chocolate = 7

Total number of gifts = viii

Probability of getting a ball = Number of balls in bike/ full number of gifts

= seven/viii

112. Sonia picks up a card from the given cards.

Calculate the probability of getting

(a) an odd number

Solution:-

Total number of odd numbers in given cards = 5

Full number of cards = ten

Probability of getting an odd number = Total number of odd numbers/ total number of

cards

= v/ten

= ½

(b) a Y menu

Solution:-

Total number of y cards = three

Total number of cards = ten

Probability of getting a Y carte = Total number of Y cards/ total number of cards

= three/10

(c) a G card

Total number of Chiliad cards = 2

Full number of cards = x

Probability of getting a Y carte du jour = Total number of Y cards/ total number of cards

= 2/x

= 1/five

(d) B card begetting number > vii

Full number B bill of fare bearing number > seven= 0

Total number of cards = ten

Probability of getting a Y card = Total number B card bearing number > 7/ full number

of cards

= 0/10

= 0

113. Identify which symbol should announced in each sector in 113, 114.

Solution:-

From the given figure,

Total number of symbol = 800 + 700 + 550 + 450 = 2500

Then,

(i) = (32/100) × 2500

114.

Solution:-

From the given effigy,

Full number of water ice-creams = 192 + 228 + 180 = 600

Then,

(i) = (38/100) × 600

= 228

∴
yellow color icecream should appear in sector having 38%.

(ii) = (32/100) × 600

= 192

∴
red colour icecream should appear in sector having 32%.

(3) = (30/100) × 600

= 180

∴
pinkish colour icecream should announced in sector having thirty%.

115. A financial counselor gave a client this pie chart describing how to upkeep his income. If the customer brings home Rs. l,000 each month, how much should he spend in each category?

Solution:-

From the question information technology is given that,

Total money spent on 7 categories = Rs. 50,000

And then,

(1) Coin spent on Housing = (xxx/100) × 50,000

= Rs. 15,000

(2) Money spent on Food (including eating out) = (20/100) × 50,000

= Rs. 10,000

(iii) Money spent on Car Loan and Maintenance = (25/100) × 50,000

= Rs. 12,500

(four) Money spent on Utilities = (x/100) × 50,000

= Rs. 5,000

(5) Coin spent on Phone = (five/100) × 50,000

= Rs. 2,500

(six) Coin spent on Habiliment = (v/100) × 50,000

= Rs. 2,500

(7) Money spent on Entertainment = (5/100) × 50,000

= Rs. 2,500

116. Following is a pie chart showing the corporeality spent in rupees (in thousands) past a company on diverse modes of advertising for a product.

Now answer the post-obit questions.

1. Which type of media advertizing is the greatest amount of the total?

Solution:-

From the given figure,

The blazon of media advertising is the greatest amount of the total is Newspaper.

two. Which type of media advertising is the least amount of the total?

Solution:-

From the given figure,

The type of media ad is the least amount of the total is Radio.

3. What per cent of the total advert amount is spent on direct mail campaigns?

Solution:-

From the given figure,

Full amount spent past a company on various modes of advertisement for a product,

= 40 + 42 + 23 + 7 + 11 + 39 + fourteen + 15 + nine

= Rs. 200

So,

Per cent of the total advertising amount is spent on straight mail campaigns

= (39/200) × 100

= 19.5%

4. What per cent of the advertising corporeality is spent on newspaper and magazine advertisements?

Solution:-

From the given effigy,

Total amount spent by a company on various modes of advertizing for a production,

= 40 + 42 + 23 + 7 + eleven + 39 + 14 + 15 + 9

= Rs. 200

And then,

Per cent of the total advertising amount is spent on newspaper and magazine

= ((42 + 23)/200) × 100

= (65/200) × 100

= 32.5%

5. What media types exercise you think are included in miscellaneous?

Why aren't those media types given their own category?

Solution:-

Internet and webmedia are included in miscellaneous.

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Frequently Asked Questions on NCERT Exemplar Solutions for Class 8 Maths Chapter ii

List out the topics present in NCERT Exemplar Solutions for Class 8 Maths Affiliate 2?

The topics present in NCERT Exemplar Solutions for Class 8 Maths Chapter 2 are
1. Represent the data in Pictorial grade, using bar graphs and double bar graph
two. Larn, organizing and group of data
3. Representation of data via histogram and pie chart
4. Probability basic concepts
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