NCERT Solutions For Class 8 Maths Chapter 5

NCERT Solutions Class 8 Maths Data Handling

Ncert Solutions For Class 8 Maths Chapter 5 PDF Free Download

NCERT Solutions for class 8 Maths Chapter 5 Data Handling is crucial for the students of class 8 to excel in their examination. These solution help students to frame a better understanding of the topic. To ease the fear of maths, we at BYJU’S provide NCERT Solution for Class 8 Maths Chapter 5 Data Handling. Students can download the NCERT Solution for class 8 Maths Chapter 5 pdf or can view it online by following the link. These solutions are provided in a detailed manner, where one can find a step-by-step solution to all the questions of class 8 maths chapter 5 NCERT Solutions.

NCERT Solutions Class 8 Maths Chapter 5 Exercises

 

Exercise 5.1

1. For which of these would you use a histogram to show the data:

(a) The number of letters for different areas in a postman’s bag.

(b) The height of competitors in an athletics meet.

(c) The number cassettes produced by 5 companies.

(d) The number of passengers boarding trains from 7.00 a.m. to 7.00 p.m. at a station.

Give a reason for each of the following.

Answer:

Since histogram is graphical representation of data, if data, if data represented in the form of class interval.

Therefore, for case (b) and (d), we would use a histogram to show the data, because in these cases, data can be divided into class-intervals. In case (b), a group of competitions having different heights in athletics meets. In case (d), the number of passengers boarding trains in an interval of one hour at a station.

In case (b), a group of competitions having different heights in athletics meets. In case (d), the number of passengers boarding trains in an interval of one hour at a station.

In case (d), the number of passengers boarding trains in an interval of one hour at a station.

 

2. Shoppers who come to a grocery store are identified as: Boy (B), Man (M), Girl (G) or Women (W). The list given below gives the shoppers who had come in the morning time slot:

W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W M W G W M G W M M B G G W 

Construct a frequency distribution table using the tally marks. Sketch a bar graph to represent it.

Answer:

The representation of the above data by bar graph is as follows:

 

3. The weekly salary (in Rs) of 30 workers in are:

830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840.

By referring the tally marks, make a frequency table with following intervals as 800 – 810 , 810 – 820 and so on.

Answer:

The illustration of data by frequency distribution table by using the tally marks is as follows:

 

4. Construct a histogram for frequency table made for the data that is given in question 3 and solve the following questions.

(i) Which group has highest workers?

(ii) Tell the number of workers who earn Rs. 850/- and more?

(iii) Identify the number of workers who earn less than Rs. 850/-

Answer:

Histogram of the above frequency distribution table is as follows:

(i) Highest number of workers are present in 830 – 840 group.

(ii) The workers who earn more than Rs. 850, these workers fall in the category of 880 – 890 or 850 – 860 or 860 – 870. Therefore, the number of workers who are earning more than 850 will be the sum of the numbers of these workers i.e., 1 + 3 + 1 +1 + 4 = 10

(iii) The workers are being paid less than Rs.850 are the number of workers who come under the category 840 – 850 or 800 – 810 or 820 – 830 or 810 – 820 .Therefore, the number of workers who paid less than 850 will be the sum of the numbers of all these categories i.e.,

3 + 2 + 1 + 9 + 5 = 20

 

5. The students of particular class watched their lecture on television for different number of hours during the holidays; the results represented through graph.

Calculate the following:

(i) What is the number of hours did the maximum number of students watch TV?

(ii) What is the number of students who watched TV for less than 4 hours?

(iii) What is the total number of students who spent more than 5 hours in watching TV?

Answer:

(i) It can be observed from the graph, that the maximum number of students (i.e., 32) watched TV for 4 – 5 hours.

(ii) The number of students who watched TV for less than 4hours are the students who watched TV for 3 – 4 hours or 2 – 3 hours or 1 – 2 hours.

Therefore, the total number of students = 4 + 8 + 22 = 34

(iii) The number of students who watched TV for more than 5 hours falls in the following category 5 – 6 hours or 6 – 7 hours.

Therefore, Sum of the above category of students = 8 + 6 = 14

 

Exercise 5.2

1. A survey was made to find the type of music a group of people liked in a town. The pie chart given below shows the information regarding that survey. From the pie chart, give the answer for the following questions.

(a) If 60 people liked semi classical music then how many people were surveyed?

(b) Which music is liked by the minimum number of people?

(c) If a company make 2000 CD’s then how many of each type would they make?

Answer:

(a) People who like semi – classical music = 30%

This 30% represents 60 people.

100% represents = \(\frac{60 \times 100}{30}\) = 200 people

Hence, 200 people were surveyed.

 

(b) From the pie chart we can see that light music is represented by the minimum part of the pie chart. Therefore, light music is least liked by people.

 

(c) Number of Classical music CDs = 20% of 2000

= \(\frac{20 \times 2000}{100}\)

= 400 CDs

 

Number of Semi – Classical music CDs = 30% of 2000

= \(\frac{30 \times 2000}{100}\)

= 600 CDs

 

Number of Light music CDs = 10% of 2000

= \(\frac{10 \times 2000}{100}\)

= 200 CDs

 

Number of Folk music CDs = 40% of 2000

= \(\frac{40 \times 2000}{100}\)

= 800 CDs

 

2. In a survey 720 people were asked about their favorite season from summer, winter and monsoon. Answer the following questions according to the given information.

(a) Which season got the minimum votes?

(b) What will be the central angle of each season?

(c) Draw pie chart for the given information.

Festival Votes
Summer 160
Winter 340
Monsoon 220

 

Answer:

(a) Winter season got the maximum number of votes.

 

(b) Total number of votes = 160 + 340 220 = 720

Central angle for summer = \(\frac{160 \times 360^{\circ}}{720}\)

= \(\frac{160}{2}\)

= \(80^{\circ}\)

 

Central angle for winter    = \(\frac{340 \times 360^{\circ}}{720}\)

= \(\frac{340}{2}\)

= \(170^{\circ}\)

 

Central angle for monsoon = \(\frac{220 \times 360^{\circ}}{720}\)

= \(\frac{220}{2}\)

= \(110^{\circ}\)

(c) Pie chart for the above information

 

3. According to the information given below, draw the pie chart representing it. The table below shows the colors liked by a group of people.

Colors Number of People
Blue 36
Red 18
Green 12
Yellow 6
Total 72

 

Answer:

Total number of votes = 72

Central angle for Blue = \(\frac{36 \times 360^{\circ}}{72}\)

= \(\frac{360}{2}\)

= \(180^{\circ}\)

 

Central angle for Red    = \(\frac{18 \times 360^{\circ}}{72}\)

= \(\frac{360}{4}\)

= \(90^{\circ}\)

 

Central angle for Green = \(\frac{12 \times 360^{\circ}}{72}\)

= \(\frac{360}{6}\)

= \(60^{\circ}\)

 

Central angle for Yellow = \(\frac{6 \times 360^{\circ}}{72}\)

= \(\frac{360}{12}\)

= \(30^{\circ}\)

 

Pie chart as above information

 

4. The pie chart given below gives the information about the marks scored by a student in 5 subjects which are English, Math, Science, Hindi and Social Science. The total marks obtained by the student were 480.

(a) The student scored 108 marks in which subject?

(b) What is the difference between the marks of Math and Hindi? Which subject has more marks?

(c) Check whether the summation of Math and Hindi is more than that of Science and Social Science.

Answer:

(a) 480 marks is \(360^{\circ}\). Therefore, 108 marks give the following central angle.

= \(\frac{108 \times 360}{480}\)

= \(81^{\circ}\)

Therefore, the subject is Science.

 

(b) Marks obtained in Math = 90% of 480

= \(\frac{90 \times 480}{360}\)

= 120 marks

 

Marks obtained in Hindi =57% of 480

= \(\frac{57 \times 480}{360}\)

= 76 marks

 

The difference between Math and Hindi marks is 120 – 76 = 44 marks.

In Math, student scored 44 marks more compared to Hindi.

 

(c) Marks obtained in Social Science = 69% of 480

= \(\frac{69 \times 480}{360}\)

= 92 marks

 

Marks obtained in Science = 81% of 480

= \(\frac{81 \times 480}{360}\)

= 108 marks

 

Summation of Math and Hindi = 120 + 76 = 196 marks

Summation of Science and Social Science = 92 + 108 = 200 marks

Summation of Science and Social Science is more than that of Math and Hindi.

 

5. The number of students in a school who are speaking various languages is as given below. Draw the pie chart according to the given information.

Language English Hindi Tamil Telugu Marathi Total
Number of

Students

20 6 5 3 2 36

 

Answer:

The central angle of each language is calculated as given below.

 

Central angle of English language = \(\frac{20 \times 360}{36}\)

= \(200^{\circ}\)

Central angle of Hindi language    = \(\frac{6 \times 360}{36}\)

= \(60^{\circ}\)

Central angle of Tamil language = \(\frac{5 \times 360}{36}\)

= \(50^{\circ}\)

Central angle of Telugu language = \(\frac{3 \times 360}{36}\)

= \(30^{\circ}\)

Central angle of Marathi language = \(\frac{2 \times 360}{36}\)

= \(20^{\circ}\)

 

 

 EXERCISE – 5.3

1. List the outcomes in details, which you see in these types of experiments -:

(i) Tossing two coins at a time

 

Answer:

(i) By tossing two coins each other, we will find four possible outcomes i.e. HH, HT, TH, TT respectively.

[Here H = Head and T = Tails]

 

2. List the outcomes of an event when a dice is thrown:

(a) (i) A prime number      (ii) Not a prime number

(b) (i) A number greater than 6     (ii) A number not greater than 6

 

Answer:

When a dice is thrown, there are seven possible outcomes, i.e., 1,2,3,4,5,6 and 7.

(a)

(i) Outcomes of event of getting a prime number are 2, 3, 5 and 7.

Hence, these are the outcomes of an event of getting a prime number on the face of a dice.

(ii) Outcomes of event of not getting a prime number are 1, 4 and 6.

Hence, these are the outcomes of an event of not getting a prime number on the face of a dice.

(b)

(i) Outcomes of event of getting a number greater than 6 is 7, which comes on the face of the dice.

(ii) Outcomes of event of not getting a number greater than 6 are 1, 2, 3, 4, 5 and 6.

 

3. Find out:-

 (i) Prospect of getting an ace from a well shuffled deck of 52 playing cards.

 (ii) Prospect of getting a red apple which is shown in the figure below.

 

Answer:

(i) There are 4 aces in a deck of 52 playing cards. So, there are four events of getting an ace.

So, the probability of getting an ace card = \(\frac{4}{52}=\frac{1}{13}\)

(ii) There are a total of 7 apples, out of which, 4 are red and 3 are green.

So, probability of getting a red apple = \(\frac{4}{7}\)

 

4. Numbers 1 to 12 are written on ten separate paper slips, one number on one slips kept in a box and mixed well. A single paper slip is picked from the box without looking into it. What is the probability of:

(a) Obtaining a number 8?

(b) Obtaining a number less than 8?

(c) Obtaining a number greater than 8?

(d) Obtaining a 1-digit number?

Answer:

(a) There are 12 paper slips in the box. However, 8 is written only on a single slip.

Therefore, probability of getting a number 8 = \(\frac{1}{12}\)

 

(b) The numbers less than 8 are 1,2,3,4,5,6,7. So there are 7 outcomes.

Therefore, probability of getting a number less than 8 = \(\frac{7}{12}\)

 

(c) The numbers greater than 8 are 9, 10, and 11. So there are 3 possible outcomes.

Therefore, probability of getting a number greater than 8 = \(\frac{3}{12}\) = \(\frac{1}{4}\)

 

(d) There are 9 numbers which are single digit numbers.

1 2,3,4,5,6,7,8 and 9.

Therefore, probability of getting a 1-digit number = \(\frac{9}{12}\) = \(\frac{3}{4}\)

 

5. A spinning wheel consisting of 1 blue sector, 3 green sectors and 1 red sector, Find out the probability of getting a green sector and also the probability of getting a none-blue sector.

Answer:

Total sectors = 3 + 1 + 1 = 5

There are five sectors. Three sectors are green out of five sectors.

Hence, probability of getting a green sector = \(\frac{3}{5}\)

We will get a non-blue sector when we will get either a green sector or a red sector.

There is one blue sector out of five sectors.

Hence, Non-blue sectors = 5 – 1 = 4 sectors

Therefore, probability of getting a non-blue sector = \(\frac{4}{5}\)

 

6. Find out the prospect of the events given in Question 2.

Answer:

When a dice is thrown, there are seven possible outcomes, i.e., 1,2,3,4,5,6 and 7.

(a)

(i) 2, 3, 5  and 7 are prime numbers. So there are 4 outcomes out of 7.

Therefore, probability of getting a prime number = \(\frac{4}{7}\)

(ii) 1, 4, 6 are not the prime numbers. So there are 3 outcomes out of 7.

Therefore, probability of getting a prime number = \(\frac{3}{7}\)

(b)

(i) Out of 7 possible outcomes, a number greater than 6 can be obtained in only 1 case.

So there is one outcome out of 7.

Therefore, probability of getting a number greater than 6 = \(\frac{1}{7}\)

(ii) Out of 7 possible outcomes, a number not greater than 6 can be obtained in only 6 cases.

Numbers not greater than 6 are 1, 2, 3, 4, 5 and 6. So there are 6 outcomes out of 7.

Therefore, probability of not getting a number greater than 6 = \(\frac{6}{7}\)