NCERT Solutions Class 8 Maths Chapter 4 Data Handling

The fourth chapter of Class 8 Maths, Data Handling, focuses on essential topics such as Looking for Information, Circle Graph or Pie Chart and Chance and Probability. This chapter plays a crucial role in improving students' analytical skills, as it teaches them how to organize, interpret, and present data effectively. By solving NCERT Solutions for Class 8 Maths Chapter 4, students can strengthen their understanding of data representation, enhance their speed and accuracy, and gain a solid foundation in topics that are vital for higher-level studies. 

1.0Download Data Handling Class 8 Exercise Answers

This article offers data handling class 8 solutions as per NCERT. Practicing these solutions will help students develop a strong foundation in Maths and gain clarity on how to approach related problems effectively, ultimately aiding in securing good scores in board exams. For a detailed understanding, students can download the NCERT Class 8 Maths Chapter 4 PDF solution below, curated by ALLEN’s experts.

2.0Class 8 Maths Chapter 4 Data Handling: Exercise Solutions

There are two exercises in chapter 4 (Data Handling) of class 8 Maths students can find the split below:

Explore Data Handling and learn how to solve various problems only on NCERT Solutions For Class 8.

3.0Class 8 Maths Chapter 4 Data Handling Overview

Before discussing the specifics of NCERT Solutions for Class 8 Maths Chapter 4: Data Handling, let's quickly review the key topics and subtopics included in this chapter of the NCERT Class 8 Maths book. 

Topics covered in this chapter:

1. Looking for Information
2. Circle Graph or Pie Chart
3. Chance and Probability

4.0NCERT Questions with Solutions Class 8 Chapter 4

EXERCISE : 4.1

• 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 of cassettes produced by 5 companies. (d) The number of passengers boarding trains from 7:00 am to 7:00 pm at a station. Give reasons for each. Sol. In (b) and (d), data can be divided into class intervals. So, their histograms can be drawn.
• The shoppers who come to a departmental store are marked as : man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning. 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 W M W G W M G W M M B G G W Sol. We arrange the data in a table using tally marks as

The bar graph of the above data is as under

3. The weekly wages (in ₹) of 30 workers in a factory 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. Using tally marks make a frequency table with intervals as 800-810, 810-820 and so on. Sol. Let us make a grouped frequency table as under:

• Draw a histogram for the frequency table made for the data in question 3 and answer the following questions. (i) Which group has the maximum number of workers? (ii) How many workers earn ₹ 850 and more? (ii) How many workers earn less than ₹ 850?

Sol.

(i) 830-840 group has max. no. of workers. (ii) 10 (iii) 20 5.

The number of hours for which students of a particular class watched television during holidays is shown through the given graph.

Answer the following : (i) For how many hours did the maximum number of students watch TV ? (ii) How many students watched TV for less than 4 hours ? (iii) How many students spend more than 5 hours in watching TV ? Sol. Clearly from the given histogram, we find that (i) The maximum number of students watch TV for 4-5 hours. (ii) The number of students who watch TV for less than 4 hours are $4+8+22=34$. (iii) The number of students who spend more than 5 hours in watching TV are $8+6=$ 14.

EXERCISE : 4.2

• A survey was made to find the type of music that a certain group of young people likes in a city. Following pie chart shows the findings of this survey. From this pie chart answer the following questions:

• (i) If 20 people like classical music, how many young people were surveyed? (ii) Which type of music is liked by the maximum number of people? (iii) If a cassette company were to make 1000 CD's, how many of each type would it make? Sol. (i) Let x be the number of young people surveyed. $\therefore 10\%$ of $\mathrm{x}=20$ $\Rightarrow \frac{10}{100}\times \mathrm{x}=20\Rightarrow \mathrm{x}=20\times 10=200$ Thus, number of young people surveyed $=$ 200 (ii) Light music is liked by the maximum number of people. (iii) Number of CD's in respect of

Classical music $=\frac{10}{100}\times 1000=100$ Semi-classical music $=\frac{20}{100}\times 1000=200$ Folk music $=\frac{30}{100}\times 1000=300$ Light music $=\frac{40}{100}\times 1000=400$ 2. A group of 360 people were asked to vote for their favourite season from the three seasons-Rainy, winter and summer.

(i) Which season got the most votes? (ii) Find the central angle of each sector. (iii) Draw a pie chart to show this information.

Sol. (i) The winter season got the maximum votes. (ii) The proportion of sectors (winter, summer and rainy seasons) are as $\frac{150}{360}=\frac{5}{12},\frac{90}{360}=\frac{1}{4}$, and $\frac{120}{360}=\frac{1}{3}$ respectively. (iii) Central angle for winter season $={\left(\frac{150}{360}\times 360\right)}^0=150^{\circ }$ Central angle for summer season $={\left(\frac{90}{360}\times 360\right)}^{\circ }=90^{\circ }$ and central angle for rainy season $={\left(\frac{120}{360}\times 360\right)}^{\circ }=120^{\circ }$ Now, the various components may be shown by the adjoining pie chart.

• Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Sol.

Blue $=\frac{18}{36}\times 360^{\circ }=180^{\circ }$ Green $=\frac{9}{36}\times 360^{\circ }=90^{\circ }$ Red $=\frac{6}{36}\times 360^{\circ }=60^{\circ }$ Yellow $=\frac{3}{36}\times 360^{\circ }=30^{\circ }$

• The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions. (i) In which subject did the student score 105 marks? (Hint: for 540 marks, the central angle $=$ $360^{\circ }$. So, for 105 marks, what is the central angle?

• (ii) How many more marks were obtained by the student in Mathematics than in Hindi? (iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi. (Hint : Just study the central angles.) Sol. (i) For 540 marks, the central angle $=$ $360^{\circ }$ $\therefore$ For 1 mark, the central angle $={\left(\frac{360}{540}\right)}^0={\left(\frac{2}{3}\right)}^0$ $\therefore$ For 105 marks, the central angle $={\left(\frac{2}{3}\times 105\right)}^0=70^{\circ }$ Hence, from the given pie chart, the subject is Hindi. (ii) Difference between the central angles made by the subject of Mathematics and Hindi $=90^{\circ }-70^{\circ }=20^{\circ }$. For the central angle of $360^{\circ }$, marks obtained $=540$ For the central angle of $1^{\circ }$, marks obtained $=\frac{540}{360}=\frac{3}{2}$ For the central angle of $20^{\circ }$, marks obtained $=\frac{3}{2}\times 20=30$ Marks Hence, 30 more marks were obtained by the student in Mathematics than Hindi. (iii) Sum of the central angles made by Social Science and Mathematics $=65^{\circ }+90^{\circ }=$ $155^{\circ }$. and the sum of the central angles made by Science and Hindi $=80^{\circ }+70^{\circ }=150^{\circ }$. Since, Science $155^{\circ }>150^{\circ }$, therefore, the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
• The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.

Sol. Hindi $=\frac{40}{72}\times 360^{\circ }=200^{\circ }$ English $=\frac{12}{72}\times 360^{\circ }=60^{\circ }$ Marathi $=\frac{9}{72}\times 360^{\circ }=45^{\circ }$ Tamil $=\frac{7}{72}\times 360=35^{\circ }$ Bengali $=\frac{4}{72}\times 360=20^{\circ }$

EXERCISE : 5.3

• List the outcomes you can see in these experiments. (i) Spinning a wheel,

• (ii) Tossing two coins together. Sol. (i) List of outcomes of spinning the given wheel are A, B, C and D. (ii) When two coins are tossed together, the possible outcomes of the experiment are HH, HT, TH and TT.
• When a die is thrown, list the outcomes of an event of getting. (i) (a) A prime number, (b) Not a prime number. (ii) (a) A number greater than 5, (b) A number not greater than 5. Sol. (i) In a throw of die, list of the outcome of an event of getting (a) A prime number are 2,3 and 5. (b) Not a prime number are 1, 4 and 6. (ii) In a throw of die, list of the outcomes of an event of getting (a) A number greater than 5 is 6 . (b) A number not greater than 5 are 1, 2, 3,4 and 5 .
• Find the (i) Probability of the pointer stopping on $D$ in (Q. 1 (i)). (ii) Probability of getting an ace from a well shuffled deck of 52 playing cards. (iii) Probability of getting a red apple (see adjoining figure).

• Sol. (i) Out of 5 sectors, the pointer can stop at any of sectors in 5 ways. $\therefore$ Total number of elementary events $=$
• There is only one ' D ' on the spinning wheel. $\therefore$ Favourable number of outcomes $=1$ $\therefore$ Required probability $=1/5$ (ii) Out of 52 cards, one card can be drawn in 52 ways. $\therefore$ Total number of outcomes $=52$ There are 4 aces in a pack of 52 cards, out of which one ace can be drawn in 4 ways. $\therefore$ Favourable number of cases $=4$. So, the required probability $=\frac{4}{52}=\frac{1}{13}$ (iii) Out of 7 apples, one apple can be drawn in 7 ways. $\therefore$ Total number of outcomes $=7$ There are 4 red apples, in a bag of 7 apples, out of which 1 red apple can be drawn in 4 ways. $\therefore$ Favourable number of cases $=4$ So, the required probability $=4/7$
• Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is choosen from the box without looking into it. What is the probability of (i) Getting a number 6? (ii) Getting a number less than 6? (iii) Getting a number greater than 6 ? (iv) Getting a 1-digit number? Sol. Out of 10 slips, 1 slip can be drawn in 10 ways. So, the total number of outcomes $=10$ (i) An event of getting a number 6, i.e., if we obtain a slip having number 6 as an outcome. So favourable number of outcomes $=1$ $\therefore$ Required probability $=1/10$ (ii) An event of getting a number less than 6, i.e., if we obtain a slip having any of numbers $1,2,3,4,5$ as an outcome. So, favourable number of cases $=5$ $\therefore$ Required probability $=\frac{5}{10}=\frac{1}{2}$ (iii) An event of getting a number greater than 6, i.e., if we obtain a slip having any of numbers 7, 8, 9, 10 as an outcome. So, favourable number of cases $=4$ $\therefore$ Required probability $=\frac{4}{10}=\frac{2}{5}$ (iv) An event of getting a one-digit number, i.e., if we obtain a slip having any of numbers $1,2,3,4,5,6,7,8,9$ as an outcome. So, favourable number of cases $=9$. $\therefore$ Required probability $=9/10$.
• If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector? Sol. Out of 5 sectors, the pointer can stop at any of sectors in 5 ways. $\therefore$ Total number of outcomes $=5$ There are 3 green sectors in the spinning wheel, out of which one can be obtained in 3 ways. $\therefore$ Favourable number of outcomes $=3$ So, the required probability $=3/5$ Further, there are 4 non-blue sectors in the spinning wheel, out of which one can be obtained in 4 ways. So, the required probability $=4/5$.
• Find the probabilities of the events given in question 2. Sol. In a single throw of a die, we can get any one of the six numbers $1,2,\dots ,6$ marked on its six faces. Therefore, the total number of outcomes $=6$. (i) Let A denote the event "getting a prime number". Clearly, event A occurs if we obtain $2,3,5$ as an outcome $\therefore$ Favourable number of outcomes $=3$. Hence, $P(A)=\frac{3}{6}=\frac{1}{2}$ (ii) Let A denote the event "not getting a prime number". Clearly, event A occurs if we obtain $1,4,6$ as an outcomes $\therefore$ Favourable number of outcomes $=3$. Hence, $P(A)=\frac{3}{6}=\frac{1}{2}$ (iii) The event "getting a number greater than 5 " will occur if we obtain the number 6 . $\therefore$ Favourable number of outcomes $=1$. Hence, required probability $=1/6$ (iv) The event "getting a number not greater than 5 " will occur if we obtain one of the numbers $1,2,3,4,5$. $\therefore$ Favourable number of outcomes $=5$. Hence, required probability $=5/6$