# NCERT Solutions for Class 7 Maths Chapter 3 Data Handling (EX 3.1) Exercise 3.1

Humans place a great deal of importance on Mathematics in their daily lives. There are numerous lucrative career opportunities, as well as research opportunities, associated with this academic field. There is no doubt that mathematics is one of the most organised and vital scientific disciplines. Indian mathematicians have a long history of impressing the world with their ingenuity and exceptional abilities. The subject of Mathematics is taught as a prominent scientific discipline from the earliest stages of school education itself, and students are encouraged to constantly strive to improve their understanding of it. The importance of Mathematics as an academic subject cannot be overstated. Mathematics has always piqued the curiosity of students. While some students struggle with it, others relish the challenges. Students may find it difficult to finish the exercises provided in their prescribed NCERT textbooks because they may find it difficult to comprehend the themes on their first attempt. Despite the fact that learning something new is never simple, once they have mastered it, they will feel tremendously satisfied.

With the help of the NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, students may complete Class 7 Maths Exercise 3.1. The exercises in the NCERT textbooks are always solved by students before they start a new chapter or theory. Students should practice NCERT solutions because doing so will help them understand the chapter’s concept and how it applies. Problem-solving and practice have a high priority in Mathematics. Statistics is a new subject for students in classes 10 and 11, as it is different from other chapters they have previously read. Students who are having trouble with Class 7 Maths Chapter 3 Exercise 3.1 can use the NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 to their advantage.

Students have a variety of possibilities if they want to pursue a more in-depth study of Mathematics. To make a sensible choice, it is, however, always advisable to be aware of one’s interests and skills. Before making a decision, students should research a variety of career options. Although each choice students make will provide them with a new opportunity, they will learn about themselves through fresh encounters and discoveries. No matter what they choose to pursue, students need to recognise that their efforts are always valuable. Students have a variety of possibilities if they want to pursue a more in-depth study of Mathematics. To make a sensible choice, it is, however, always advisable to be aware of one’s interests and skills. Students must ask for help if they feel they are answering a question too slowly. Instead of concentrating on the questions, they might ask their classmates and professors for help. In order to find the answers, they can also look in the NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, on the Extramarks mobile application and website. The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 are provided to students in order to help them get past any challenges they may experience when learning data handling.

**NCERT Solutions for Class 7 Maths Chapter 3 Data Handling (EX 3.1) Exercise 3.1 **

The book published by the National Council for Educational Research and Teaching (NCERT) is the best source of information or textbook available to students for preparation for term-end or half-yearly exams of any class. Students can learn the fundamentals of each concept and subject matter via NCERT books. Teachers frequently prescribe the NCERT books as recommendations for test preparation. Many schools that are affiliated with the Central Board of Secondary Education (CBSE) include NCERT textbooks in their curricula. These schools use NCERT textbooks to instruct their students. NCERT is a government organisation created to offer educational resources for teachers and students to use in their studies and classroom instruction. The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, provide answers to Exercise 3.1 of Chapter 3 of the NCERT textbook for Class 7.

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Students frequently skim over some subjects when studying for exams. This might be because, when students rapidly check their books before the test, some topics might be overlooked without understanding because they did not have enough time to completely read over each chapter. Students may experience difficulties if the subject turns out to be important. Students may encounter difficulties if a question from that subject is subsequently asked in an exam. Students should always review their notes and go over them to avoid this because the NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 include a synopsis of all the chapter’s subjects as well as questions from the exercise that may be asked in the tests. To avoid getting confused during the test, students should practice using the NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1. Students should not take studying from NCERT textbooks lightly because it is an important part of their education. Various factors make solving NCERT textbooks important, and they can practice the exercises given in them using the NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1. NCERT textbooks are essential to study because they provide the foundations for the more challenging questions on entrance examinations. This is something that students preparing for entrance exams like JEE or CUET should keep in mind. Students will not be able to respond to challenging questions if they have not completely read the chapter’s introductory information. The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, must consequently be pursued by students.

The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 provide an overview of the chapter. The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 contain various important topics like the Introduction to Data Handling, Collecting Data, Organisation of Data, Representation of Values, Arithmetic Mean, Range, Mode, Mode of Large Data, Median, Use Of Bar Graphs with a Different Purpose, Choosing a Double Bar Graph, Drawing a Double Bar Graph, Chance and Probability. The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 will help students find solutions to these questions in exercises.

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**Exercise 3.1**

The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, give an overview of the chapter as well as the exercise. Chapter 3 Data Handling of Class 7 is centred around the theme of Statistics. This exercise is based on the topics of data collection, data organistaion, representation of values, arithmetic mean, and range. The first exercise comprises questions related to this chapter. The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 will help students find answers to these questions and strengthen their knowledge. This exercise has 9 questions that help students understand the above-mentioned concepts. It will help students get better at solving these questions more accurately and faster in their exams. Some examples with similar concepts are also given in the NCERT textbook to help students understand the procedure of answering the questions. Solving and studying similar kinds of problems repeatedly will help them get better at those concepts and be aware of their mistakes.

**Exercise 3.2**

The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1 ,give an overview of the chapter as well as the exercise. The topics on which this exercise is based include mode, mode of large data, and median. The first exercise comprises questions related to this chapter. The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, would help students find answers to these questions and strengthen their understanding. There are five questions in this exercise that will help students understand the concepts above. Students will be able to solve these questions more accurately and faster during their exams as a result of this course. Some examples with similar concepts are also given in the NCERT textbook to help students understand the procedure for answering the questions. Solving and studying similar kinds of problems repeatedly will help them get better at those concepts and be aware of their weak areas.

**Exercise 3.3**

The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, give an overview of the chapter as well as the exercise. The topics on which this exercise is based include uses of bar graphs for different purposes, choosing a double bar graph, and drawing a double bar graph. The first exercise consists of questions related to this chapter. With the aid of NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, students will learn how to answer these questions. In this exercise, students will be asked to answer six questions related to the concepts discussed above. In their exams, students will be able to answer these questions more accurately and faster. Students can also find examples with similar concepts in the NCERT textbook to help them understand how to answer the questions.

**Exercise 3.4**

The NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, provide a comprehensive view of the chapter as well as the exercise. This exercise is based on the concepts of chance and probability, as well as what is meant by probability. There are questions related to this chapter in the first exercise. Using the NCERT Solutions For Class 7 Maths Chapter 3 Exercise 3.1, students will be able to find answers to these questions and will be able to strengthen their knowledge of the subject matter. Three questions are included in this exercise to help students understand the concepts described above. Students will be able to solve these questions more accurately and quickly in exams with the help of the reference materials for this exercise. In order to help students understand how to answer questions, the NCERT textbook also provides examples with similar concepts. Repeatedly solving and studying similar problems will help them understand those concepts and identify their weak areas.

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**NCERT Solutions for Class 7**

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**Q.1 **

$\begin{array}{l}\mathrm{Organise}\mathrm{the}\mathrm{following}\mathrm{marks}\mathrm{in}\mathrm{a}\mathrm{class}\mathrm{assessment},\mathrm{in}\\ \mathrm{at}\mathrm{a}\mathrm{bular}\mathrm{form}.\\ \end{array}$

4 | 6 | 7 | 5 | 3 | 5 | 4 | 5 | 2 | 6 |

2 | 5 | 1 | 9 | 6 | 5 | 8 | 4 | 6 | 7 |

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}\mathrm{Which}\mathrm{number}\mathrm{is}\mathrm{the}\mathrm{greatest}?\\ \left(\mathrm{ii}\right)\mathrm{Which}\mathrm{number}\mathrm{is}\mathrm{the}\mathrm{lowest}?\\ \left(\mathrm{iii}\right)\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{data}?\\ \left(\mathrm{iv}\right)\mathrm{Find}\mathrm{the}\mathrm{arithmetic}\mathrm{mean}.\end{array}$

**Ans.**

\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\text{9}\\ \text{(ii)}\text{\hspace{0.17em}}1\\ (iii)\text{\hspace{0.17em}}\text{Range=9-1}=\overline{)\text{8}}\\ (iv)\text{\hspace{0.17em}}\text{Arithmetic}\text{\hspace{0.17em}}\text{mean=}\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{100}{20}=\overline{)5}\end{array}

**Q.2** Find the mean of the first five whole numbers.

**Ans.**

$\begin{array}{l}\text{First five whole numbers are 0, 1, 2, 3 and 4}\\ \text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ \text{So, Mean =}\frac{0+1+2+3+4}{5}=\frac{10}{2}=\overline{)5}\\ \text{Thus, mean of first five whole number is 5.}\end{array}$

**Q.3** A cricket scores the following runs in eight innings:

58, 76, 40, 35, 46, 45, 0, 100.

Find the mean score.

**Ans.**

$\begin{array}{l}\text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\text{}\\ \text{So},\text{\hspace{0.17em}}\text{Mean}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{58+76+40+35+46+45+0+100}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{400}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\overline{)50}\\ \text{Thus, the mean score is 50}\end{array}$

**Q.4** Following table shows the points of each player scored in four game.

Player | Game | Game | Game | Game |

A | 14 | 16 | 10 | 10 |

B | 0 | 8 | 6 | 4 |

C | 8 | 11 | Did not Play | 13 |

Now answer the following questions:

(i) find the mean to determine A’ save range number of points scored per game.

(ii) To find mean number of points per game of C,

would you divide the total points by 3 or by 4? Why?

(iii) B played in all four games. How would you find the mean?

(iv) Who is the best performer?

**Ans.**

\begin{array}{l}\text{(i)}\\ \text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\text{}\\ \text{So,}\text{\hspace{0.17em}}\text{Mean}=\frac{14+16+10+10}{4}\\ =\frac{50}{4}=\overline{)12.5}\\ \text{Thus, A}\u2019\text{s average number of points scored per game is 12}\text{.5}\\ \text{(ii)}\\ \text{Since C played only 3 games, so to find the mean number of}\\ \text{points per game for C},\text{divide the total points by 3}\text{.}\\ \text{(iii)}\\ \text{Mean of B =}\frac{0+8+6+4}{4}=\frac{18}{4}=4.5\\ \text{(iv)}\\ \text{Since, the mean of A}\text{}\text{\hspace{0.17em}}\text{is better and bigger than B and C, so A is}\\ \text{the best performer}\text{.}\end{array}

**Q.5 **The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75.

Find the :

(i) Highest and the lowest marks obtained by the students.

(ii) Range of the marks obtained.

(iii) Mean marks obtained by the group.

**Ans.**

\begin{array}{l}\text{(i) Clearly, the from the given data, the highest and the lowest}\\ \text{marks are 95 and 39 respectively}\text{.}\\ \text{(ii) Range}=\text{Highest}-\text{Lowest}\\ =95-39=56\\ \text{(iii)}\text{}\text{\hspace{0.17em}}\text{Mean}=\text{\hspace{0.17em}}\frac{\text{Sum of all observations}}{\text{Total number of observations}}\text{}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{85+76+90+58+39+48+56+95+81+75}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\overline{)70.3}\end{array}

**Q.6 **The enrolment in a school during six consecutive years was as follows:

1555, 1670, 1750, 2013, 2540, 2820

Find the mean enrolment of the school for this period.

**Ans.**

\begin{array}{c}\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{1555+1670+1750+2013+2540+2820}{6}\\ =\frac{12348}{6}\\ =\overline{)2058}\end{array}

**Q.7** The enrolment in a school during six consecutive years was as follows:

1555, 1670, 1750, 2013, 2540, 2820

Find the mean enrolment of the school for this period.

**Ans.**

\begin{array}{c}\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{1555+1670+1750+2013+2540+2820}{6}\\ =\frac{12348}{6}\\ =\overline{)2058}\end{array}

**Q.8 **The rain fall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |

Rainfall
(in mm) |
0.0 | 12.2 | 2.1 | 0.0 | 20.5 | 5.5 | 1.0 |

(

**Ans.**

\begin{array}{l}\text{(i) Range}=\text{Highest-Lowest}\\ =20.\text{5 mm}\u20130.\text{0 mm}=\overline{)20.5}\text{\hspace{0.17em}}\text{mm}\\ \text{(ii)}\text{\hspace{0.17em}}\text{Mean rainfall}=\text{}\frac{0.0+12.2+2.1+0.0+20.5+5.5+1.0}{7}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{41.3}{7}=5.9\text{\hspace{0.17em}}\text{mm}\\ \text{(iii)}\text{\hspace{0.17em}}\text{For five days}\text{\hspace{0.17em}}\text{(Monday, Wednesday, Thursday, Saturday}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and Sunday), rainfall was less than the mean rainfall}\text{.}\end{array}

**Q.9 **

$\begin{array}{l}\begin{array}{l}\text{The\hspace{0.33em}heights\hspace{0.33em}of\hspace{0.33em}10\hspace{0.33em}girls were measured in cm and the}\\ \text{results are as follows:}\\ \text{135,150,139,128,151,132,146,149,143,141}\end{array}\\ \begin{array}{l}\left(\text{i}\right)\text{What\hspace{0.33em}is\hspace{0.33em}the\hspace{0.33em}height\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}tallest\hspace{0.33em}girl?}\\ \left(\text{ii}\right)\text{What\hspace{0.33em}is\hspace{0.33em}the\hspace{0.33em}height\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}shortest\hspace{0.33em}girl?}\\ \left(\text{iii}\right)\text{What\hspace{0.33em}is\hspace{0.33em}the\hspace{0.33em}range\hspace{0.33em}of the\hspace{0.33em}data?}\\ \left(\text{iv}\right)\text{What\hspace{0.33em}is the\hspace{0.33em}mean\hspace{0.33em}height\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}girls?}\\ \left(\text{v}\right)\text{How\hspace{0.33em}many\hspace{0.33em}girls\hspace{0.33em}have\hspace{0.33em}heights\hspace{0.33em}more\hspace{0.33em}than\hspace{0.33em}the\hspace{0.33em}mean}\end{array}\end{array}$

**Ans.**

\begin{array}{l}\text{(i) The height of the tallest girl is 151 cm}\text{.}\\ \text{(ii) The height of the shortest girl is 128}\text{.}\\ \text{(iii) Range}=\text{Highest}-\text{Lowest}=\left(151-128\right)\text{cm}=\overline{)23\text{\hspace{0.17em}}\text{cm}}\\ \text{(iv) Mean height}\\ =\frac{135+150+139+128+151+132+146+149+143+141}{10}\\ =\frac{1414}{10}\\ =\overline{)141.4\text{\hspace{0.17em}}\text{cm}}\\ \text{Thus, five girls have more height than the mean height}\text{.}\end{array}

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