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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1 - FREE PDF Download

NCERT class 7 maths chapter 3 exercise 3.1 solutions - Data handling introduces students to the fundamental concepts like arithmetic mean, Range. Working through this chapter , students will learn how to collect data, represent it using various methods like bar graphs and pictographs, and analyze it to make informed decisions. Students can access the revised Class 7 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.

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Table of Content
1. NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 3 Exercise 3.1 Class 7 | Vedantu
3. Formulas Used in Class 7 Chapter 3 Exercise 3.1
4. Access NCERT Solutions for Maths Class 7 Chapter 3 - Data Handling
    4.1Exercise 3.1
5. Conclusion
6. Class 7 Maths Chapter 3: Exercises Breakdown
7. CBSE Class 7 Maths Chapter 3 Other Study Materials
8. Chapter-Specific NCERT Solutions for Class 7 Maths
9. Important Related Links for NCERT Class 7 Maths
FAQs


The exercise emphasizes practical examples to help you visualize data clearly and accurately. Pay close attention to the steps involved in creating and interpreting these graphical representations. Vedantu’s NCERT Solutions for Class 7 Chapter 3 provide clear explanations and step-by-step solutions by our expert masters and updated according to the latest CBSE Class 7 Maths Syllabus.


Glance on NCERT Solutions Maths Chapter 3 Exercise 3.1 Class 7 | Vedantu

  • The NCERT Class 7 Maths Exercise 3.1 solutions Data Handling, deals with Arithmetic Mean and where does arithmetic mean lies and Range.

  • The arithmetic mean (simply called the mean or average) is the sum of a collection of numbers divided by the count of numbers in the collection.

  • The difference between the highest and the lowest observation gives us an idea of the spread of the observations, is called range.

  • This can be found by subtracting the lowest observation from the highest observation.

    • ie, Range=Maximum Value−Minimum Value

  • There are 9 fully solved questions in Data Handling Class 7 Exercise 3.1.


Formulas Used in Class 7 Chapter 3 Exercise 3.1

  • Arithmetic Mean= $\frac{\sum_{i=1}^{n} x_{i}}{n}$

  • Range=Maximum Value−Minimum Value

Access NCERT Solutions for Maths Class 7 Chapter 3 - Data Handling

Exercise 3.1

1.Find the range of heights of any ten students in your class.

Ans:

S. No.

Name of Students

Height (in feet)

1.

Gunjan

4.2 

2.

Aditi

4.5

3.

Nikhil

4.

Akhil

5.1

5.

Ria

5.2

6.

Akshat

5.3

7.

Abhishek

5.1

8.

Mayank

4.7

9.

Rahul

4.9

10.

Ayush

4.5

We will find the range of heights by getting the difference between the highest height and lowest height.

Range= Highest height – Lowest Height

 $ =5.3-4.2\dfrac{z}{3} $ 

$ =1.1 $ 

Hence, the range of height of ten students of the class is 1.1 feet.


2. Organize the following marks in a class assessment, in a tabular form: 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7.

(i) Which number is highest?

(ii) Which number is lowest?

(iii) What is the range of the lowest?

(iv) Find the arithmetic mean

Ans: 

Marks

Tally Marks

Frequency

1

|

1

2

||

2

3

|

1

4

|||

3

5

||||

5

6

||||

4

7

||

2

8

|

1

9

|

1

  1. The highest number among the given data is 9.

  2. The lowest number among the given data is 1.

  3. We will find the range of marks by getting the difference between the highest value and lowest value.

Range= Highest value – Lowest value

$ =9-1 \\ $

$ =8 \\ $

Hence, the range of marks is 8.

  1. We know that, arithmetic mean is the sum of all observations divided by the total number of observations.

$ \text{Mean=}\dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\ $

$ \text{=}\dfrac{4+6+7+5+3+5+4+5+2+6+2+5+1+9+6+5+8+4+6+7}{20} \\ $

$ \text{=}\dfrac{100}{20} \\ $

$ \text{=5} \\ $


3.Find the mean of the first five whole numbers.

Ans: The first five whole numbers are 0, 1, 2, 3, 4.

Arithmetic mean is the sum of all observations divided by total number of observations.

$\text{Mean=}\dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\ $

$ \text{=}\dfrac{0+1+2+3+4}{5} \\ $

$ \text{=}\dfrac{10}{5} \\ $

$ \text{=2} \\ $

Hence, the mean of the first five natural numbers is 2.


4. A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

Ans: Arithmetic mean is sum of all observations divided by total number of observations.

Number of Observations= 8

$ \text{Mean=}\dfrac{\text{Sum of all observations}}{\text{Number of observations}} \\ $

$ \text{=}\dfrac{58+76+40+35+46+45+0+100}{8} \\ $

$ \text{=}\dfrac{400}{8} \\ $

$ \text{=50} \\ $

Hence, the mean score of the cricketer is 50.


5. Following table shows the points each player scored in four games.

Player

Game 1

Game 2

Game 3

Game 4

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’s average number of points scored per game.

(ii) To find the mean number of points per game for C, would you divide the total points by 3 or 47? Why? 

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

(iv) Who is the best performer?

Ans:

(i) The average of player A can be determined by dividing the sum of all scores of A by the number of games he played. 

 $ \text{Mean of player A=}\dfrac{\text{Sum of scored by A}}{\text{Number of games played by A}} \\ $

 $ \text{=}\dfrac{14+16+10+10}{4} \\ $

$ \text{ =}\dfrac{50}{4} \\ $

$ \text{ =12}\text{.5} \\ $

Hence, the average number of points scored by A is 12.5.

(ii)For finding average points scored, we divide the total scores by the total number of games played. In this case, C has played 3 games and therefore we will divide the total points by 3 and not by 4.

(iii)B played all 4 games. So, we will divide the total points scored by 4.

$ \text{Mean of player B=}\dfrac{\text{Sum of scored by B}}{\text{Number of games played by B}} \\ $

$ \text{=}\dfrac{0+8+6+4}{4} \\ $

$ \text{=}\dfrac{18}{4} \\ $

$ \text{=4}\text{.5} \\ $

(iv)Best performer is the one who has the highest average score among all the other players.

$ \text{Mean of player A=}\dfrac{\text{Sum of scored by A}}{\text{Number of games played by A}} \\ $

$ \text{=}\dfrac{14+16+10+10}{4} \\$ 

$ \text{=}\dfrac{50}{4} \\ $

$ \text{=12}\text{.5} \\ $

$ \text{Mean of player B=}\dfrac{\text{Sum of scored by B}}{\text{Number of games played by B}} \\ $

$ \text{=}\dfrac{0+8+6+4}{4} \\ $

$ \text{=}\dfrac{18}{4} \\ $

$ \text{=4}\text{.5} \\ $

$  \text{Mean of player C=}\dfrac{\text{Sum of scored by C}}{\text{Number of games played by C}} \\ $

$ \text{=}\dfrac{8+11+13}{3} \\ $

$ \text{=}\dfrac{32}{3} \\ $

$ \text{=10}\text{.67} \\ $

On comparing means of all players, we can see that the average score of player A is the highest, i.e., 12.5.

Hence, player A is the best performer.


6. 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) The highest and the lowest marks obtained by the students. 

(ii) Range of the marks obtained. 

(iii) Mean marks obtained by the group.

Ans:

(i) Highest marks obtained by the student = 95

Lowest marks obtained by the student = 39

(ii) We will find the range of marks by getting the difference between the highest value and lowest value.

Range= Highest value – Lowest value

$ =95-39 \\ $

$ =56 \\ $

Hence, the range of marks obtained is 56.

(iii) For finding average marks scored, we divide the total scores by the total number of observations.

 $ \text{Mean of obtained marks=}\dfrac{\text{Sum of marks}}{\text{Total number of marks}} \\ $

$ \text{=}\dfrac{85+76+90+85+39+48+56+95+81+75}{10} \\ $

$ \text{=}\dfrac{730}{10} \\$ 

$ \text{=73} \\ $

Hence, the mean marks obtained by the group of students is 73.


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: To find the mean enrolment, we will divide the sum of numbers of enrolment by the number of years of enrolment period.

$ \text{Mean enrolment=}\dfrac{\text{Sum of number of enrolment}}{\text{Total number of years}} \\$ 

$ \text{=}\dfrac{1555+1670+1751+2013+2540+2820}{6} \\ $

$ \text{=}\dfrac{12348}{6} \\ $

$ \text{=2058} \\ $

Hence, the mean enrolment of school over 6 years is 2058.


8. The rainfall (in mm) in a city on 7 days of a certain week recorded as follows

Day

Mon

Tue

Wed

Thurs

Fri

Sat

Sun

Rainfall

 (in mm)

0.0

12.2

2.1

0.0

20.5

5.5

1.0


(i) Find the range of the rainfall in the above data.

(ii) Find the mean rainfall for the week.

(iii) On how many days was the rainfall less than the mean rainfall.

Ans

(i) Range of rainfall = Highest rainfall - lowest rainfall

= 20.5 - 0.0

= 20.5 mm


(ii) Mean of a rainfall = $\dfrac{\text{Sum of all Observations}}{\text{Number of observations}}\\ $

= $\dfrac{0.0+12.2+2.1+0.0+20.5+5.5+1.0}{7}$

=$\dfrac{41.3}{7}$

=5.9 mm


(iii) From the mean calculated in (ii) we may observe that for 5 days those are Monday, Wednesday, Thursday, Saturday and Sunday, the rainfall was less than the average rainfall.


9. The height of 10 girls were measured in cm and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141 

(i) What is the height of the tallest girl? 

(ii) What is the height of the shortest girl? 

(iii) What is the range of data? 

(iv) What is the mean height of the girls? 

(v) How many girls have heights more than the mean height?

Ans: 

(i)As seen from the above given data, the height of the tallest girl is 151cm.

(ii)As seen from the above given data, the height of the shortest girl is 128cm.

(iii)We will find the range of heights by getting the difference between the highest value and lowest value.

$ \text{Range= Highest Value}-\text{Lowest Value} \\ $

$ \text{=151-128} \\ $

$ \text{=23} \\ $

Hence, the range of heights is 23cm.

(iv)To find the mean height, we will divide the sum of heights of all girls by the number of girls.

$ \text{Mean of height=}\dfrac{\text{Sum of heights of all girls}}{\text{Total number of girls}} \\ $

$ \text{=}\dfrac{135+150+139+128+151+132+146+149+143+141}{10} \\$ 

$ \text{=}\dfrac{1414}{10} \\ $

$ \text{=141}\text{.4} \\ $

Hence, the mean height of girls is 141.4cm.

(v) Mean height is 141.4cm. From the above data, we can see that heights 150, 151, 146, 149 and 143 are more than the mean value. 

Hence, five girls have heights more than the mean height.


Conclusion

The NCERT Class 7 Maths Chapter 3 Exercise 3.1 Solutions, provided by Vedantu, offer a clear understanding of Data Handling concepts. This data handling class 7 exercise 3.1  focuses on calculating the arithmetic mean, range, mode, and median, which are crucial for analyzing data. Students should pay special attention to understanding these basic statistical measures as they form the foundation for more advanced topics. Practicing these solutions will enhance students' problem-solving skills and help them to score more in exams.


Class 7 Maths Chapter 3: Exercises Breakdown

Exercises

Number of Questions

Exercise 3.2

5 Questions and Solutions 

Exercise 3.3

6 Questions and Solutions



CBSE Class 7 Maths Chapter 3 Other Study Materials



Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.




Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.


FAQs on NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

1. How can one download the NCERT Solutions of Exercise 3.1 of Chapter 3 Class 7 Maths?

The following procedures can help you to download NCERT Solutions for Exercise 3.1 of Chapter 3 Class 7 Maths;

  • Click on the NCERT Solutions for Exercise 3.1 of Chapter 3 Class 7 Maths.

  • Vedantu's page will open after clicking the link.

  • On that page, you will see that there is an option of "Download PDF".

  • Tap that option.

  • After all this, your PDF file will get downloaded.

  • Now, you can solve the questions of NCERT Solutions for Exercise 3.1 of Chapter 3 Class 7 Maths, even offline.

2. What are the topics in Chapter 3 of Class 7 Maths?

The topics are:

  • Introduction

  • Collecting Data

  • Organisation of Data

  • Representative Values

  • Arithmetic Mean

  • Range 

  • Mode

  • Mode of Large Data

  • Median

  • Using Bar Graphs For Different Purpose

  • Choosing a Scale

  • Drawing Double Bar Graphs

  • Chance and Probability

3. What are the privileges of studying Chapter 3 of Class 7 Maths from Vedantu?

The perks that one achieves after studying Chapter 3 of Class 7 Maths from Vedantu include:

  • Vedantu is one of the best online learning platforms that provide all the NCERT Solutions of all chapters of Maths Class 7 as it provides free solutions.

  • Each question of every chapter is well explained.

  • You can have the required content in the form of a PDF.

  • The contents are of high quality and are prepared by well-qualified teachers.

  • You can also download these contents if you want to study offline and also study on the Vedantu Mobile app.

  • Question papers for each chapter are also available so that students can practice them.

  • The notes and solutions are written in understandable language.

4. How can I top in Chapter 3 of Class 7 Maths?

Many students find it difficult to score good marks in Chapter 3 of Class 7 Maths. If they want to top in this chapter, then they have to prefer the NCERT book. As this book is prescribed by the CBSE, students can easily understand the concepts of the chapter. By solving the NCERT questions, they can comprehend the chapter easily. They should solve previous years question papers and mock tests to know about the type of questions asked in the exam. Do not miss out on any school lectures and clarify your doubts with the help of your teachers. Prepare notes so that you learn formulas for solving questions.

5. What technique should I acquire to make an effective study plan for Chapter 3 of Class 7 Maths?

The following technique will be helpful:

  • Make a well-organized timetable.

  • The timetable must include all the subjects including Maths subject.

  • Study Chapter 3 of Class 7 Maths for an hour.

  • Do not extend your study time.

  • Make sure that your timetable also includes playtime.

  • To build a strong grip on every concept, practice questions related to that topic.

  • Make notes as it becomes easy for you to learn the chapter.

6. What is the main focus of data handling class 7 exercise 3.1?

The main focus of Exercise 3.1 is to understand and calculate key statistical measures. These measures include the arithmetic mean, range, mode, and median. Grasping these concepts is crucial as they form the basis of data analysis. Students learn how to summarize and interpret data effectively.

7. How do you calculate the arithmetic mean for class 7 maths chapter 3 exercise 3.1 solutions?

To calculate the arithmetic mean, you first add all the observations in the data set. Next, divide the total sum by the number of observations. This gives you the average value of the data set. The arithmetic mean helps to understand the central tendency of the data.

8. What is the range class 7 maths chapter 3.1?

The range is the difference between the highest and lowest values in a data set. To find the range, subtract the smallest observation from the largest one. This measure gives an idea of the spread or dispersion of the data. It indicates the variability within the data set.

9. How do you find the mode in class 7 math exercise 3.1?

The mode is the value that appears most frequently in a data set. To identify the mode, count how many times each value occurs and find the one with the highest frequency. Understanding the mode is useful for recognizing patterns and common occurrences in data.

10. Why is it important to understand these class 7 math exercises 3.1 ?

Understanding these concepts is fundamental for analyzing and interpreting data accurately. They provide a foundation for more advanced statistical methods. These skills are useful in various real-life situations and other academic subjects. Mastery of these concepts enhances critical thinking and problem-solving abilities.

11. How many questions from class 7 maths chapter 3 exercise 3.1  were asked in previous years' exams?

In previous years' exams, 2 questions from Exercise 3.1 have been asked. These questions often test the understanding of arithmetic mean, range, mode, and median. Knowing this can help students focus their preparation on frequently tested areas. It highlights the importance of mastering these fundamental concepts.