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NCERT Solutions For Class 7 Maths Chapter 3 Data Handling Exercise 3.1 - 2025-26

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Data Handling Questions and Answers - 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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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.




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FAQs on NCERT Solutions For Class 7 Maths Chapter 3 Data Handling Exercise 3.1 - 2025-26

1. What key concepts are covered in the NCERT Solutions for Class 7 Maths Chapter 3, Data Handling?

The NCERT Solutions for Class 7 Maths Chapter 3 primarily focus on methods for interpreting and organising data. Key concepts covered include:

  • Calculating the arithmetic mean (average).

  • Determining the range of a dataset.

  • Finding the mode and median of ungrouped data.

  • Constructing and interpreting single and double bar graphs.

  • Understanding the basic principles of chance and probability.

2. How do you correctly calculate the arithmetic mean for a set of observations as per the NCERT method?

To calculate the arithmetic mean, you need to follow a simple two-step process as outlined in the NCERT solutions. First, find the sum of all the given observations or values in the dataset. Second, divide this sum by the total number of observations. The formula is: Mean = (Sum of all observations) / (Number of observations). This gives you the central or average value of the data.

3. What is the step-by-step process for finding the mode of a given dataset in Chapter 3?

The mode is the observation that occurs most frequently in a dataset. To find it, follow these steps:

  • Step 1: Arrange the given data in ascending order or create a frequency table to count the occurrences of each number.

  • Step 2: Identify the observation that appears the highest number of times.

  • Step 3: This most frequent observation is the mode of the dataset. A dataset can have more than one mode.

4. How is the range of a dataset calculated in Class 7 Maths?

The range of a dataset indicates its spread. To calculate it, you subtract the lowest (minimum) value from the highest (maximum) value in the dataset. The formula is simple: Range = Highest Observation - Lowest Observation. Finding the range helps you understand the variation within your data.

5. How do you find the median of a dataset, and what is the difference in method for an odd versus an even number of observations?

The median represents the middle value of an ordered dataset. The first step is always to arrange the data in ascending order. After that, the method differs:

  • For an odd number of observations, the median is simply the middle value in the list.

  • For an even number of observations, the median is the average of the two middle values. You find these two values, add them together, and divide the sum by 2.

6. What is a common mistake students make when solving for the median of a dataset with an even number of values?

A very common mistake is forgetting to calculate the average of the two middle numbers when there is an even number of observations. Students often mistakenly list both middle numbers as the median or just pick one of them. The correct method is to add the two central values and divide by two to find the single value that represents the median.

7. When is the median a more appropriate measure of the central tendency of data than the mean?

The median is often a better measure than the mean when the dataset contains outliers—extremely high or low values that can skew the average. Because the mean incorporates every value, an outlier can significantly raise or lower it, giving a misleading sense of the centre. The median, being just the middle value, is not affected by these extreme outliers and thus provides a more accurate representation of the typical value.

8. What is the purpose of a double bar graph, and how does it differ from a single bar graph in representing data?

A double bar graph is used to compare two sets of data on the same graph. For example, you could compare the marks of students in Maths and Science across two terms. A single bar graph, on the other hand, represents only one set of data. The key difference is that a double bar graph uses two bars side-by-side for each category to facilitate direct comparison, while a single bar graph has only one bar per category.

9. How are the NCERT Solutions for Data Handling useful for preparing for the 2025-26 exams?

The NCERT Solutions for Data Handling are crucial for exam preparation as they provide step-by-step methods for solving every problem in the textbook, which is prescribed by CBSE. By practising these solutions, you learn the correct format for presenting answers, understand the logic behind each formula (mean, median, mode), and master the construction of bar graphs, all of which are common topics in exams.

10. Why is it important to organise data before calculating measures like mean, median, or mode?

Organising data, typically by arranging it in ascending order or in a frequency table, is a critical first step for several reasons. For calculating the median, it is essential to find the middle value. For finding the mode, organisation makes it easier to spot the most frequent value. It also helps in identifying the highest and lowest values needed for the range and reduces the chance of making calculation errors for the mean.

11. How is probability defined in Chapter 3, and how do you calculate the probability of a simple event?

In Chapter 3, probability is defined as the measure of how likely an event is to occur. It is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The formula is: Probability of an event = (Number of outcomes that make an event happen) / (Total number of possible outcomes). The value of probability always lies between 0 (impossible event) and 1 (certain event).

12. Beyond school exams, where are the data handling concepts from Chapter 3 used in the real world?

The concepts from Data Handling are widely used in many real-world scenarios. For example, the mean is used to calculate average rainfall or average temperature. The median is used to find the median household income to avoid skewing by extremely wealthy individuals. Bar graphs are used in business reports and news articles to compare sales figures or election results. Probability is fundamental to weather forecasting, sports analytics, and games of chance.