NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

NCERT Solutions for Class 7 Maths Chapter 3 

Data Handling Class 7 is an important topic covered in Class 7 Maths Chapter 3, helping students understand how to collect, organise, represent, and interpret data. The chapter builds a strong foundation for real-life applications such as reading graphs, analysing information, and making logical conclusions based on numbers.

In Class 7th Maths Chapter 3 Data Handling, students learn concepts like range, mean, bar graphs, double bar graphs, probability basics, and organising data using tables. These concepts are explained step by step in Data Handling Class 7 NCERT Solutions, making it easier for students to follow and apply formulas correctly.

Data Handling Class 7 solutions provide clear explanations for each question, helping students understand the logic behind every step. Practising Data Handling Class 7 questions with answers improves problem-solving skills and boosts confidence for exams. The solutions are designed strictly according to the latest NCERT syllabus, ensuring accuracy and relevance.

Using Class 7 Maths Chapter 3 NCERT solutions regularly helps students revise quickly and prepare effectively for tests. With proper practice, students can easily master Data Handling and score well in mathematics while developing analytical thinking skills useful beyond the classroom.

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Data Handling Class 7 Maths Questions with Answers

Exercise 3.1 Page: 62

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

Solution:-

Let us assume the heights (in cm) of 10 students in our class be = 130, 132, 135, 137, 139, 140, 142, 143, 145, 148 By observing the above-mentioned values, the highest value is = 148 cm By observing the above-mentioned values, the lowest value is = 130 cm Then, Range of Heights = Highest value – Lowest value = 148 – 130 = 18 cm

2. Organise 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 the highest? (ii) Which number is the lowest?

(iii) What is the range of the data? (iv) Find the arithmetic mean.

Solution:-

First, we have to arrange the given marks in ascending order. = 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9 Now, we will draw the frequency table of the given data.

Marks	Tally Marks	Frequency
1		1
2		2
3		1
4		3
5		5
6		4
7		2
8		1
9		1

(i) By observing the table clearly, the highest number among the given data is 9.

(ii) By observing the table clearly, the lowest number among the given data is 1.

(iii) We know that Range = Highest value – Lowest value

= 9 – 1 = 8

(iv) Now, we have to calculate Arithmetic Mean,

Arithmetic mean = (Sum of all observations)/ (Total number of observations) Then, Sum of all observation = 1 + 2 + 2 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 7 + 7 + 8 + 9 = 100 Total Number of Observations = 20 Arithmetic mean = (100/20) = 5

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

Solutions:-

The first five Whole numbers are 0, 1, 2, 3, and 4. Mean = (Sum of first five whole numbers)/ (Total number of whole numbers) Then, Sum of five whole numbers = 0 + 1 + 2 + 3 +4 = 10 Total Number of whole numbers = 5 Mean = (10/5) = 2 ∴ The mean of the first five whole 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.

Solution:-

Mean score = (Total runs scored by the cricketer in all innings)/ (Total number of innings played by the cricketer) Total runs scored by the cricketer in all innings = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400 Total number of innings = 8 Then, Mean = (400/8) = 50 ∴ The mean score of the cricketer is 50.

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

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 by 4? Why?

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

(iv) Who is the best performer?

Solution:-

(i) A’s average number of points scored per game = Total points scored by A in 4 games/

Total number of games = (14 + 16 + 10 + 10)/ 4 = 50/4 = 12.5 points

(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played only 3 games.

(iii) B played in all four games, so we will divide the total points by 4 to find out the mean.

Then, Mean of B’s score = Total points scored by B in 4 games/ Total number of games = (0 + 8 + 6 + 4)/ 4 = 18/4 = 4.5 points (vi) Now, we have to find the best performer among the 3 players. So, we have to find the average points of C = (8 + 11 + 13)/3 = 32/3 = 10.67 points By observing, the average points scored A is 12.5, which is more than B and C. Clearly, we can say that A is the best performer among the three.

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) Highest and lowest marks obtained by the students.

(ii) Range of the marks obtained.

(iii) Mean marks obtained by the group.

Solution:-

First, we have to arrange the marks obtained by a group of students in a science test in ascending order. = 39, 48, 56, 75, 76, 81, 85, 85, 90, 95 (i) The highest marks obtained by the student = 95 The lowest marks obtained by the student = 39 (ii) We know that Range = Highest marks – Lowest marks = 95 – 39 = 56 (iii) Mean of Marks = (Sum of all marks obtained by the group of students)/ (Total number of marks) = (39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95)/ 10 = 730/10 = 73

7. The enrolment in a school for six consecutive years was as follows:

1555, 1670, 1750, 2013, 2540, 2820.

Find the mean enrolment of the school for this period.

Solution:-

Mean enrolment = Sum of all observations / Number of observations = (1555 + 1670 + 1750 + 2013 + 2540 + 2820)/ 6 = (12348/6) = 2058 ∴ The mean enrolment of the school for this given period is 2058.

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

(i) Find the range of 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?

Solution:-

(i) Range of rainfall = Highest rainfall – Lowest rainfall = 20.5 – 0.0 = 20.5 mm (ii) Mean of rainfall = Sum of all observations / Number of observations = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)/ 7 = 41.3/7 = 5.9 mm (iii) We may observe that for 5 days, i.e. Monday, Wednesday, Thursday, Saturday and Sunday, the rainfall was less than the average rainfall.

9. The heights 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 the data? (iv) What is the mean height of the girls?

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

Solution:-

First, we have to arrange the given data in ascending order. = 128, 132, 135, 139, 141, 143, 146, 149, 150, 151 (i) The height of the tallest girl is 151 cm. (ii) The height of the shortest girl is 128 cm. (iii) Range of given data = Tallest height – Shortest height = 151 – 128 = 23 cm (iv) Mean height of the girls = Sum of the height of all the girls / Number of girls = (128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151)/ 10 = 1414/10 = 141.4 cm (v) 5 girls have heights more than the mean height (i.e. 141.4 cm).

Exercise 3.2 Page: 68

1. The scores on the Mathematics test (out of 25) of 15 students are as follows:

19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20

Find the mode and median of this data. Are they the same?

Solution:-

Arranging the given scores in ascending order, we get 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25 Mode Mode is the value of the variable which occurs most frequently. Clearly, 20 occurs a maximum number of times. Hence, the mode of the given sores is 20. Median The value of the middle-most observation is called the median of the data. Here, n = 15, which is odd. Where n is the number of students. ∴ median = value of ½ (n + 1) ^th observation = ½ (15 + 1) = ½ (16) = 16/2 = 8 Then, the value of the 8 ^th term = 20 Hence, the median is 20. Yes, both values are the same.

2. The runs scored in a cricket match by 11 players are as follows:

6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15

Find the mean, mode and median of this data. Are the three same?

Solution:-

Arranging the runs scored in a cricket match by 11 players in ascending order, we get 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120 Mean Mean of the given data = Sum of all observations / Total number of observations = (6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120)/ 11 = 429/11 = 39 Mode, Mode is the value of the variable which occurs most frequently. Clearly, 15 occurs a maximum number of times. Hence, the mode of the given sores is 15. Median, The value of the middle-most observation is called the median of the data. Here n = 11, which is odd. Where n is the number of players. ∴ median = value of ½ (n + 1) ^th observation. = ½ (11 + 1) = ½ (12) = 12/2 = 6 Then, the value of the 6 ^th term = 15 Hence, the median is 15. No, these three are not the same.

3. The weights (in kg.) of 15 students of a class are:

38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47

(i) Find the mode and median of this data.

(ii) Is there more than one mode?

Solution:-

Arranging the given weights of 15 students of a class in ascending order, we get 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50 (i) Mode and Median Mode Mode is the value of the variable which occurs most frequently. Clearly, 38 and 43 both occur 3 times. Hence, the modes of the given weights are 38 and 43. Median The value of the middle-most observation is called the median of the data. Here, n = 15, which is odd. Where n is the number of students. ∴ median = value of ½ (n + 1) ^th observation = ½ (15 + 1) = ½ (16) = 16/2 = 8 Then, the value of the 8 ^th term = 40 Hence, the median is 40. (ii) Yes, there are 2 modes for the given weights of the students.

4. Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14

Solution:-

Arranging the given data in ascending order, we get = 12, 12, 13, 13, 14, 14, 14, 16, 19 Mode Mode is the value of the variable which occurs most frequently. Clearly, 14 occurs the maximum number of times. Hence, the mode of the given data is 14. Median The value of the middle-most observation is called the median of the data. Here, n = 9, which is odd. Where n is the number of students. ∴ median = value of ½ (9 + 1) ^th observation = ½ (9 + 1) = ½ (10) = 10/2 = 5 Then, the value of the 5 ^th term = 14 Hence, the median is 14.

5. Tell whether the statement is true or false.

(i) The mode is always one of the numbers in a data.

Solution:-

The statement given above is true. Because Mode is the value of the variable which occurs most frequently in the given data. Hence, a mode is always one of the numbers in the data.

(ii) The mean is one of the numbers in the data.

Solution:-

The statement given above is false. Because mean may or may not be one of the numbers in the data.

(iii) The median is always one of the numbers in a data.

Solution:-

The statement given above is true. Because the median is the value of the middle-most observation in the given data while arranged in ascending or descending order. Hence, the median is always one of the numbers in a data

(iv) The data 6, 4, 3, 8, 9, 12, 13, and 9 have the mean 9.

Solution:-

Mean = Sum of all given observations / Number of observations = (6 + 4 + 3 + 8 + 9 + 12 + 13 + 9)/8 = (64/8) = 8 Hence, the given statement is false.

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Mastering Data Handling Class 7 Questions with Answers through NCERT

Learning how to manage information is a vital part of modern mathematics. In Chapter 3, we dive into how you can take a messy pile of numbers and turn them into a clear story. Whether you're looking for data handling class 7 questions with answers or step-by-step exercise breakdowns, the goal remains the same: accuracy. We start by looking at the "Range," which is simply the difference between the highest and the lowest observations in your set. This helps you understand the spread of your data immediately.

When you tackle data handling class 7 questions with answers,, the focus is primarily on the Arithmetic Mean. You find this by adding all your observations together and dividing that sum by the total number of observations. It sounds simple, but it's the foundation for everything else you'll learn in statistics.

Exercise 3.1: Organizing Data and Finding the Mean

In this section, we focus on the initial stages of data processing. Before you can analyze anything, you have to organize it. Most students find that using a tally mark table is the most efficient way to ensure no number is left behind.

• Step 1: Arrange your data in ascending or descending order.

• Step 2: Use tally marks to count the frequency of each specific value.

• Step 3: Calculate the Range to see the extent of variation.

By working through these steps, you'll find that the data handling class 7 questions with answers 3.1 become much more manageable. For instance, if you're asked to find the mean of the first five whole numbers (0, 1, 2, 3, 4), you simply add them to get 10 and divide by 5, resulting in a mean of 2. It’s these small, consistent calculations that build your confidence for more complex chapters later on.

Exercise 3.2: Exploring Mode and Median

While the mean gives you an average, it doesn't always tell the whole story. This is where the Mode and Median come into play. These are also known as representative values.

• The Mode: This is the value that appears most frequently in your data set. A set of numbers can have more than one mode if multiple values appear the same number of times.

• The Median: This is the middle value when your data is arranged in order. If you have an odd number of observations, the middle one is easy to spot.

When you solve data handling class 7 questions with answers, you'll often see problems that ask you to find all three: mean, median, and mode. Comparing these three values helps you see if the "average" is actually a good representation of the group. If the mode and median are the same, your data is often quite symmetrical.

Exercise 3.3: Visualizing Information with Bar Graphs

Numbers can be boring on their own. We use bar graphs to make them visual. In data handling class 7 questions with answers 3.3, you'll learn how to choose a proper scale. Choosing a scale is a vital part of the process because if your scale is too large, your bars will look like tiny stubs. If it's too small, they won't fit on the page.

Double bar graphs are particularly useful. They let you compare two sets of data at a glance, like the marks of students in two different terms. When you draw these, remember to keep the width of the bars uniform and the gaps between them equal. This isn't just for aesthetics; it's for mathematical accuracy.

Exercise 3.4: Chance and Probability

The final part of the chapter introduces the concept of "Chance." Some things are certain to happen, like the sun rising in the east. Other things are impossible, like a turtle flying. Then there are things that "may or may not" happen, like it will rain tomorrow.

In mathematics, we quantify this as Probability. The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. If you toss a coin, there are 2 possible outcomes (Heads or Tails). The chance of getting a Head is 1/2. Understanding this helps you predict patterns in data that aren't fixed.

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