NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

Data handling is an essential part of mathematics that helps us make sense of the world by organising, interpreting, and analysing information. Whether it’s tracking daily temperatures, recording scores in a game, or understanding survey results, data handling skills are used in real-life situations every day. This chapter introduces students to collecting and representing data using graphs, charts, and other methods like finding the arithmetic mean, median, mode and range. With step-by-step explanations, the NCERT Solutions act as a reliable resource for exam preparation.

Latest: NCERT Class 7 Maths: Chapterwise Important Formulas with Examples and Points

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1. NCERT Solutions for Maths Chapter 3 Data Handling Class 7 - Important Points
2. NCERT Solutions for Maths Chapter 3 Data Handling Class 7
3. NCERT Solutions for Class 7 Maths Chapter 3 Data Handling(Exercise)
4. Data Handling Class 7 Maths Chapter 3-Topics
5. NCERT Solutions for Class 7 Maths Chapter 3 Data Handling - Points to Remember
6. NCERT Solutions for Class 7 Maths Chapter Wise

These solutions are prepared by subject matter experts at Careers360, aiming to strengthen these skills and help students practice different types of problems, understand key concepts clearly, and build confidence in answering questions in exams. To explore more chapters and enhance your exam preparation, click on NCERT Solutions for Class 7 Maths.

NCERT Solutions for Maths Chapter 3 Data Handling Class 7 - Important Points

Arithmetic Mean (Average): The sum of all observations divided by the total number of observations.

Mean = $\frac{X_1+X_2+X_3+X_4+\dots +X_n}{n}$

Range: The difference between the largest and smallest values in a dataset.

Range = Largest value - smallest value

Mode: The value(s) that appear most frequently in a given dataset.

Mode = Value or values that appear most frequently in a given data set.

Median: The middle value of a dataset when arranged in ascending or descending order. For odd datasets, it's the middle value; for even datasets, it's the average of the two middle values.

Frequency: The count of how many times a specific number appears in a dataset.

Bar Graph: A visual representation using rectangular bars, where each bar's height corresponds to the frequency of a category.

NCERT Solutions for Maths Chapter 3 Data Handling Class 7

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

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.

Answer:

(i) $9$ is the highest number.

(ii) $1$ is the lowest number

(iii) Range of the data = Highest number - Lowest number

$9-1=8$

(iv) Arithmetic mean = $\frac{sumofobservations}{numberofobservation}$

$=\frac{(1\times 1)+(2\times 2)+(3\times 1)+(4\times 3)+(5\times 5)+(6\times 4)+(7\times 2)+(8\times 1)+(9\times 1)}{20}$

$$\begin{gathered} =\frac{1+4+3+12+25+24+14+8+9}{20} \\ =\frac{100}{20} \\ =5 \end{gathered}$$

Therefore, the mean of the data is 5

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

Answer: The first five whole numbers are $0,1,2,3,4$

We know the Arithmetic mean

$=\frac{sumofobservations}{numberofobservation}$

$$\begin{gathered} =\frac{0+1+2+3+4}{5} \\ =\frac{10}{5} \\ =2 \end{gathered}$$

Therefore, 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.

Answer: Given,

Runs scored in eight innings = $58,76,40,35,46,45,0,100$

We know the Arithmetic mean

$=\frac{sumofobservations}{numberofobservation}$

$$\begin{gathered} =\frac{58+76+40+35+46+45+0+100}{8} \\ =\frac{400}{8} \\ =50 \end{gathered}$$

Therefore, the mean score is 50

5. The 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?

Answer: Given,

We know,

$Arithmeticmean=\frac{sumofobservations}{numberofobservation}$

(i) Therefore, A's average number of points scored per game

$$\begin{gathered} =\frac{14+16+10+10}{4} \\ =\frac{50}{4} \\ =12.5 \end{gathered}$$

(ii) Number of games C played = 3

Therefore, to find the mean, we will divide by 3

Mean of C =

$$\begin{gathered} =\frac{8+11+13}{3} \\ =\frac{32}{3} \\ =10.67 \end{gathered}$$

(iii) Although he scored 0 in a match, he played all 4 games. Therefore, we will divide by 4.

Mean of B =

$$\begin{gathered} =\frac{0+8+6+4}{4} \\ =\frac{18}{4} \\ =4.5 \end{gathered}$$

(iv) Mean of A = $12.5$

Mean of B = $4.5$

Mean of C = $10.67$

Since A has the highest mean, 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.

Answer: Given,

Marks obtained by the students = $85,76,90,85,39,48,56,95,81,75$

Arranging in ascending order = $39,48,56,75,76,81,85,85,90,95$

(i) Highest marks obtained = $95$

Lowest marks obtained = $39$

(ii) Range of marks = Highest marks - Lowest marks

$=95-39=56$

(iii) We know the Arithmetic mean

$=\frac{sumofobservations}{numberofobservation}$

$$\begin{gathered} =\frac{39+48+56+75+76+81+85+85+90+95}{10} \\ =\frac{730}{10} \\ =73 \end{gathered}$$

Therefore, mean marks = 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.

Answer: Given,

Enrolment during 6 consecutive years = $1555,1670,1750,2013,2540,2820$

We know the Arithmetic mean

$=\frac{sumofobservations}{numberofobservation}$

$$\begin{gathered} =\frac{1555+1670+1750+2013+2540+2820}{6} \\ =\frac{12348}{6} \\ =2058 \end{gathered}$$

Therefore, the mean enrolment per year 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 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?

Answer:

Given,

(i) Range = Maximum rainfall - Minimum rainfall

$=12.2-0.0$

$=12.2$

(ii) We know, Arithmetic mean = $\frac{sumofobservations}{numberofobservation}$

$$\begin{gathered} =\frac{0.0+12.2+2.1+0.0+20.5+5.5+1.0}{7} \\ =\frac{41.3}{7} \\ =5.9 \end{gathered}$$

Therefore, the mean rainfall is $5.9mm$

(iii) Here, mean = $5.9mm$

Mon(0.0), Wed(2.1), Thu(0.0), Sat(5.5), Sun(1.0) are less than the mean

Therefore, on 5 days, rainfall was less than the mean.

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?

Answer: Given,

Heights of 10 girls (in cm) = $135,150,139,128,151,132,146,149,143,141$

Arranging the above data in ascending order = $128,132,135,139,141,143,146,149,150,151$

(i) Height of the tallest girl = $151cm$

(ii) Height of the shortest girl = $128cm$

(iii) Range of the data = $(151-128)cm$

$=23cm$

(iv) We know, Arithmetic mean = $\frac{sumofobservations}{numberofobservation}$

$$\begin{gathered} =\frac{128+132+135+139+141+143+146+149+150+151}{10} \\ =\frac{1414}{10} \\ =141.4cm \end{gathered}$$

Therefore, the mean height of the girls is $141.4cm$

(iv) Here, mean height = $141.4cm$

And, $143,146,149,150,151$ are more than $141.4$

Therefore, 5 girls have a height greater than the average height.

1. The scores in 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?

Answer: Scores of 15 students are: $19,25,23,20,9,20,15,10,5,16,25,20,24,12,20$

Arranging the data in ascending order = $5,9,10,12,15,16,19,20,20,20,20,23,24,25,25$

We know,

Mode: The number that occurs most frequently.

Median: The Median is the middle value of the set of numbers.

Here, 20 occurs 4 times, therefore, 20 is the mode of the data.

Also, the middle value is 20.

Therefore, the median of the data is 20.

Yes, the median and mode of this data 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 the same?

Answer: Runs scored in the match = $6,15,120,50,100,80,10,15,8,10,15$

Arranging in ascending order = $6,8,10,10,15,15,15,50,80,100,120,$

Here, 15 occurs the maximum number of times. Hence, 15 is the mode of the data

Now,

$Mean=\frac{Sumofscores}{Numberofplayers}$

$$\begin{gathered} =\frac{6+8+10+10+15+15+15+50+80+100+120}{11} \\ =\frac{429}{11}=39 \end{gathered}$$

Therefore, the mean score is $39$

Now, the median is the middle observation of the data.

There are 11 terms. Therefore the middle observation is $\frac{11+1}{2}=6^{th}term$

Therefore, 15 is the median of the data.

No, they 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?

Answer: Given, weights of 15 students (in kg)= $38,42,35,37,45,50,32,43,43,40,36,38,43,38,47$

Arranging the data in ascending order = $32,35,36,37,38,38,38,40,42,43,43,43,45,47,50$

(i) Clearly, 38 and 43 both occur thrice. So, they both are the mode of the data.

Now, there are 15 values, so the median is the $\frac{15+1}{2}=8^{th}term$

Hence, 40 is the median value.

(ii) Yes, there are two modes of the data given.

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

Answer: Given, $13,16,12,14,19,12,14,13,14$

Arranging in ascending order = $12,12,13,13,14,14,14,16,19$

We know,

The mode is the observation that occurs the maximum number of times.

Median is the middle observation when the data is arranged in ascending or descending order.

Now,

Clearly, $14$ occurs thrice. So, $14$ is the mode of the data.

Now, there are 9 values, so the median is the $\frac{9+1}{2}=5^{th}term$

Hence, $14$ is the median value.

5. Tell whether the statement is true or false:
(i) The mode is always one of the numbers in a dataset.
(ii) The mean is one of the numbers in a dataset.
(iii) The median is always one of the numbers in a dataset.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has a mean of 9.

Answer: (i) True.

The mode is the observation that occurs the maximum number of times.

Hence, the mode will always be one of the numbers in the data.

(ii) False.

(iii) False.

For an even number of observations, the median is the mean of the ${\frac{n}{2}}^{th}$ and ${\left(\frac{n+1}{2}\right)}^{th}$ values.

(iv) Mean = $\frac{Sumofobservations}{Numberofobservations}$

$$\begin{gathered} =\frac{6+4+3+8+9+12+13+9}{8} \\ =\frac{64}{8} \\ =8 \end{gathered}$$

Hence, the statement is false.

1. Use the bar graph (Fig. 3.3) to answer the following questions.
(a) Which is the most popular pet? (b) How many students have a dog as a pet?

Answer:

(a) The bar graph represents the pets owned by the students of class seven.

And, the bar of the cat is the tallest

Hence, the cat is the most popular pet.

(b) The bar of the dog reaches the value 8

8 students have a dog as a pet.

2. Read the bar graph (Fig. 3.4) which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:
(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?

(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?

Answer:

(i) Observing the graph,

Number of books sold in 1989 = 180

Number of books sold in 1992 = 220

Number of books sold in 1990 = 480

(ii) 475 books were sold in the year 1990

225 books were sold in the year 1992

(iii) The years in which the total number of books sold was less than 250 are 1989 and 1992.

(iv) Around 175 books were sold in the year 1989.

By drawing a line from the top of the 1989 bar to the y-axis.

3. The number of children in six different classes is given below. Represent the data on a bar graph.

(a) How would you choose a scale? (b) Answer the following questions: (i) Which class has the maximum number of children? And the minimum? (ii) Find the ratio of students in class six to the students in class eight.

Answer:

(a) The scale: 1 unit = 10 children.

(b) (i) Class 5 has the maximum number of children.

Whereas, the tenth class has the minimum number of children

(ii) Number of students in class six = 120

Number of students in class eight = 100

Therefore, the required ratio = $\frac{120}{100}$

$=\frac{6}{5}$

Required ratio is $6:5$

4. The performance of a student in the 1st Term and the 2nd Term is given. Draw a double bar graph, choosing an appropriate scale, and answer the following:

(i) In which subject has the child improved his performance the most? (ii) In which subject is the improvement the least? (iii) Has the performance gone down in any subject?

Answer:

(i) According to the graph, Maths had the maximum increase in marks.

Hence,

The child improved his performance the most in Maths.

(ii) According to the graph, S. Science had the least increase in marks.

Hence,

The child improved his performance the least in S. Science.

(iii) According to the graph, Hindi's marks went down in the second term

Hence,

The child’s performance has gone down in Hindi.

5. Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph? (ii) Which sport is most popular? (iii) Which is preferred, watching or participating in sports?

Answer:

(i)

The double bar graph represents the number of people who like watching and participating in various sports.

We observe that,

The maximum number of people like either watching or participating in cricket.

The minimum number of people like either watching or participating in athletics.

(ii) According to the graph, the tallest bar is for cricket. Hence, cricket is the most popular sport.

(iii) The bars corresponding to watching sports are longer than the bars corresponding to participating in sports.

Hence, watching sports is preferred over participating in sports.

6. Take the data giving the minimum and the maximum temperature of various cities given at the beginning of this Chapter (Table 3.1). Plot a double bar graph using the data and answer the following:
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city, and which is the coldest city?
(iii) Name two cities where the maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city that has the least difference between its minimum and maximum temperature.

Answer:

Table 3.1 is

The double bar graph is :

(i) From the graph,

Jammu has the largest difference between the maximum and minimum temperature bars.

$\therefore$ Jammu is the city with the largest difference in its maximum temperature and minimum temperature on the given date.

(ii) The city with the maximum temperature would be the hottest.

And the city with the lowest temperature will be the coolest.

According to the graph, Jammu has the highest maximum temperature bar.

$\therefore$ Jammu is the hottest city.

Also, the lowest minimum temperature bar is in Bangalore.

$\therefore$ Bangalore is the coolest city.

(iii) Maximum temperature of Bangalore is ${28}^{\circ }C$

The minimum temperature of Jaipur is ${29}^{\circ }C$

$\therefore$ The required pair of two cities is: Bangalore and Jaipur

(iv) Mumbai is a city with the least difference in its maximum temperature and minimum temperature.

Data Handling Class 7 Maths Chapter 3-Topics

• Collecting Data
• Organisation Of Data
• Representative Values
• Arithmetic Mean
• Mode
• Median
• Use Of Bar Graphs With A Different Purpose

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling - Points to Remember

Arithmetic Mean (A.M.) - The average or arithmetic mean, or simply mean, is defined by the sum of all observations divided by the number of observations.

$Mean=\frac{Sumofobservations}{Numberofobservations}$

Range- It is the difference between the highest observation and the lowest observation.

Mode- Mode refers to the observation in a set of observations that occurs most often. For example, in groups 12, 13, 13, 13, 16, 16, 24 and 30, the number 13 is the mode.

Median- In a given dataset, arranged in order from lowest to highest, the median gives us the middle observation.

NCERT Solutions for Class 7 Maths Chapter Wise

NCERT Solutions for Class 7 Subject Wise

The NCERT Solutions for Class 7 subject-wise is one of the essential study materials that helps students to practice more problems in each subject of Class 7. Click on the link below to check out the subject-wise solutions.

Also, check the NCERT Books and the NCERT Syllabus here:

• NCERT Syllabus Class 7 Maths
• NCERT Books Class 7
• NCERT Syllabus Class 7