Our NCERT Solutions for Class 7 Maths Chapter 3 are prepared by professional tutors and provide stepbystep solutions to the questions present in various exercises of the chapter. These solutions are written in a clear and concise manner, with complete descriptions, to help students understand the concepts and solve the problems on their own. We encourage students to use our NCERT Solutions for Class 7 Maths Chapter 3 to prepare for their exams and achieve their academic goals.
The NCERT Solutions for Class 7 Maths Chapter 3  Data Handling are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 20232024. These solutions provide thorough explanations with a stepbystep approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Students can access the NCERT Solutions for Class 7 Maths Chapter 3  Data Handling. Curated by experts according to the CBSE syllabus for 2023–2024, these stepbystep solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be wellprepared for tests.
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.
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
Find the mean of the first five whole numbers.
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.
A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
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.
Find the range of heights of any ten students in your class.
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 abovementioned values, the highest value is = 148 cm
By observing the abovementioned values, the lowest value is = 130 cm
Then,
Range of Heights = Highest value – Lowest value
= 148 – 130
= 18 cm
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 by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who is the best performer?
(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.
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.
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
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.
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.
The rainfall (in mm) in a city on 7 days of a certain week was 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 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?
(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.
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?
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).
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?
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 the maximum number of times.
Hence, the mode of the given sores is 20
Median,
The value of the middlemost 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 8th term = 20
Hence, the median is 20.
Yes, both values are the same.
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?
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 the maximum number of times.
Hence, the mode of the given sores is 15.
Median,
The value of the middlemost 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 6th term = 15
Hence, the median is 15.
No, these three are not the same.
The weights (in kg.) of 15 students in 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?
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 mode of the given weights is 38 and 43.
Median
The value of the middlemost 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 8th term = 40
Hence, the median is 40.
(ii) Yes, there are 2 modes for the given weights of the students.
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14.
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 middlemost 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 5th term = 14
Hence, the median is 14.
Tell whether the statement is true or false:
(i) The mode is always one of the numbers in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 have a mean 9.
(i) The statement given above is true.
Because mode is the value of the variable which occurs most frequently in the given data.
Hence, the mode is always one of the numbers in the data.
(ii) The statement given above is false.
Because mean may or may not be one of the numbers in a data.
(iii) The statement given above is true.
Because the median is the value of the middlemost 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) 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.
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?
The bar graph represents the pets owned by the students.
(a) From the bar graph, the most popular pet is a cat. It is owned by 10 students out of 12 students.
(b) From the bar graph, 8 students have a dog as a pet out of 12 students.
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?
(i) By observing the bar graph,
175 books were sold in the year 1989.
475 books were sold in the year 1990.
225 books were sold in the year 1992.
(ii) By observing the bar graph,
475 books were sold in the year 1990.
225 books were sold in the year 1992.
(iii) By observing the bar graph,
In the years 1989 and 1992, the number of books sold was less than 250.
(iv) By observing the bar graph, we can conclude that,
The number of books sold in the year 1989 is about 1 and ¾th part of 1 cm.
WKT, Scale is taken as 1 cm = 100 books
= 100 + (¾ × 100)
= 100 + (3 × 25)
= 100 + 75
= 175
Number of children in six different classes is given below. Represent the data on a bar graph.
Class 
Fifth 
Sixth 
Seventh 
Eighth 
Ninth 
Tenth 
Number of Children 
135 
120 
95 
100 
90 
80 
(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 of class sixth to the students of class eight.
(a) We will take the scale as 1 unit = 10 children because we can represent a bigger and clear difference among the number of students in class 7th and class 9th.
(b) (i) Class 5th has the maximum number of children, i.e., 135, and class 7th has the minimum number of children, i.e., 95.
(ii) The total number of students in class 6th is 120, and the total number of students in class 8th is 100.
Then,
Ratio between the number of students in class 6th and 8th,
= (120/100)
= 6/5
= 6: 5
The performance of a student in the 1st Term and 2nd Term is given. Draw a double bar graph choosing the appropriate scale, and answer the following.
Subject 
English 
Hindi 
Maths 
Science 
S. Science 
1st Term (M.M. 100) 
67 
72 
88 
81 
73 
2nd Term (M.M. 100) 
70 
65 
95 
85 
75 
(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?
(i) By observing the double bar graph, there was a maximum mark increase in the Maths subject. So, the child has improved his performance in Maths.
(ii) By observing the double bar graph, the improvement was the least in S. Science.
(iii) By observing the double bar graph, the performance In Hindi has gone down.
Consider this data collected from a survey of a colony.
Favourite Sport 
Cricket 
Basket Ball 
Swimming 
Hockey 
Athletics 
Watching 
1240 
470 
510 
430 
250 
Participating 
620 
320 
320 
250 
105 
(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 more preferred, watching or participating in sports?
(i) The fig above is the double bar graph, which represents the people who prefer either watching or participating in different sports. By observing the double bar graph, we came to conclude that most people like watching and participating in cricket, while the least number of people like watching and participating in athletics.
(ii) By observing the double bar graph, we came to conclude that the people who like watching and participating in cricket are the tallest among all the bars. So, cricket is the most popular sport.
(iii) By observing the double bar graph, we came to conclude that watching sports has more preference, rather than participating in sports.
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 which has the least difference between its minimum and the maximum temperature.
(i) By observing the double bar graph, we came to conclude that Jammu has the largest difference in the minimum and maximum temperature on 20.6.2006.
(ii) By observing the double bar graph, we came to conclude that Jammu is the hottest city and Bangalore is the coldest city.
(iii) By observing the double bar graph, Bangalore and Jaipur, Bangalore and Ahmedabad, for Bangalore, the maximum temperature was 28oC, while the minimum temperature of both Ahmadabad and Jaipur was 29oC.
(iv) By observing the double bar graph, Mumbai has the least difference between its minimum and maximum temperature.
Tell whether the following is certain to happen, impossible, or can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A die, when tossed, shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
(i) It is certain to happen.
(ii) It can happen but not certain.
(iii) It is impossible because there are only six faces on a die, marked as 1, 2, 3, 4, 5, and 6 on it.
(iv) It can happen but not certain.
(v) It can happen but not certain.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
A coin has two faces, one is the Head, and the other one is the Tail.
Now, one team can choose either Head or Tail.
The probability of our team starting the first= Number of favourable outcomes/
Number of possible outcomes
= ½
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with the number 2?
(ii) What is the probability of drawing a marble with the number 5?
(i) From the question, it is given that
There are 6 marbles in the box with numbers from 1 to 6 marked.
The probability of drawing a marble with the number 2 = Number of favourable outcomes/
Number of possible outcomes
= (1/6)
(ii) From the question, it is given that
There are 6 marbles in the box with numbers from 1 to 6 marked.
The probability of drawing a marble with the number 5 = Number of favourable outcomes/
Number of possible outcomes
= (1/6)
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The NCERT solution for Class 7 Chapter 3: Data Handling is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 7 Chapter 3: Data Handling is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Data Handlingally, they can solve the practice questions and exercises that allow them to get examready in no time.
You can get all the NCERT solutions for Class 7 Maths Chapter 3 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 3: Data Handling as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 7 Maths Chapter 3 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.