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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling are formulated by our professional tutors to help you with your exam preparation to attain good marks in Maths. The NCERT book is one of the top materials when it comes to providing a question bank to practice. Our main focus is on helping students to understand and crack these problems. We at Swiflearn have prepared the NCERT Solutions for Class 7 Maths Chapter 3 wherein problems are solved step by step with complete descriptions.

NCERT Solutions for Class 7 Maths Chapter 3 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 7 Maths Chapter 3 are designed as per the CBSE Class 7 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 7 Chapter 3. NCERT Solutions for Class 7 Maths Chapter 3 Data Handling: EX 3.1 PDF

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

Solution:-
Taking random heights of 10 students in the class = 129, 132, 134, 136, 139, 140, 141, 143, 146, 149
By observing, the highest value is = 149 cm
By observing, the lowest value is = 129 cm
Then,
Range of Heights = Highest value – Lowest value
= 149 cm – 129 cm
= 20 cm

Question 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:-
Firstly, 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
Frequency Table for the given data is as follows,  (i) By observing, the highest number is 9.
(ii) By observing, the lowest number is 1.
(iii) As we know, Range = Highest value – Lowest value
= 9 – 1
= 8
(iv) Arithmetic Mean,
Arithmetic mean = {(Sum of all observations)/ (Total number of observation)}
Then,
Sum of the observation = {1 + 2 + 2 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 5 + 6 + 6 + 6 + 6 + 7 + 7 + 8 + 9}
= 100
Number of Observation = 20
Arithmetic mean = (100/20)
= 5

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

Solutions:-
0, 1, 2, 3, and 4 = these are the first five whole numbers.
Mean = (Sum of first five whole numbers)/ (Total number of whole numbers)
Then,
Sum of first five whole numbers = 0 + 1 + 2 + 3 +4
= 10
Total Number of whole numbers = 5
Mean = (10/5 )
= 2 {Mean of first five whole numbers}

Question 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)
Overall 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 {Mean score of the cricketer}

Question 5:Following table shows the points of each player scored in four games: Now answer the following questions:(i) Find the mean to determine A’s average number of points scored pergame.(ii) To find the mean number of points per game for C, would you dividethe total points by 3 or by 4? Why?(iii) B played in all the four games. How would you find the mean?(iv) Who is the best performer?

Solution:-
(i) A’s average points scored per game = Total points by A in 4 games/Total number of games = (14 + 16 + 10 + 10)/4

= 50/4
= 12.5 points

(ii) Mean points per game for C, total points will be divided by 3. As C played only 3 games.
(iii) B played all the four games, so it will be divided by 4 to find out the mean.
Then, Mean score of B = Total points scored by B in 4 games/ Total number of games
= (0 + 8 + 6 + 4)/4 = 18/4
= 4.5 points

(iv) Now, The best performer among 3 players.
So, we have to find the average of C = (8 + 11 + 13)/3
= 32/3
= 10.67 points

By observing, the average points
A scored 12.5 which is more than B and C.
Therefore, A is the best performer among three.

Question 6:The marks (out of 100) obtained by a group of students in a science test are85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the:(i) Highest and the lowest marks obtained by the students.(ii) Range of the marks obtained.(iii) Mean marks obtained by the group.

Solution:-
Firstly, arrange the obtained marks in ascending order.
= 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i)
The highest marks obtained = 95
The lowest marks obtained = 39
(ii)
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

Question 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.

Solution:-
Mean enrolment = Sum of all observations/ Number of observations
= (555 + 1670 + 1750 + 2013 + 2540 + 2820)/6
= (12348/6 )
= 2058

Question 8:The rainfall (in mm) in a city on 7 days of a certain week was recorded asfollows: (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?

Solution:-
(i) Range of rainfall = Highest rainfall – Lowest rainfall
= 20.5 – 0.0
= 20.5 mm
(ii) Mean of rainfall = Sum of all the observations / Number of observation
= (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)/7
= 41.3/7
= 5.9 mm
(iii) By observing it was concluded that on. 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:128, 149, 139, 135, 151, 146, 150, 143, 141, 132.(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:-

Firstly, arrange the obtained marks in ascending order,
= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) Tallest girl is 151 cm
(ii) Shortest girl is 128 cm
(iii) Range of given data = Tallest height of the girl – Shortest height
= 151 – 128
= 23 cm
(iv) Mean height of the girls = Combined (sum) 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) Total of 5 girls have heights more than the mean height (i.e. 141.4 cm)

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling: EX 3.2 PDF

Question 1:
The scores in mathematics test (out of 25) of 15 students is as follows:
19, 25, 24, 20, 12, 20, 20, 10, 5, 16, 25, 15, 23, 9, 20
Find the mode and median of this data. Are they same?

Solution:-
Firstly, arrange the obtained marks in ascending order,
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode: Mode is a variable in the data which occurs most frequently.
By observing, 20 occurs most number of times.
Hence, mode = 20
Median: It is the middle most value in the given data.
Here n = 15 {odd value}
∴Median = value of (n + 1) observation.
= ½ (15 + 1)th
= ½ (16)
=
16
2
= 8
So, value of 8th term is 20
Therefore, the median is 20.
Yes, both the values are exact same.

Question 2:The runs scored in a cricket match by 11 players is as follows:6, 15, 15 , 50, 100,15 , 10, 120, 8, 10, 80Find the mean, mode and median of this data. Are the three same?

Solution:-
Firstly, arrange the runs scored by the players in ascending order,
6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
Mean, = Sum of all observations of the given data / Total number of observations
= (6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120)/11
= 429/11
= 39
Mode: Mode is a variable in the data which occurs most frequently.
By observing, 15 occurs most number of times.
Hence, mode = 15
Median: It is the middle most value in the given data.
Here n = 11 {odd values}
∴Median = value of (n + 1)th observation = ½ (11 + 1)
= ½ (12)
= 12/2
= 6
Then, value of 6th term = 15
Therefore, the median is 15.
No, the three values are not same.

Question 3:The weights (in kg.) of 15 students of a class are:32, 42, 35, 37, 45, 47, 38, 43, 43, 40, 36, 38, 43, 38, 50(i) Find the mode and median.(ii) Is there more than one mode?

Solution:-
Firstly, arrange weights of the students in ascending order,
32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
(i) Mode and Median
Mode: Mode is a variable in the data which occurs most frequently.
Since, 38 and 43 both occurs 3 times.
Therefore, mode of the given weights is 38 and 43.
Median: It is the middle most value in the given data.
Here n = 15 {odd value}
∴Median = value of (n + 1)th observation
= ½ (15 + 1)
= ½ (16)
=16/2
= 8
So, value of 8th term = 40
Therefore, the median is 40.
(ii) Yes, mode has 2 values in the given data.

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

Solution:-
Arranging the given data in an ascending order,
= 12, 12, 13, 13, 14, 14, 14, 16, 19
Mode: Mode is a variable in the data which occurs most frequently.
By observing, 14 occurs most number of times.
So, mode= 14.
Median: It is the middle most value in the given data.
Here n = 9 {odd value}
∴Median = value of (n + 1)th observation
= ½ (9 + 1)
= ½ (10)
=
10/2
= 5
Since, value of 5th term = 14
Therefore, the median is 14.

Question 5: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 has mean 9.

Solution:-
(i)
True,
Mode: Mode is a variable in the data which occurs most frequently.
(ii)
False,
As it can be a number from the data, or another average of the data.
(iii)
True,
Median: It is the middle most value in the given data.
(iv)
Mean = Sum of all observations of the given data / Total number of observations
=
(6 + 4 + 3 + 8 + 9 + 12 + 13 + 9)/8
= (64/8 )
= 8
So, False

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling: EX 3.3 PDF

Question 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 dog as a pet? Solution:-
(a) According to the graph, Cat is the most popular pet.
(b) According to the graph, 8 students have dog as a pet.

Question 2:Read the bar graph (Fig 3.4) which shows the number of books sold by abookstore during five consecutive years and answer the followingquestions:(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 in1989? Solution:-
(i)
According to the graph,
In the year 1989, 175 books were sold.
In the year 1990, 475 books were sold.
In the year 1992, 225 books were sold.
(ii)
According to the graph,
In the year 1990, 475 books were sold.
In the year 1992, 225 books were sold.
(iii) According to the graph, Books sold in the year 1989 and 1992, were less than 250.
(iv)
By observing the bar graph, we can conclude that,
The y-axis line is divided into 10 small parts for 10 books each. So, we can say the number of books sold in 1989 were about 180.

Question 3:Number of children in six different classes are given below. Represent thedata on a bar graph. (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.

Solution:- (a) Taking scale as 1 unit = 10 children
(b)
(i) The maximum number of children is in 5th class that is 135 and class 7th has the minimum number of children that is 95.
(ii) The total students in class 6th is 120 and the total students in class 8th is 100
Now, Ratio of class 6th and 8th students
= ( 120/100 )
= 6/5
= 6: 5

Question 4:The performance of a student in 1st Term and 2nd Term is given. Draw adouble bar graph choosing 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?

Solution:- (i) By observation, the child has improved his marks in Maths.
(ii) By observation, the improvement was the least in Social Science.
(iii) By observation, the performance In Hindi has gone down.

Question 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 more preferred, watching or participating in sports?

Solution:- (i) The above bar graph shows the number of people who are watching and who are participating in sports.
(ii) By observing the graph, we get to know that the people like watching and participating in cricket. So, cricket is the most popular sport.
(iii) By observing the graph, we get to know that people like watching sports has more preference, rather than participating in sports.

Question 6:Take the data giving the minimum and the maximum temperature ofvarious cities. Plot a double bar graph using the data and answer thefollowing: (i) Which city has the largest difference in the minimum and maximumtemperature on the given date?(ii) Which is the hottest city and which is the coldest city?(iii) Name two cities where maximum temperature of one was less than theminimum temperature of the other.(iv) Name the city which has the least difference between its minimum andthe maximum temperature.

Solution:- (i) By observing the graph, we came to know that Jammu has the maximum difference in the minimum and maximum temperature as on 20.6.2006.
(ii) By observing the graph, we came to know that Jammu is the hottest city and Bangalore is the coldest.
(iii) By observing the graph, Bangalore and Jaipur, Bangalore and Ahmedabad.
For Bangalore, 28oC as maximum temperature while Ahmadabad and Jaipur was 29oC as minimum temperature
(iv) By observing the graph, Mumbai has the least difference between its minimum and the maximum temperature.

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling: EX 3.4 PDF

Question 1:Tell whether the following is certain to happen, impossible, 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.

Solution:-
(i)
Yes, it is certain to happen.
(ii)
Yes, It can happen but not certain.
(iii)
It is impossible as, there are only six faces on a die.
(iv)
Yes, It can happen but not certain.
(v)
Yes, It can happen but not certain.

Question 2:There are 6 marbles in a box with numbers from 1 to 6 marked on each ofthem.(i) What is the probability of drawing a marble with number 2?(ii) What is the probability of drawing a marble with number 5?

Solution:-
From the question,
(i)Number of marbles in the box: 6
Probability of drawing number 2 = Number of favorable outcomes
Number of possible outcomes
= (1/6 )
(ii)
From the question,
Number of marbles in the box: 6
Probability of drawing number 5 = Number of favorable outcomes
Number of possible outcomes
= (1/6 )

Question 3:
A coin is flipped to decide which team starts the game. What is the
probability that your team will start?

Solution:-
A coin has two faces, Head and Tail.
Probability of head or tail = Number of favorable outcomes
Number of possible outcomes =1/2 