NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

 

NCERT solutions for class 7 maths chapter 3 Data Handling- In everyday life people deal with various types of data. The chapter data handling discusses how to represent and organize data in different ways so that their interpretation and analysis become easier. The solutions of NCERT class 7 maths chapter 3 data handling contains 4 exercises with 23 questions. Also along with these, there is an explanation to topic wise questions in the CBSE NCERT solutions for class 7 maths chapter 3 data handling. Collection of data and representing it using a bar graph are dealt with in class 6. More types of data and graphs are dealt with in class 7 chapter 3 data handling. Certain examples of data are temperatures of cities on a particular day, weekly absentees in a class, etc. and some basic terms like arithmetic mean, range, mode, and median are discussed. Questions on these topics are explained in the  NCERT solutions for class 7 maths chapter 3 data handling. To perform well in the exam the NCERT solutions are helpful.

Topics of NCERT grade 7 maths chapter 3 Data Handling-

3.1  Introduction

3.2  Collecting Data

3.3  Organisation of Data

3.4  Representative Values

3.5  Arithmetic Mean

3.5.1  Range

3.6  Mode

3.6.1  Mode of Large Data

3.7  Median

3.8  Use of Bar Graphs withe a Different Purpose

3.8.1 Choosing a Scale

3.9  Chance and Probability

3.9.1  Chance

NCERT solutions for class 7 maths chapter 3 data handling topic 3.5

Q. Find at least 5 numbers between 1/2 and 1/3 

Answer:

Given number are:  \frac{1}{2} and \frac{1}{3}

Their average = \frac{\frac{1}{2} + \frac{1}{3}}{2} = \frac{3+2}{2\times6} = \frac{5}{12}

Average between \frac{5}{12}\ and\ \frac{1}{3} =  \frac{\frac{5}{12} + \frac{1}{3}}{2} = \frac{5+4}{2\times12} = \frac{9}{24}= \frac{3}{8}

Average between \frac{5}{12}\ and\ \frac{1}{2} =   \frac{\frac{5}{12} + \frac{1}{2}}{2} = \frac{5+6}{2\times12} = \frac{11}{24}

Average between \frac{11}{24}\ and\ \frac{1}{2} =  \frac{\frac{11}{24} + \frac{1}{2}}{2} = \frac{11+12}{2\times24} = \frac{23}{48}

Average between  \frac{9}{24}\ and\ \frac{1}{3}\frac{\frac{9}{24} + \frac{1}{3}}{2} = \frac{9+8}{2\times24} = \frac{17}{48}

Therefore, 5 numbers between  \frac{1}{2} and \frac{1}{3} are:

\frac{17}{48},\frac{23}{48},\frac{5}{12},\frac{11}{24},\frac{9}{24}

 

Solutions of NCERT class 7 maths chapter 3 data handling topic 3.6 

Find the mode of 

(i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

(ii) 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Answer:

(i) Given,

2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Arranging in ascending order:

0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6

Here, 2, 3 and 4 all occur three times. Therefore, all three are modes of data.

(ii) Arranging in ascending order:

2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18

Here, 14 occurs 6 times. Therefore, 14 is the mode of the data.

Find the mode of 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Answer:

Arranging in ascending order:

2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18

Here, 14 occurs 6 times. Therefore, 14 is the mode of the data.

Find the mode of the following data:
12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15,
17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14

Answer:

Data Tally bars Number of times
12    ||| 3
13    |||| 4
14   \cancel{||||} 5
15   \cancel{||||}\cancel{||||} 10
16   \cancel{||||}| 6
18    | 1
19    \ \ |   1

Clearly, 15 occurs 10 times. Therefore, 15 is the mode of the data.

Solutions for NCERT class 7 maths chapter 3 data handling topic 3.8.1

2 a) Sale of English and Hindi books in the years 1995, 1996, 1997 and 1998 are given below:

 Draw a double bar graph and answer the following questions:

(a) In which year was the difference in the sale of the two language books least?.

Answer:

In 1998, the difference in the sale of the two language books was the least

NCERT solutions for class 7 maths chapter 3 data handling exercise 3.1

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:

Numbers Frequency
1 1
2 2
3 1
4 3
5 5
6 4
7 2
8 1
9 1

(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{sum\ of\ observations}{number\ of\ observation}

= \frac{(1\times1) + (2\times2) + (3\times1)+ (4\times3 )+ (5\times5 )+ (6\times4) + (7\times2)+(8\times1)+(9\times1)}{20}

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

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, Arithmetic mean

 =\frac{sum\ of\ observations}{number\ of\ observation}

\\ = \frac{0+1+2+3+4}{5} \\ = \frac{10}{5} \\ \\ = 2

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, Arithmetic mean 

=\frac{sum\ of\ observations}{number\ of\ observation}

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

Therefore, the mean score is 50

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 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 the four games. How would you find the mean?
(iv) Who is the best performer?

Answer:

Given,

We know, 

Arithmetic\: mean =\frac{sum\ of\ observations}{number\ of\ observation}

 

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

\\ = \frac{14+16+10+10}{4} \\ = \frac{50}{4} \\ \\ = 12.5

(ii) Number of games C played = 3

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

Mean of C =

\\ = \frac{8+11+13}{3} \\ = \frac{32}{3} \\ \\ = 10.67

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

Mean of B = 

\\ = \frac{0+8+6+4}{4} \\ = \frac{18}{4} \\ \\ = 4.5

(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) 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 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, Arithmetic mean

 =\frac{sum\ of\ observations}{number\ of\ observation}

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

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, Arithmetic mean

=\frac{sum\ of\ observations}{number\ of\ observation}

\\ = \frac{1555+ 1670+ 1750+ 2013+ 2540+ 2820}{6} \\ = \frac{12348}{6} \\ \\ = 2058

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{sum\ of\ observations}{number\ of\ observation}

\\ = \frac{0.0+12.2+2.1+0.0+20.5+5.5+1.0}{7} \\ = \frac{41.3}{7} \\ \\ = 5.9

Therefore, mean rainfall is 5.9\ mm

(iii) Here, mean = 5.9\ mm

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 = 151\ cm

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

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

= 23\ cm

(iv)  We know, Arithmetic mean = \frac{sum\ of\ observations}{number\ of\ observation}

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

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

(iv) Here, mean height = 141.4\ cm

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

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

NCERT Solutions for class 7 maths chapter 3 data handling exercise 3.2

1.  The scores in mathematics test (out of 25) of 15 students is 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 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 which occurs most frequently. 

Median: Median is the mid-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 is 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?

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{Sum\ of\ scores}{Number\ of\ players}

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

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 which occurs a maximum number of times.

Median is the middle observation when that 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 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.

Answer:

(i) True.

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

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

(ii) False.

(iii) False.

For 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{Sum\ of\ observations}{Number\ of\ observations}

\\ = \frac{6+ 4+ 3+ 8+ 9+ 12+ 13+ 9}{8} \\ = \frac{64}{8} \\ \\ = 8

Hence, the statement is false.

CBSE NCERT solutions for class 7 maths chapter 3 data handling exercise 3.3

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?

 

 

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 dog as 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 1989 bar to the y-axis.

3. Number of children in six different classes are 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 of class sixth to the students of class eight.

 

Answer:

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

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

Whereas, class tenth has the minimum number of children

(ii) Number of students in class sixth = 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 1st Term and 2nd Term is given. Draw a double 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?

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 more 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 of 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 in 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 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.

 

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 temperatures on the given date.

 

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

And the city with the least 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 of 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 are: Bangalore and Jaipur

 

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

NCERT solutions for class 7 maths chapter 3 data handling exercise 3.4

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.

Answer:

(i) You are older today than yesterday.

-This event is certain to happen.

(ii) A tossed coin will land heads up.

- This event can happen but not certain.     

(iii) A die when tossed shall land up with 8 on top.

- This event is impossible. A die can only show one of the numbers (1,2,3,4,5,6)

(iv) The next traffic light seen will be green.

- This event can happen but not certain

(v) Tomorrow will be a cloudy day.

- This event can happen but not certain

2. 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 number 2?
(ii) What is the probability of drawing a marble with number 5?

Answer:

There are 6 marbles in the box numbered from 1 to 6

(i) Number of marble with number 2 on it = 1

\therefore Probability of getting a marble with number 2

=\frac{number\ of\ favourable\ outcomes}{number\ of\ possible\ outcomes}

= \frac{1}{6}

(ii) Number of marble with number 5 on it = 1

\therefore Probability of getting a marble with number 5 

 =\frac{number\ of\ favourable\ outcomes}{number\ of\ possible\ outcomes}

= \frac{1}{6}

 

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

Answer:

We know, a coin has two faces: Head and Tail

One can choose either Head or Tail.

Therefore, the number of favourable outcome = 1

We know, Probability of any outcome

=\frac{number\ of\ favourable\ outcomes}{number\ of\ possible\ outcomes}

\therefore P(our\ team\ will\ start\ first) = \frac{1}{2}

NCERT Solutions for Class 7 Maths Chapter-wise 

Chapter No.

Chapter Name

Chapter 1

Solutions of NCERT for class 7 maths chapter 1 Integers

Chapter 2

CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals

Chapter 3

NCERT solutions for class 7 maths chapter 3 Data Handling

Chapter 4

Solutions of NCERT for class 7 maths chapter 4 Simple Equations

Chapter 5

CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles

Chapter 6

NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties

Chapter 7

Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles

Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities

Chapter 9

CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers

Chapter 10

NCERT solutions for class 7 maths chapter 10 Practical Geometry

Chapter 11

Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area

Chapter 12

CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions

Chapter 13

NCERT solutions for class 7 maths chapter 13 Exponents and Powers

Chapter 14

Solutions of NCERT for class 7 maths chapter 14 Symmetry

 NCERT Solutions for Class 7 Subject-wise 

Solutions of NCERT for class 7 maths

CBSE NCERT solutions for class 7 science

Some important terms:

Following are the basic terms delt in the chapter. These terms are important and are used in the NCERT solutions for class 7 maths chapter 3 data handling. Also in higher classes, these terms are studied in detail.

  • 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{Sum \:of\:observations}{Number\:of\:observations}

  • Range- It is the difference between the highest observation and the lowest observation.
  • Mode- Mode refers to the observation in a set of observation that occurs most often. For example, in group 12, 13, 13, 13, 16, 16, 24 and 30, the number 13 is the mode.
  • Median- In a given data, arranged in order from lowest to highest, the median gives us the middle observation.
 

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