# 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. Here you will get solutions to four exercises of this chapter.

Exercise:3.1

Exercise:3.2

Exercise:3.3

Exercise:3.4

## 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

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

(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

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

 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

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?.

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

Draw a double bar graph and answer the following questions:

(b) Can you say that the demand for English books rose faster? Justify.

## 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.

 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 $5$

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 $2$

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 $50$

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.

(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.

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 = $73$

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 $2058$

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.

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.

## 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?

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.

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.

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.

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.

(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.

## 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?

(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.

(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) 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.

(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 $6:5$

(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.

(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.

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

(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

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

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

$= \frac{1}{6}$

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

## 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.