NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

# NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

Edited By Ramraj Saini | Updated on Feb 07, 2024 04:56 PM IST

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling are discussed here. These NCERT solutions are created by the expert team at Careers360 keeping in mind the latest CBSE syllabus 2023. These solutions are simple and comprehensive and cover step by step solutions to each problem. 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 Maths chapter 3 Data Handling class 7. Questions based on temperatures of cities on a particular day, weekly absentees in a class, etc are explained in the NCERT solutions for Class 7 Maths chapter 3 Data Handling.

You must go through the NCERT Syllabus for Class 7 Maths before checking the NCERT Class 7 solutions. To perform well in the exam the NCERT Solutions are helpful. Here you will get solutions to four exercises of this chapter. NCERT provides practise questions to enhance the knowledge of the students and the NCERT Solutions for Class 7 is useful.

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

Arithmetic mean (Average) = (Sum of all the observations)/(Total number of observations)

Mean = (X1 + X2 + X3 + X4 + . . . . . . . +Xn)/n

Range = Largest value - smallest values

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

P(E) = (Number of Trials in favour)/(Total number of Trials)

## 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.

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

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

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 two middle values.

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

Random Experiment: An experiment whose outcome cannot be predicted before it occurs.

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

Trial: An action that produces one or more outcomes in an experiment.

Event: A collection of outcomes associated with a random experiment.

Empirical Probability: The probability of an event E favoring the outcomes, calculated as the number of favorable trials divided by the total number of trials.

Free download NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF for CBSE Exam.

## NCERT Solutions for Maths Chapter 3 Data Handling class 7 Topic 3.5

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:

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

## NCERT Soltions for Class 7 Maths Chapter 3 Data Handling Topic 3.6

(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

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.

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

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

 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.

## NCERT Solutions for Maths Class 7 Chapter 3 Data Handling Topic 3.8.1

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.

Yes, the demand for English books rose faster. It can be seen that the difference in height of the two bars keeps on decreasing.

## NCERT Solutions for Maths Class 7 Chapter 3 Data Handling Exercise 3.1

 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

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

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

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.

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

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.

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.

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 Chapter 3 Maths Data Handling Exercise 3.2

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.

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.

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.

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 .

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.

## NCERT Solutions for Class 7 Chapter 3 Maths Data Handling Exercise 3.3

(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.

(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) 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) 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) 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) 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) 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 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 Chapter 3 Maths Data Handling Exercise 3.4

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.

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

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

$= \frac{1}{6}$

One can choose either Head or Tail.

Therefore, the number of favourable outcomes = 1

We know, Probability of any outcome

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

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

## 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 Difference Purpose
• Chance And Probability

### NCERT Solutions for Class 7 Maths Chapter Wise

 Chapter No. Chapter Name Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangle and its Properties Chapter 7 Congruence of Triangles Chapter 8 Comparing quantities Chapter 9 Rational Numbers Chapter 10 Practical Geometry Chapter 11 Perimeter and Area Chapter 12 Algebraic Expressions Chapter 13 Exponents and Powers Chapter 14 Symmetry Chapter 15 Visualising Solid Shapes

### NCERT Solutions for Class 7 Subject Wise

 NCERT Solutions for Class 7 Maths 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 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 data, arranged in order from lowest to highest, the median gives us the middle observation.

Also Check NCERT Books and NCERT Syllabus here:

1. What is the importance of learning all the concepts covered in the NCERT Solutions for chapter 3 maths class 7?

It is essential to learn all the concepts presented in Chapter 3 of the NCERT Solutions for Class 7 Maths. These concepts are aligned with the exam pattern and model question paper, ensuring that students are well-prepared for their exams. Therefore, acquiring a thorough understanding of all the concepts in NCERT Solutions for chapter 3 maths class 7 is crucial.

2. List out the important topics present in NCERT Solutions for Class 7 Maths Chapter 3.

The NCERT Solutions for Class 7 Maths Chapter 3 encompass the following topics:

1. Data collection
2. Data organization
3. Arithmetic mean
4. Mode
5. Mode of large data sets
6. Median
7. Utilizing bar graphs for different purposes
8. Probability and chance

These concepts hold significant importance from an examination standpoint. They align with the latest CBSE board syllabus and are designed with reference to CBSE question papers and marking schemes.

3. What is the total number of exercises included in the NCERT Solutions for Class 7 Maths Chapter 3?

The NCERT Solutions for Class 7 Maths Chapter 3 consist of a total of 4 exercises. The first exercise comprises 9 questions, the second exercise consists of 5 questions with sub-questions, the third exercise includes 5 questions with sub-questions, and the fourth exercise contains 9 questions focusing on the concepts of data handling.

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