case study class 7 maths data handling

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. It helps the students to understand slowly and to get practice well to become perfect and again a good score in their examination. Below we have listed NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Exercise 3.1, Ex 3.2, Ex 3.3 and Ex 3.4.

These materials are prepared based on Class 7 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 7 Solutions Maths Chapter 3 Data Handling are in accordance with the latest CBSE guidelines and marking schemes

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.1

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.2

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.3

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling Ex 3.4

Class 7 Maths Chapter 3 Data Handling Textbook Solutions

Chapter 3 of NCERT Solutions Class 7 Maths consists of 4 exercises and provided here are the solutions to the questions present in various exercises of the chapter. Enlisted below are the different concepts covered in this chapter.

• Collection of Data
• Organisation of Data
• Arithmetic Mean
• Mode of Large Data
• Use of Bar Graphs with a Different Purpose
• Chance and Probability

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• Important Questions for CBSE Class 7 Maths Chapter 3 - Data Handling

CBSE Class 7 Maths Chapter 3: Data Handling- Important Questions Free PDF Download

Data handling is an important chapter for Class 7 students which has a weightage of around 6-8 marks in their board exams. These PDF solutions for Class 7 Maths Chapter 3 Important Questions are prepared by the experts in order to ease the students' work to look for solutions. Here students can find top-notch solutions prepared according to the NCERT curriculum and easily prepare for their exams.

Free PDF download of Important Questions with solutions for CBSE Class 7 Maths Chapter 3 - Data Handling prepared by expert Mathematics teachers from the latest edition of CBSE(NCERT) books. Register Online for NCERT Solutions Class 7 Science tuition on Vedantu.com to score more marks in CBSE board examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for better solutions can download Class 7 Math s NCERT Solutions to help you to revise the complete syllabus and score more marks in your examinations. 

The main topics from the Data Handling Important Questions For Class 7 Maths Chapter 3 are as follows:

Introduction to Data

Collecting Data

Organisation of Data

Representative Values

 Arithmetic Mean

I. Mode of Large Data

Use of Bar Graphs with a Different Purpose

Choosing a Scale

Drawing Double Bar Graph

Chance and Probability

II. What is Probability?

The PDF also has Class 7 Maths Chapter 3 Extra Questions where students can practice to prepare for their exams with more confidence. These extra questions are carefully designed in such a way that they are unique questions and solutions which test the students' basic understanding of the subject.

Study Important Questions for Class7 Mathematics Chapter 3 – Data Handling

1 Mark 

1. Insert a number between \[\dfrac{1}{4}{\text{ and }}\dfrac{1}{7}{\text{ }}{\text{. }}\]

Ans. To insert a number between two numbers, first we add both numbers and divide the sum by two.

$\dfrac{{\dfrac{1}{4} + \dfrac{1}{7}}}{2} = \dfrac{{\dfrac{{7 + 4}}{{28}}}}{2}$

$= \dfrac{{11}}{{2 \times 28}} = \dfrac{{11}}{{56}}$

2. Give the formula to find mean.

Ans. Formula to find mean is given by, 

${\text{ Mean }} = \dfrac{{{\text{ Sum of all observations }}}}{{{\text{ Number of observations }}}}$

Example: Mean of \[2,4,6,8,10\;\] = $\dfrac{{2 + 4 + 6 + 8 + 10}}{5}\, = \;\dfrac{{30}}{5}\; = \,6$

3. Define Mode. 

Ans . The most frequently occurring value in a data set Is called Mode.

For example, in set \[2,3,4,5,5,5,7,7\]

$5$ occurs $3$ times. Hence $5$ will be the mode of the given data. 

4. Define Average.

Ans. In mathematics, an average is referred to as a mean. It may be obtained by adding the numbers together and then dividing the result by the total number of numbers.

Example average of \[8,10\;\] = $\dfrac{{8 + 10}}{2}\, = \;\dfrac{{18}}{2}\; = \,9$

5. Insert a number between \[-3{\text{ and}}-4{\text{ }}.\]

  $\dfrac{{ - 3 - 4}}{2} = \dfrac{{ - 7}}{2} =  - 3.5$

2 Mark Questions

6. Find the mode for the given set of data : \[1,{\text{ }}2,{\text{ }}3,{\text{ }}5,{\text{ }}6,{\text{ }}7,{\text{ }}2,{\text{ }}1,{\text{ }}4,{\text{ }}1,{\text{ }}6,{\text{ }}1\]

Ans . Given data is \[1,{\text{ }}2,{\text{ }}3,{\text{ }}5,{\text{ }}6,{\text{ }}7,{\text{ }}2,{\text{ }}1,{\text{ }}4,{\text{ }}1,{\text{ }}6,{\text{ }}1\]

Arranging this data in ascending order 

\[{\text{1,1,1,1,2,2,3,4,5,6,6,7,7,8}}\]

$1$  is repeating most frequently that is $4$ times

Therefore, mode of the given data is $1$

7. Find the median of the data \[1,{\text{ }}2,{\text{ }}23,{\text{ }}48,{\text{ }}26,{\text{ }}33,{\text{ }}4\]

Ans. Given data is \[1,{\text{ }}2,{\text{ }}23,{\text{ }}48,{\text{ }}26,{\text{ }}33,{\text{ }}4\]

\[{\text{1,}}\;{\text{2,}}\;{\text{23,}}\;{\text{26,}}\;{\text{35,}}\;{\text{45,}}\;{\text{48}}\]

We know that, median of a data is middle term.

Here, we have a total of $7$ terms. 

And here the middle term is the 4 th   term i.e.  26.

Therefore, the median of the data is 26.

8. Theja studies 8 hours, 6 hours and 2 hours on three consecutive days. How many hours did he study daily? 

Ans.   Given, Theja studied  \[8{\text{ hours}},{\text{ }}6{\text{ hours and 2 hours}}\] .

Theja studied daily = Average of  $8,\;6,\;2$ =  $\dfrac{{{\text{sum}}\;{\text{of}}\;{\text{given}}\;{\text{data}}}}{{{\text{Number}}\;{\text{of}}\;{\text{days}}}}$

$\dfrac{{8 + 6 + 2}}{3}\; = \;\dfrac{{16}}{3}\; = \;5.33\;\;{\text{hours}}$

Theja studied $5.33\;\;{\text{hours}}$ .

9. Answer the following by observing the given data The age’s of cricket players of ICC is as follows: \[21,{\text{ }}23,{\text{ }}25,{\text{ }}26,{\text{ }}32,{\text{ }}34\]

a) What is the age of the eldest players?

b) What is the age of the youngest players in the team?

Ans.   Given data is \[21,{\text{ }}23,{\text{ }}25,{\text{ }}26,{\text{ }}32,{\text{ }}34\]

a) Here the highest term is $34$ .

Therefore, the age of the eldest player is $34\;{\text{years}}$

b) Here the lowest term is $21$ .

Therefore, the age of the youngest player is $21\;{\text{years}}$ .

10. The goals scored by Lakshya in five matches are 3, 1, 2, 3, 1. Find the mean goal scored by her. 

Ans.   Goals scored by Lakshya in five matches = 3, 1, 2, 3, 1

${\text{Mean goal }} = \dfrac{{{\text{Sum of}}\;{\text{goal's}}\;{\text{scored}}\;{\text{in}}\;{\text{all}}\;{\text{matches}}}}{{{\text{ Number of Matches}}}}$

$= \dfrac{{3 + 1 + 2 + 3 + 1}}{5}$

$ = \dfrac{{10}}{5}\; = 2$

Mean goal is $2$ .

11. Find the range of heights of cricket players (in cm) the heights are follows \[158,{\text{ }}150,{\text{ }}156,{\text{ }}162,{\text{ }}155,{\text{ }}152,{\text{ }}154,{\text{ }}164,{\text{ }}153,{\text{ }}165,{\text{ }}160\]

Ans.   Range = Highest value – Lowest Value 

Here, Highest Value = $165$

Lowest Value = $150$

 Range = $165 - 150$ = $15\;{\text{cm}}$

12. Runs scored by a cricket player is in few matches is as follows: \[28,{\text{ }}25,{\text{ }}38,{\text{ }}46,{\text{ }}58,{\text{ }}32\]. Find the average runs. 

Ans.   ${\text{Average}}\;{\text{Runs}}\;{\text{ = }}\dfrac{{{\text{Sum}}\;{\text{of}}\;{\text{all}}\;{\text{runs}}}}{{{\text{Number}}\;{\text{of}}\;{\text{Matches}}}}$

Here Number of Matches = $6$

${\text{Average}}\;{\text{Runs}}\;{\text{ = }}\dfrac{{28 + 25 + 38 + 46 + 58 + 32}}{6}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \;\dfrac{{277}}{6}\;\; = \;37.38$

${\text{Average}}\;{\text{Runs}} = \;37.38\;{\text{Runs}}$

13. Kids borrow books from a library from Monday to Friday. A week value is recorded and is as follows. Find the average \[65,{\text{ }}148,{\text{ }}351,{\text{ }}481,{\text{ }}390.\]

Ans . ${\text{Average}}\;\;{\text{ = }}\dfrac{{{\text{Sum}}\;{\text{of}}\;{\text{all}}\;{\text{books}}\;{\text{issued}}}}{{{\text{Number}}\;{\text{of}}\;{\text{days}}}}\;$

${\text{ = }}\dfrac{{5 + 148\; + 351{\text{  + }}\;481\; + {\text{ }}390.}}{5}$

$= \;\dfrac{{1435}}{5}\;\; = \;287$

14. The score of 10 kids in mathematics shows below. Find mean \[33,{\text{ }}48,{\text{ }}27,{\text{ }}49,{\text{ }}30,{\text{ }}32,{\text{ }}19,{\text{ }}25,{\text{ }}37,{\text{ }}35\]

Ans. ${\text{Mean}}\;\;{\text{ = }}\dfrac{{{\text{Sum}}\;{\text{of}}\;{\text{all}}\;{\text{scores}}}}{{{\text{Number}}\;{\text{of students}}}}\;$

${\text{ = }}\dfrac{{33\; + \;\;48\; + 27\; + \;49\; + \;30\; + {\text{ }}32\; + 19\; + {\text{ }}25\; + {\text{ }}37\; + {\text{ }}35}}{5}$

$= \;\dfrac{{335}}{{10}}\;\; = \;33.5$

Thus, the average score of kids In Mathematics is $33.5$ .

Chapter 3 - Data Handling

Data handling refers to the process of collecting, recording and displaying information, for example in graphs or charts, in a way that is beneficial to others. Data handling is also known as statistics.

Steps in Data Handling

Data collection using a scheduled technique.

Recording data with accuracy and precision.

Data analyzing to draw conclusions.

Sharing data in a way that is helpful to everyone.

Data Collection

Data collection is a term used to describe a process of planning and gathering data. The aim of the collection of data is to obtain the information to be registered, to make decisions on significant issues, to pass on information to others.

Ex: A teacher has assigned students to collect the average temperature data for the month of February. The only thing the teacher wants is Average data.

So, students have to decide at what time daily they have to measure the temperature data. So for these, they have to do proper planning and also they have to gather suitable temperature measuring devices. 

This process of collecting temperature data for 28/29 days in the month of February is called Data collection.

Data organization refers to the organized arrangement of raw data so that the data can be readily interpreted and further statistical treatment is more convenient.

In the data organisation, we will use tables and charts to show the collected data rather than just giving the raw numbers. This is because the person who is collecting data knows more about the raw data representation if we are showing it to someone they have to easily understand the data representation. So we use tables, graphs and charts to represent the data.

The representative value of a set of measurements is the one that we assume is nearest to the measurement's real value. If we carry out our series of calculations, their average will be the indicative value, except those values that we have shown to be far from the true value.

Arithmetic Mean

The arithmetic mean is the simplest and most commonly used measurement system used to calculate the mean or average of a data. It simply involves taking the sum of a number group, then dividing that sum by the number of numbers that are used in the sequence.

Ex: The data of height(cm) of students in the class is given as follows. Calculate the arithmetic mean.

150, 155, 152, 150, 158, 154, 152, 156, 154, 152.

Ans: So to find Arithmetic mean, first we have to calculate the sum of the data.

Sum = 150 + 155 + 152 + 150 + 158 + 154 + 152 + 156 + 154 + 152 = 1533

Total number of students = 10

The arithmetic mean = Sum of heights / Total number of students

Arithmetic mean = 1533 / 10 = 153.3

In Statistics, the difference between the largest and the smallest values in the range of a set of data.

Ex: The data of height(cm) of students in the class is given as follows. Calculate the range.

150, 155, 152, 150, 158, 154, 148, 156, 154, 152

Ans: L To calculate range first we have to find the height of taller students and shorter students in the class. So the taller student height is 158 cm and shorter student height is 148 cm. 

So the range will be the difference between these two heights.

Range = 158 - 148 = 10 cm

The mode is the value that appears in a data set most frequently. A data set can have one mode, more than one mode, or no mode at all.

Ex: The data of height(cm) of students in the class is given as follows. Calculate the mode.

Ans: Here 152 cm is the most frequent height of 3 students. Therefore the Mode of the following height data set will be 152 cm.

In a sorted, ascending or descending, list of numbers, the median is the middle number and may be more representative of that data set than the average.

In comparison to the mean, the median is often used when there are outliers in the series that could distort the average of the values.

If there are odd numbers, the median value is the number in the centre, with numbers below and above the same sum.

The middle pair must be estimated, added together, and divided by two to find the median value if there is an even number in the set.

1. The Data of Height(cm) of Students in the Class is Given as Follows. Calculate the Median Height.

Ans: The total number of students in the class is 10 which is even, so we have added the middle terms i.e 5th and 6th term and divide them by 2 to find the median.

Median = (158 + 154) / 2 = 156 cm.

Therefore the median height of the class is 156 cm.

2. The Weight(Kg) of the Students of the Class is Given as Follows. Calculate the Median Weight.

45, 48, 52, 47, 46, 50, 44

Ans: The total number of students in the class is 7. So the middle term of this weight data set will be our Median weight.

Median = 47 Kg.

Representation of Data Using Bar Graphs

A bar chart or bar graph is a chart or graph that provides rectangular bars with categorical data with heights or lengths proportional to the values they represent. It is possible to plot the bars vertically or horizontally. Sometimes, a vertical bar map is called a column chart.

Ex: The Height Data of a Class is Represented in the Table Below. Plot the Bar Graph to Show a Pictorial Representation of the Data.

Ans: The bar graph representation for the following data is as follows:

(Image will be uploaded soon)

Probability

Probability is the numerical representations of how likely an occurrence is to occur or how likely a proposition is to be valid. The likelihood of an occurrence is a number between 0 and 1, where 0 indicates the impossibility of the event, roughly speaking, and 1 indicates certainty.

Benefits of Referring to the Important Questions of Class 7 Chapter 3 Data Handling

Students must go through the important questions of each chapter of Mathematics, but why? Here are some benefits:

These questions help you to evaluate your knowledge and understanding of any particular chapter.

If you refer to other questions that are not given in your NCERT book, then you will prepare yourself for the examination, where you might be asked some tricky questions.

It also helps you to understand the topic in depth.

These important questions and answers are given in easy-to-understand language and a step-by-step process. 

It helps students’ thinking ability and problem-solving abilities, which enable students to secure good marks not only in the current class but in all academic sessions. 

Vedantu's Important Questions for CBSE Class 7 Maths Chapter 3 - Data Handling offers a comprehensive and valuable resource for students to master this fundamental topic. With a vast array of meticulously curated questions, the platform ensures a thorough understanding of data handling concepts. The well-structured and engaging material enables students to strengthen their analytical and problem-solving skills. By focusing on real-life scenarios and practical applications, Vedantu empowers learners to grasp the relevance of data handling in daily life. With this resource, students can build a strong foundation, instilling confidence to tackle more complex challenges in mathematics. Vedantu's dedication to enhancing learning experiences proves instrumental in nurturing confident and capable mathematicians.

FAQs on Important Questions for CBSE Class 7 Maths Chapter 3 - Data Handling

1. What are the important questions for Chapter 3 of Class 7 Maths?

The important questions for Chapter 3 of Class 7 Maths are the ones that are likely to be asked in the exams. These questions cover the important concepts of data handling, such as collecting data, organizing data, representing data, and interpreting data.

2. How can I prepare for the important questions for Chapter 3 of Class 7 Maths?

You can prepare for the important questions for Chapter 3 of Class 7 Maths by doing the following:

Read the chapter carefully and understand the important concepts.

Solve the practice questions given in the textbook and in the NCERT Solutions.

Practice solving the important questions given in the Vedantu website.

Take a mock test on the important questions.

3. What are the different types of data handling questions?

There are different types of data handling questions, such as:

Questions on collecting data

Questions on organizing data

Questions on representing data

Questions on interpreting data

4. How can I solve the data handling questions?

To solve the data handling questions, you need to understand the following steps:

Collect the data.

Organize the data in a table or a graph.

Represent the data in a suitable way.

Interpret the data.

5. What are the benefits of studying data handling?

Studying data handling has the following benefits:

It helps you to understand how to collect, organize, represent, and interpret data.

It helps you to develop problem-solving skills.

It helps you to make informed decisions.

It helps you to be more competitive in the job market.

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NCERT Solutions Class 7 Maths Chapter 3

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NCERT Solutions Class 7 Mathematics Chapter 3

Ncert solutions for class 7 mathematics chapter 3 data handling .

In this world, data is the fuel that allows us to explore and interpret a wide range of possibilities. The concept of data and its importance are introduced in the NCERT Class 7 Mathematics Chapter 3 data handling. This chapter covers everything from data collection to data storage and display, as well as how to determine the central tendency of the data collected. In Chapter 3, the relevance of the average, median, and mode to various data sets is also discussed, with examples. It also introduces the concept of probability and its theorem, which can be used to determine the likelihood of an event occurring. 

Quick Links

To understand the chapter in a better way, students should practise in-text and end-text exercises.They can solve these questions on their own or refer to NCERT Solutions for Class 7 Mathematics Chapter 3 to get answers to all the questions. Extramarks provides these solutions to help students to strengthen their preparation for the tests, exams or Olympiad .

NCERT Solutions for Class 7 Mathematics Chapter 3  

Chapter 3 Data Handling 

Chapter 3 of Class 7 Mathematics Data handling is a chapter that describes  the procedures for working with data and storing it in various formats. According to the CBSE board, this is an important chapter because it teaches us how to analyse data from various fields. 

NCERT Solutions Class 7 Mathematics Chapter 3 

The answers given in NCERT Solutions for Class 7 Mathematics Chapter 3 cover a variety of important concepts that are necessary for understanding higher-level mathematics. The concept of calculating mean, median, and mode is thoroughly explained, and students will be able to recall it with ease. The average value is referred to as ‘mean,’ the midpoint value is referred to as ‘median,’ and the most common value is referred to as ‘mode.’ 

All the exercises from 3.1 to 3.9 are included in the NCERT Solutions for Class 7 Mathematics Chapter 3 . In this article, you will find detailed solutions prepared by Extramarks experienced subject matter experts. These solutions are provided here in accordance with the NCERT textbook and CBSE guidelines. 

3.1 Introduction 

Students can learn what data is in the introduction to the data handling chapter. Data is simply well-organised information. Students will learn about the various steps in handling specific data, such as collecting the data, properly analysing it, evaluating it to obtain an average, and finally, how to represent the data. 

3.2 Collecting Data 

This section explains how students can learn about data collection from various sources. It also explains how to extract data from it and how to find answers by correlating with the situation. Consider some of the following scenarios: 

• Gujarat  literacy rate-year?
• Bangalore  current weather report
• The number of females in a school  

3.3 Organisation of Data 

Students will learn how to properly organise data collected in this section. Students can learn how to organise and record data in a specific format. NCERT Solution for Class 7 Mathematics Chapter 3 explains all of the different types of data that are used to represent forms.  The tabular format is a widely used and basic format for accurately organising data.  Proper organisation of data is very important since it ensures that data is easy to understand and interpret.

3.4 Representative Values 

Students become aware of these values, which are referred to as representative values, in this lesson. These are used to explain how the data is organised. In the data handling chapter, students must remember the word ‘average.’ The term can be applied to a variety of situations, such as average temperature, average time, average distance, average number of boys, and so on. As a result, students should bear in mind  the approximate  value  rather than an exact value. In NCERT Solutions for Class 7 Mathematics Chapter 3 , it is explained with examples and tables. 

3.5 Arithmetic Mean 

Students will learn about an important aspect of data management in this section. The average value of a group of units is known as the arithmetic mean or mean. It is computed using the formula: 

The sum of a unit’s total values/The total number of units 

The NCERT Solutions contains examples with solutions applying the formula. If they ever get stuck on a question, they can always refer to the solutions prepared by subject matter experts. 

3.5.1 Range 

Students are taught the sub-topic of arithmetic mean. They must take an average for the mean, but must subtract the minimum from the maximum to obtain the range of complete data or observation. 

Students can learn about another important topic known as a mode in this section. The mode, like the mean, is another type of symbolic value. Students must use various types of values depending on the type of data. The mode is a value that appears repeatedly in a set of observations. It means that the value that appears the most times can be considered the mode. 

3.6.1 Mode of Large Data 

We tabulate the large data to find the mode, and the frequency of the data helps us find the mean of the data. Students may be sceptical of the mode. How can the problem be solved if the data is too large? The problem is addressed in this part.

3.7 Median 

Students will learn about yet another type of representative value in this section. The ascending order should be remembered here. Find the exact middle value by writing all the observations in ascending order. It is capable of accurately classifying them into two groups. It’s known as the median because it’s located in the centre. In other words, the Median is the mid-value that divides a unit into two equal halves. 

3.8 Use of Bar Graphs with a Different Purpose 

The use of bar graphs  explores the various applications of the bar graph, including how it can be used to present data on frequency of mode on a graph. Students will learn about the concept of a bar graph and its variations in this lesson. It’s a simple way to represent data in a structured way. 

3.8.1 Choosing a Scale 

After learning about bar graphs, students will be aware of using the scale to draw the bar graph. Students can choose a larger scale if the data is too large, and vice versa. Students can choose one unit that equals 1,10,100, and so on, depending on the values and size. The choosing scale may vary, but the plotting must be on point. 

Drawing Double Bar Graphs 

Students can learn more about bar graphs in this section. NCERT Mathematics Class 7 Chapter 3 clearly explains the concepts when data is given for two or more people to compare. Students must create bar Graphs for two groups, but must differentiate them by shading or colouring. 

3.9 Chance and Probability 

Students must estimate and expect chance and probability in the final section. Simply put, it is the possibility of something occurring. Probability is the term used when chance is defined mathematically. The ratio of favourable cases to the total number of possible cases determines the probability of an event occurring. Each topic is explained separately in NCERT Solutions for Class 7 Mathematics Chapter 3 so that the concept is clearly understood. 

3.9.1 Chance 

Chance is nothing more than a forecast. Students will be able to predict and express the data’s outcome. It is necessary to predict the outcome, whether it is correct or incorrect. This concept is explained clearly with solved examples in these NCERT Solutions . 

3.9.2. Probability 

Probability is the final topic that students can learn in this chapter. Probability is the estimation of the chances of something happening, such as tossing a coin or rolling dice. 

Key Factors of NCERT Solutions for Class 7 Mathematics Chapter 3 

While solving these NCERT Solutions , students gain a variety of benefits, some of   them are listed below: 

• The solutions were created and compiled by subject matter experts , hence their authenticity and accuracy is guaranteed. They are useful for both the teachers and the students, and they can be accessed anywhere without much hassle.  
• The solutions are written in clear, simple language and include easily understandable charts and diagrams to ensure that students fully comprehend the subject. 
• The solutions have been created with the goal of enhancing the students’ potential and achieving better results. 

NCERT Solutions for Class 7 Mathematics 

All textbook sums based on triangles, area, and perimeter of different shapes, data handling, integers, and other topics are covered in NCERT Solutions Class 7 Mathematics. As a result, CBSE students will be able to develop Mathematical knowledge and analytical skills. 

These solutions allow students of  grade 7 to develop  an in-depth understanding of Mathematics. Students from other boards refer to the NCERT Solutions for Class 7 Mathematics. They are reliable and accurate. You can access these chapter-wise solutions from the link given below.

NCERT Solutions for Class 7 

Students who have a good grasp on the answers to all of the questions in the NCERT textbook can ace the exams easily . We have provided NCERT Solutions Class 7 to assist students in finding the most suitable solutions to their problems. Subject experts have written the answers based on the CBSE curriculum. Every question in the most recent NCERT books has 100% accurate step-by-step solutions. NCERT Solutions will help you imbibe your concepts to the core, ensuring that you retain them in the long run. It encourages students to practise daily without getting stressed and to improve their performance significantly. 

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

Let the heights (in cm) of any ten students be 124, 126, 133,135, 136, 139, 140, 142, 144, 147 Highest value among these observations = 147   cm Lowest value among these observations = 124 cm Range = Highest value-Lowest value = ( 147 − 124 ) cm = 23 cm

Q.2 Organise the following marks in a class assessment , in at a bular form .

( i )   Which number is the greatest ? ( ii ) Which number is the lowest ? ( iii ) What is the range of the data ? ( iv ) Find the arithmetic mean .

(i)   9 (ii)   1 ( i i i )   Range=9-1 = 8 ( i v )   Arithmetic   mean= Sum of all observations Total number of observations                                                 = 100 20 = 5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaaMe8Uaaeyo aaqaaiaabIcacaqGPbGaaeyAaiaabMcacaaMe8UaaGymaaqaaiaacI cacaWGPbGaamyAaiaadMgacaGGPaGaaGjbVlaabkfacaqGHbGaaeOB aiaabEgacaqGLbGaaeypaiaabMdacaqGTaGaaeymaiabg2da9maaL4 babaGaaeioaaaaaeaacaGGOaGaamyAaiaadAhacaGGPaGaaGjbVlaa bgeacaqGYbGaaeyAaiaabshacaqGObGaaeyBaiaabwgacaqG0bGaae yAaiaabogacaaMe8UaaeyBaiaabwgacaqGHbGaaeOBaiaab2dadaWc aaqaaiaabofacaqG1bGaaeyBaiaabccacaqGVbGaaeOzaiaabccaca qGHbGaaeiBaiaabYgacaqGGaGaae4BaiaabkgacaqGZbGaaeyzaiaa bkhacaqG2bGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaqGZbaaba Gaaeivaiaab+gacaqG0bGaaeyyaiaabYgacaqGGaGaaeOBaiaabwha caqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaab+gacaqGMbGaaeiiai aab+gacaqGIbGaae4CaiaabwgacaqGYbGaaeODaiaabggacaqG0bGa aeyAaiaab+gacaqGUbGaae4CaaaaaeaacaaMe8UaaGjbVlaaysW7ca aMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaa ysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaG jbVlaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpdaWcaaqaaiaaigda caaIWaGaaGimaaqaaiaaikdacaaIWaaaaiabg2da9maaL4babaGaaG ynaaaaaaaa@C41C@

Q.3 Find the mean of the first five whole numbers.

First five whole numbers are 0, 1, 2, 3 and 4 Since Mean = Sum of all observations Total number of observations So, Mean = 0 + 1 + 2 + 3 + 4 5 = 10 2 = 5 Thus, mean of first five whole number is 5.

Q.4 A cricket scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

Since Mean = Sum of all observations Total number of observations So ,   Mean         = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 8                             = 400 8                             = 50 Thus, the mean score is 50

Q.5 Following table shows the points of each player scored in four game.

Now answer the following questions: (i) find the mean to determine A’ save range number of points scored per game. (ii) To find mean number of points per game of 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) Since Mean = Sum of all observations Total number of observations So,   Mean = 14 + 16 + 10 + 10 4 = 50 4 = 12.5 Thus, A ’ s average number of points scored per game is 12 .5 (ii) Since C played only 3 games, so to find the mean number of points per game for C , divide the total points by 3 . (iii) Mean of B = 0 + 8 + 6 + 4 4 = 18 4 = 4.5 (iv) Since, the mean of A ​   is better and bigger than B and C, so A is the best performer . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcaaeaacaqGtbGa aeyAaiaab6gacaqGJbGaaeyzaiaabccacaqGnbGaaeyzaiaabggaca qGUbGaeyypa0ZaaSaaaeaacaqGtbGaaeyDaiaab2gacaqGGaGaae4B aiaabAgacaqGGaGaaeyyaiaabYgacaqGSbGaaeiiaiaab+gacaqGIb Gaae4CaiaabwgacaqGYbGaaeODaiaabggacaqG0bGaaeyAaiaab+ga caqGUbGaae4CaaqaaiaabsfacaqGVbGaaeiDaiaabggacaqGSbGaae iiaiaab6gacaqG1bGaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqG VbGaaeOzaiaabccacaqGVbGaaeOyaiaabohacaqGLbGaaeOCaiaabA hacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaaaGaaeiiaaqa aiaabofacaqGVbGaaeilaiaaysW7caqGnbGaaeyzaiaabggacaqGUb Gaeyypa0ZaaSaaaeaacaaIXaGaaGinaiabgUcaRiaaigdacaaI2aGa ey4kaSIaaGymaiaaicdacqGHRaWkcaaIXaGaaGimaaqaaiaaisdaaa aabaGaeyypa0ZaaSaaaeaacaaI1aGaaGimaaqaaiaaisdaaaGaeyyp a0ZaauIhaeaacaaIXaGaaGOmaiaac6cacaaI1aaaaaqaaiaabsfaca qGObGaaeyDaiaabohacaqGSaGaaeiiaiaabgeacaGGzaIaae4Caiaa bccacaqGHbGaaeODaiaabwgacaqGYbGaaeyyaiaabEgacaqGLbGaae iiaiaab6gacaqG1bGaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqG VbGaaeOzaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabo hacaqGGaGaae4CaiaabogacaqGVbGaaeOCaiaabwgacaqGKbGaaeii aiaabchacaqGLbGaaeOCaiaabccacaqGNbGaaeyyaiaab2gacaqGLb GaaeiiaiaabMgacaqGZbGaaeiiaiaabgdacaqGYaGaaeOlaiaabwda aeaacaqGOaGaaeyAaiaabMgacaqGPaaabaGaae4uaiaabMgacaqGUb Gaae4yaiaabwgacaqGGaGaae4qaiaabccacaqGWbGaaeiBaiaabgga caqG5bGaaeyzaiaabsgacaqGGaGaae4Baiaab6gacaqGSbGaaeyEai aabccacaqGZaGaaeiiaiaabEgacaqGHbGaaeyBaiaabwgacaqGZbGa aeilaiaabccacaqGZbGaae4BaiaabccacaqG0bGaae4Baiaabccaca qGMbGaaeyAaiaab6gacaqGKbGaaeiiaiaabshacaqGObGaaeyzaiaa bccacaqGTbGaaeyzaiaabggacaqGUbGaaeiiaiaab6gacaqG1bGaae yBaiaabkgacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaaqaaiaabcha caqGVbGaaeyAaiaab6gacaqG0bGaae4CaiaabccacaqGWbGaaeyzai aabkhacaqGGaGaae4zaiaabggacaqGTbGaaeyzaiaabccacaqGMbGa ae4BaiaabkhacaqGGaGaae4qaiaacYcacaqGKbGaaeyAaiaabAhaca qGPbGaaeizaiaabwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaa bshacaqGVbGaaeiDaiaabggacaqGSbGaaeiiaiaabchacaqGVbGaae yAaiaab6gacaqG0bGaae4CaiaabccacaqGIbGaaeyEaiaabccacaqG ZaGaaeOlaaqaaiaabIcacaqGPbGaaeyAaiaabMgacaqGPaaabaGaae ytaiaabwgacaqGHbGaaeOBaiaabccacaqGVbGaaeOzaiaabccacaqG cbGaaeiiaiaab2dacaqGGaWaaSaaaeaacaaIWaGaey4kaSIaaGioai abgUcaRiaaiAdacqGHRaWkcaaI0aaabaGaaGinaaaacqGH9aqpdaWc aaqaaiaaigdacaaI4aaabaGaaGinaaaacqGH9aqpcaaI0aGaaiOlai aaiwdaaeaacaqGOaGaaeyAaiaabAhacaqGPaaabaGaae4uaiaabMga caqGUbGaae4yaiaabwgacaqGSaGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGTbGaaeyzaiaabggacaqGUbGaaeiiaiaab+gacaqGMbGa aeiiaiaabgeacaaMb8UaaGjbVlaabMgacaqGZbGaaeiiaiaabkgaca qGLbGaaeiDaiaabshacaqGLbGaaeOCaiaabccacaqGHbGaaeOBaiaa bsgacaqGGaGaaeOyaiaabMgacaqGNbGaae4zaiaabwgacaqGYbGaae iiaiaabshacaqGObGaaeyyaiaab6gacaqGGaGaaeOqaiaabccacaqG HbGaaeOBaiaabsgacaqGGaGaae4qaiaabYcacaqGGaGaae4Caiaab+ gacaqGGaGaaeyqaiaabccacaqGPbGaae4CaaqaaiaabshacaqGObGa aeyzaiaabccacaqGIbGaaeyzaiaabohacaqG0bGaaeiiaiaabchaca qGLbGaaeOCaiaabAgacaqGVbGaaeOCaiaab2gacaqGLbGaaeOCaiaa b6caaaaa@8996@

Q.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.

(i) Clearly, the from the given data, the highest and the lowest marks are 95 and 39 respectively . (ii) Range = Highest − Lowest = 95 − 39 = 56 (iii) ​   Mean =   Sum of all observations Total number of observations                         = 85 + 76 + 90 + 58 + 39 + 48 + 56 + 95 + 81 + 75 10                         = 70.3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaae4q aiaabYgacaqGLbGaaeyyaiaabkhacaqGSbGaaeyEaiaabYcacaqGGa GaaeiDaiaabIgacaqGLbGaaeiiaiaabAgacaqGYbGaae4Baiaab2ga caqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabEgacaqGPbGaaeODai aabwgacaqGUbGaaeiiaiaabsgacaqGHbGaaeiDaiaabggacaqGSaGa aeiiaiaabshacaqGObGaaeyzaiaabccacaqGObGaaeyAaiaabEgaca qGObGaaeyzaiaabohacaqG0bGaaeiiaiaabggacaqGUbGaaeizaiaa bccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiBaiaab+gacaqG3bGaae yzaiaabohacaqG0baabaGaaeyBaiaabggacaqGYbGaae4Aaiaaboha caqGGaGaaeyyaiaabkhacaqGLbGaaeiiaiaabMdacaqG1aGaaeiiai aabggacaqGUbGaaeizaiaabccacaqGZaGaaeyoaiaabccacaqGYbGa aeyzaiaabohacaqGWbGaaeyzaiaabogacaqG0bGaaeyAaiaabAhaca qGLbGaaeiBaiaabMhacaqGUaaabaGaaeikaiaabMgacaqGPbGaaeyk aiaabccacaqGsbGaaeyyaiaab6gacaqGNbGaaeyzaiabg2da9iaabI eacaqGPbGaae4zaiaabIgacaqGLbGaae4CaiaabshacqGHsislcaqG mbGaae4BaiaabEhacaqGLbGaae4CaiaabshaaeaacqGH9aqpcaaI5a GaaGynaiabgkHiTiaaiodacaaI5aGaeyypa0JaaGynaiaaiAdaaeaa caqGOaGaaeyAaiaabMgacaqGPbGaaeykaiaaygW7caaMe8Uaaeytai aabwgacaqGHbGaaeOBaiabg2da9iaaysW7daWcaaqaaiaabofacaqG 1bGaaeyBaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeiBaiaabY gacaqGGaGaae4BaiaabkgacaqGZbGaaeyzaiaabkhacaqG2bGaaeyy aiaabshacaqGPbGaae4Baiaab6gacaqGZbaabaGaaeivaiaab+gaca qG0bGaaeyyaiaabYgacaqGGaGaaeOBaiaabwhacaqGTbGaaeOyaiaa bwgacaqGYbGaaeiiaiaab+gacaqGMbGaaeiiaiaab+gacaqGIbGaae 4CaiaabwgacaqGYbGaaeODaiaabggacaqG0bGaaeyAaiaab+gacaqG UbGaae4CaaaacaqGGaaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaays W7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Uaeyyp a0ZaaSaaaeaacaaI4aGaaGynaiabgUcaRiaaiEdacaaI2aGaey4kaS IaaGyoaiaaicdacqGHRaWkcaaI1aGaaGioaiabgUcaRiaaiodacaaI 5aGaey4kaSIaaGinaiaaiIdacqGHRaWkcaaI1aGaaGOnaiabgUcaRi aaiMdacaaI1aGaey4kaSIaaGioaiaaigdacqGHRaWkcaaI3aGaaGyn aaqaaiaaigdacaaIWaaaaaqaaiaaysW7caaMe8UaaGjbVlaaysW7ca aMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlab g2da9maaL4babaGaaG4naiaaicdacaGGUaGaaG4maaaaaaaa@2CFB@

Q.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.

Mean = Sum of all observations Total number of observations = 1555 + 1670 + 1750 + 2013 + 2540 + 2820 6 = 12348 6 = 2058 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaabbeaacaqGnbGaaeyzaiaabggacaqGUbGaeyyp a0ZaaSaaaeaacaqGtbGaaeyDaiaab2gacaqGGaGaae4BaiaabAgaca qGGaGaaeyyaiaabYgacaqGSbGaaeiiaiaab+gacaqGIbGaae4Caiaa bwgacaqGYbGaaeODaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaae 4CaaqaaiaabsfacaqGVbGaaeiDaiaabggacaqGSbGaaeiiaiaab6ga caqG1bGaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqGVbGaaeOzai aabccacaqGVbGaaeOyaiaabohacaqGLbGaaeOCaiaabAhacaqGHbGa aeiDaiaabMgacaqGVbGaaeOBaiaabohaaaaabaGaeyypa0ZaaSaaae aacaaIXaGaaGynaiaaiwdacaaI1aGaey4kaSIaaGymaiaaiAdacaaI 3aGaaGimaiabgUcaRiaaigdacaaI3aGaaGynaiaaicdacqGHRaWkca aIYaGaaGimaiaaigdacaaIZaGaey4kaSIaaGOmaiaaiwdacaaI0aGa aGimaiabgUcaRiaaikdacaaI4aGaaGOmaiaaicdaaeaacaaI2aaaaa qaaiabg2da9maalaaabaGaaGymaiaaikdacaaIZaGaaGinaiaaiIda aeaacaaI2aaaaaqaaiabg2da9maaL4babaGaaGOmaiaaicdacaaI1a GaaGioaaaaaaaa@9170@

Q.8 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.

Q.9 The rain fall (in mm) in a city on 7 days of a certain week was recorded as follows:

(i ) F i n d t h e r a n g e o f t h e r a in f a l l i n t h e a b o v e d a t a . ( i i ) F i n d t h e m e a n r a i n f a l l f o r t h e w e e k . ( i i i ) O n h o w m a n y d a y s w a s t h e r a i n f a l l l e s s t h a n t h e m e a n r a i n f a l l .

(i) Range = Highest-Lowest = 20. 5 mm – 0. 0 mm = 20.5   mm (ii)   Mean rainfall = 0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0 7                                           = 41.3 7 = 5.9   mm (iii)   For five days   (Monday, Wednesday, Thursday, Saturday         and Sunday), rainfall was less than the mean rainfall . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeOu aiaabggacaqGUbGaae4zaiaabwgacqGH9aqpcaqGibGaaeyAaiaabE gacaqGObGaaeyzaiaabohacaqG0bGaaeylaiaabYeacaqGVbGaae4D aiaabwgacaqGZbGaaeiDaaqaaiabg2da9iaaikdacaaIWaGaaiOlai aabwdacaqGGaGaaeyBaiaab2gacaGGTaGaaGimaiaac6cacaqGWaGa aeiiaiaab2gacaqGTbGaeyypa0ZaauIhaeaacaaIYaGaaGimaiaac6 cacaaI1aaaaiaaysW7caqGTbGaaeyBaaqaaiaabIcacaqGPbGaaeyA aiaabMcacaaMe8UaaeytaiaabwgacaqGHbGaaeOBaiaabccacaqGYb GaaeyyaiaabMgacaqGUbGaaeOzaiaabggacaqGSbGaaeiBaiaabcca cqGH9aqpcaqGGaWaaSaaaeaacaaIWaGaaiOlaiaaicdacqGHRaWkca aIXaGaaGOmaiaac6cacaaIYaGaey4kaSIaaGOmaiaac6cacaaIXaGa ey4kaSIaaGimaiaac6cacaaIWaGaey4kaSIaaGOmaiaaicdacaGGUa GaaGynaiabgUcaRiaaiwdacaGGUaGaaGynaiabgUcaRiaaigdacaGG UaGaaGimaaqaaiaaiEdaaaaabaGaaGjbVlaaysW7caaMe8UaaGjbVl aaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8Ua aGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7ca aMe8Uaeyypa0ZaaSaaaeaacaaI0aGaaGymaiaac6cacaaIZaaabaGa aG4naaaacqGH9aqpcaaI1aGaaiOlaiaaiMdacaaMe8UaaeyBaiaab2 gaaeaacaqGOaGaaeyAaiaabMgacaqGPbGaaeykaiaaysW7caqGgbGa ae4BaiaabkhacaqGGaGaaeOzaiaabMgacaqG2bGaaeyzaiaabccaca qGKbGaaeyyaiaabMhacaqGZbGaaGjbVlaabIcacaqGnbGaae4Baiaa b6gacaqGKbGaaeyyaiaabMhacaqGSaGaaeiiaiaabEfacaqGLbGaae izaiaab6gacaqGLbGaae4CaiaabsgacaqGHbGaaeyEaiaabYcacaqG GaGaaeivaiaabIgacaqG1bGaaeOCaiaabohacaqGKbGaaeyyaiaabM hacaqGSaGaaeiiaiaabofacaqGHbGaaeiDaiaabwhacaqGYbGaaeiz aiaabggacaqG5baabaGaaGjbVlaaysW7caaMe8UaaGjbVlaabccaca qGHbGaaeOBaiaabsgacaqGGaGaae4uaiaabwhacaqGUbGaaeizaiaa bggacaqG5bGaaeykaiaabYcacaqGGaGaaeOCaiaabggacaqGPbGaae OBaiaabAgacaqGHbGaaeiBaiaabYgacaqGGaGaae4DaiaabggacaqG ZbGaaeiiaiaabYgacaqGLbGaae4CaiaabohacaqGGaGaaeiDaiaabI gacaqGHbGaaeOBaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyB aiaabwgacaqGHbGaaeOBaiaabccacaqGYbGaaeyyaiaabMgacaqGUb GaaeOzaiaabggacaqGSbGaaeiBaiaab6caaaaa@2482@

Q.10 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

(i) The height of the tallest girl is 151 cm . (ii) The height of the shortest girl is 128 . (iii) Range = Highest − Lowest = ( 151 − 128 ) cm = 23   cm (iv) Mean height = 135 + 150 + 139 + 128 + 151 + 132 + 146 + 149 + 143 + 141 10 = 1414 10 = 141.4   cm Thus, five girls have more height than the mean height . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeiv aiaabIgacaqGLbGaaeiiaiaabIgacaqGLbGaaeyAaiaabEgacaqGOb GaaeiDaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwga caqGGaGaaeiDaiaabggacaqGSbGaaeiBaiaabwgacaqGZbGaaeiDai aabccacaqGNbGaaeyAaiaabkhacaqGSbGaaeiiaiaabMgacaqGZbGa aeiiaiaabgdacaqG1aGaaeymaiaabccacaqGJbGaaeyBaiaab6caae aacaqGOaGaaeyAaiaabMgacaqGPaGaaeiiaiaabsfacaqGObGaaeyz aiaabccacaqGObGaaeyzaiaabMgacaqGNbGaaeiAaiaabshacaqGGa Gaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaaboha caqGObGaae4BaiaabkhacaqG0bGaaeyzaiaabohacaqG0bGaaeiiai aabEgacaqGPbGaaeOCaiaabYgacaqGGaGaaeyAaiaabohacaqGGaGa aeymaiaabkdacaqG4aGaaeOlaaqaaiaabIcacaqGPbGaaeyAaiaabM gacaqGPaGaaeiiaiaabkfacaqGHbGaaeOBaiaabEgacaqGLbGaeyyp a0JaaeisaiaabMgacaqGNbGaaeiAaiaabwgacaqGZbGaaeiDaiabgk HiTiaabYeacaqGVbGaae4DaiaabwgacaqGZbGaaeiDaiabg2da9maa bmaabaGaaGymaiaaiwdacaaIXaGaeyOeI0IaaGymaiaaikdacaaI4a aacaGLOaGaayzkaaGaae4yaiaab2gacqGH9aqpdaqjEaqaaiaaikda caaIZaGaaGjbVlaabogacaqGTbaaaaqaaiaabIcacaqGPbGaaeODai aabMcacaqGGaGaaeytaiaabwgacaqGHbGaaeOBaiaabccacaqGObGa aeyzaiaabMgacaqGNbGaaeiAaiaabshaaeaacqGH9aqpdaWcaaqaai aaigdacaaIZaGaaGynaiabgUcaRiaaigdacaaI1aGaaGimaiabgUca RiaaigdacaaIZaGaaGyoaiabgUcaRiaaigdacaaIYaGaaGioaiabgU caRiaaigdacaaI1aGaaGymaiabgUcaRiaaigdacaaIZaGaaGOmaiab gUcaRiaaigdacaaI0aGaaGOnaiabgUcaRiaaigdacaaI0aGaaGyoai abgUcaRiaaigdacaaI0aGaaG4maiabgUcaRiaaigdacaaI0aGaaGym aaqaaiaaigdacaaIWaaaaaqaaiabg2da9maalaaabaGaaGymaiaais dacaaIXaGaaGinaaqaaiaaigdacaaIWaaaaaqaaiabg2da9maaL4ba baGaaGymaiaaisdacaaIXaGaaiOlaiaaisdacaaMe8Uaae4yaiaab2 gaaaaabaGaaeivaiaabIgacaqG1bGaae4CaiaabYcacaqGGaGaaeOz aiaabMgacaqG2bGaaeyzaiaabccacaqGNbGaaeyAaiaabkhacaqGSb Gaae4CaiaabccacaqGObGaaeyyaiaabAhacaqGLbGaaeiiaiaab2ga caqGVbGaaeOCaiaabwgacaqGGaGaaeiAaiaabwgacaqGPbGaae4zai aabIgacaqG0bGaaeiiaiaabshacaqGObGaaeyyaiaab6gacaqGGaGa aeiDaiaabIgacaqGLbGaaeiiaiaab2gacaqGLbGaaeyyaiaab6gaca qGGaGaaeiAaiaabwgacaqGPbGaae4zaiaabIgacaqG0bGaaeOlaaaa aa@1B15@

Q.11 The scores in mathmatics 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?

Arranging the score in ascending order. 5,9,10,12,15,16,19,20,20,20,20,23,24,25,25 Since, mode is that value of observation which occurs for the most number of time and median is the middle observation. There are 15 values, so median would be the 8th observation and 20 occurs 4 times. So, Mode = 20 and Median = 20 Yes both are same.

Q.12 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?

Arranging in ascending order to get 6,8,10,10,15,15,15,50,80,100,120 Mean = 6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120 11     = 429 11 = 39 Mode is that value of observation which occurs for the most number of time and median is the middle observation . There are 11 values, so median would be the 6th observation and 15 occurs 3 times . So, Mode = 15   and Median = 15 No, these three are not same . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGbbGaaeOCaiaabkhacaqGHbGaaeOB aiaabEgacaqGPbGaaeOBaiaabEgacaqGGaGaaeyAaiaab6gacaqGGa GaaeyyaiaabohacaqGJbGaaeyzaiaab6gacaqGKbGaaeyAaiaab6ga caqGNbGaaeiiaiaab+gacaqGYbGaaeizaiaabwgacaqGYbGaaeiiai aabshacaqGVbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaabAdacaqG SaGaaeioaiaabYcacaqGXaGaaeimaiaabYcacaqGXaGaaeimaiaabY cacaqGXaGaaeynaiaabYcacaqGXaGaaeynaiaabYcacaqGXaGaaeyn aiaabYcacaqG1aGaaeimaiaabYcacaqG4aGaaeimaiaabYcacaqGXa GaaeimaiaabcdacaqGSaGaaeymaiaabkdacaqGWaaabaGaaeytaiaa bwgacaqGHbGaaeOBaiabg2da9maalaaabaGaaGOnaiabgUcaRiaaiI dacqGHRaWkcaaIXaGaaGimaiabgUcaRiaaigdacaaIWaGaey4kaSIa aGymaiaaiwdacqGHRaWkcaaIXaGaaGynaiabgUcaRiaaigdacaaI1a Gaey4kaSIaaGynaiaaicdacqGHRaWkcaaI4aGaaGimaiabgUcaRiaa igdacaaIWaGaaGimaiabgUcaRiaaigdacaaIYaGaaGimaaqaaiaaig dacaaIXaaaaaqaaiaaxMaacaaMe8UaaGjbVlabg2da9maalaaabaGa aGinaiaaikdacaaI5aaabaGaaGymaiaaigdaaaGaeyypa0ZaauIhae aacaaIZaGaaGyoaaaaaeaacaqGnbGaae4BaiaabsgacaqGLbGaaeii aiaabMgacaqGZbGaaeiiaiaabshacaqGObGaaeyyaiaabshacaqGGa GaaeODaiaabggacaqGSbGaaeyDaiaabwgacaqGGaGaae4BaiaabAga caqGGaGaae4BaiaabkgacaqGZbGaaeyzaiaabkhacaqG2bGaaeyyai aabshacaqGPbGaae4Baiaab6gacaqGGaGaae4DaiaabIgacaqGPbGa ae4yaiaabIgacaqGGaGaae4BaiaabogacaqGJbGaaeyDaiaabkhaca qGZbGaaeiiaiaabAgacaqGVbGaaeOCaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaGaaeyBaiaab+gacaqGZbGaaeiDaaqaaiaab6gacaqG1b GaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabcca caqG0bGaaeyAaiaab2gacaqGLbGaaeiiaiaabggacaqGUbGaaeizai aabccacaqGTbGaaeyzaiaabsgacaqGPbGaaeyyaiaab6gacaqGGaGa aeyAaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaab2gaca qGPbGaaeizaiaabsgacaqGSbGaaeyzaiaabccacaqGVbGaaeOyaiaa bohacaqGLbGaaeOCaiaabAhacaqGHbGaaeiDaiaabMgacaqGVbGaae OBaiaab6caaeaacaqGubGaaeiAaiaabwgacaqGYbGaaeyzaiaabcca caqGHbGaaeOCaiaabwgacaqGGaGaaeymaiaabgdacaqGGaGaaeODai aabggacaqGSbGaaeyDaiaabwgacaqGZbGaaeilaiaabccacaqGZbGa ae4BaiaabccacaqGTbGaaeyzaiaabsgacaqGPbGaaeyyaiaab6gaca qGGaGaae4Daiaab+gacaqG1bGaaeiBaiaabsgacaqGGaGaaeOyaiaa bwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabAdacaqG0bGaae iAaiaabccacaqGVbGaaeOyaiaabohacaqGLbGaaeOCaiaabAhacaqG HbGaaeiDaiaabMgacaqGVbGaaeOBaaqaaiaabggacaqGUbGaaeizai aabccacaqGXaGaaeynaiaabccacaqGVbGaae4yaiaabogacaqG1bGa aeOCaiaabohacaqGGaGaae4maiaabccacaqG0bGaaeyAaiaab2gaca qGLbGaae4Caiaab6caaeaacaqGtbGaae4BaiaabYcacaqGGaWaauIh aeaacaqGnbGaae4BaiaabsgacaqGLbGaeyypa0Jaaeymaiaabwdaca aMe8Uaaeyyaiaab6gacaqGKbGaaeiiaiaab2eacaqGLbGaaeizaiaa bMgacaqGHbGaaeOBaiabg2da9iaabgdacaqG1aaaaaqaaiaab6eaca qGVbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGZbGaaeyzaiaa bccacaqG0bGaaeiAaiaabkhacaqGLbGaaeyzaiaabccacaqGHbGaae OCaiaabwgacaqGGaGaaeOBaiaab+gacaqG0bGaaeiiaiaabohacaqG HbGaaeyBaiaabwgacaqGUaaaaaa@7346@

Q.13 The weights  inkg. 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?

Arranging the weight in ascending order . 32,35,36,37,38,38,38,40,42,43,43,43,45,47,50 Mode of given data is that value of observation which occurs for the most number of time and median is the middle observation . There are 15 values, so median would be the 8th observation So, Median = 40 Also 38 and 43 both occurs 3 times . So, Mode = 38   and 43 Yes there are two mode for given data . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGbbGaaeOCaiaabkhacaqGHbGaaeOB aiaabEgacaqGPbGaaeOBaiaabEgacaqGGaGaaeiDaiaabIgacaqGLb GaaeiiaiaabEhacaqGLbGaaeyAaiaabEgacaqGObGaaeiDaiaabcca caqGPbGaaeOBaiaabccacaqGHbGaae4CaiaabogacaqGLbGaaeOBai aabsgacaqGPbGaaeOBaiaabEgacaqGGaGaae4BaiaabkhacaqGKbGa aeyzaiaabkhacaqGUaaabaGaae4maiaabkdacaqGSaGaae4maiaabw dacaqGSaGaae4maiaabAdacaqGSaGaae4maiaabEdacaqGSaGaae4m aiaabIdacaqGSaGaae4maiaabIdacaqGSaGaae4maiaabIdacaqGSa GaaeinaiaabcdacaqGSaGaaeinaiaabkdacaqGSaGaaeinaiaaboda caqGSaGaaeinaiaabodacaqGSaGaaeinaiaabodacaqGSaGaaeinai aabwdacaqGSaGaaeinaiaabEdacaqGSaGaaeynaiaabcdaaeaacaqG nbGaae4BaiaabsgacaqGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabE gacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabsgacaqGHbGaaeiD aiaabggacaqGGaGaaeyAaiaabohacaqGGaGaaeiDaiaabIgacaqGHb GaaeiDaiaabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzaiaabcca caqGVbGaaeOzaiaabccacaqGVbGaaeOyaiaabohacaqGLbGaaeOCai aabAhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqG3bGa aeiAaiaabMgacaqGJbGaaeiAaiaabccacaqGVbGaae4yaiaabogaca qG1bGaaeOCaiaabohacaqGGaGaaeOzaiaab+gacaqGYbaabaGaaeiD aiaabIgacaqGLbGaaeiiaiaab2gacaqGVbGaae4CaiaabshacaqGGa GaaeOBaiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaab+ga caqGMbGaaeiiaiaabshacaqGPbGaaeyBaiaabwgacaqGGaGaaeyyai aab6gacaqGKbGaaeiiaiaab2gacaqGLbGaaeizaiaabMgacaqGHbGa aeOBaiaabccacaqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwgaca qGGaGaaeyBaiaabMgacaqGKbGaaeizaiaabYgacaqGLbGaaeiiaiaa b+gacaqGIbGaae4CaiaabwgacaqGYbGaaeODaiaabggacaqG0bGaae yAaiaab+gacaqGUbGaaeOlaaqaaiaabsfacaqGObGaaeyzaiaabkha caqGLbGaaeiiaiaabggacaqGYbGaaeyzaiaabccacaqGXaGaaeynai aabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzaiaabohacaqGSaGa aeiiaiaabohacaqGVbGaaeiiaiaab2gacaqGLbGaaeizaiaabMgaca qGHbGaaeOBaiaabccacaqG3bGaae4BaiaabwhacaqGSbGaaeizaiaa bccacaqGIbGaaeyzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae ioaiaabshacaqGObGaaeiiaiaab+gacaqGIbGaae4CaiaabwgacaqG YbGaaeODaiaabggacaqG0bGaaeyAaiaab+gacaqGUbaabaGaae4uai aab+gacaqGSaGaaeiiaiaab2eacaqGLbGaaeizaiaabMgacaqGHbGa aeOBaiabg2da9maaL4babaGaaeinaiaabcdaaaaabaGaaeyqaiaabY gacaqGZbGaae4BaiaabccacaqGZaGaaeioaiaabccacaqGHbGaaeOB aiaabsgacaqGGaGaaeinaiaabodacaqGGaGaaeOyaiaab+gacaqG0b GaaeiAaiaabccacaqGVbGaae4yaiaabogacaqG1bGaaeOCaiaaboha caqGGaGaae4maiaabccacaqG0bGaaeyAaiaab2gacaqGLbGaae4Cai aab6caaeaacaqGtbGaae4BaiaabYcacaqGGaWaauIhaeaacaqGnbGa ae4BaiaabsgacaqGLbGaeyypa0Jaae4maiaabIdacaaMe8Uaaeyyai aab6gacaqGKbGaaeiiaiaabsdacaqGZaaaaaqaaiaabMfacaqGLbGa ae4CaiaabccacaqG0bGaaeiAaiaabwgacaqGYbGaaeyzaiaabccaca qGHbGaaeOCaiaabwgacaqGGaGaaeiDaiaabEhacaqGVbGaaeiiaiaa b2gacaqGVbGaaeizaiaabwgacaqGGaGaaeOzaiaab+gacaqGYbGaae iiaiaabEgacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabsgacaqG HbGaaeiDaiaabggacaqGUaaaaaa@7978@

Q.14 Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14

Arrange in ascending order to get: 12,12,13,13,14,14,1416,19 Mode of given data is that value of observation which occurs for the most number of time and median is the middle observation . There are 9 values, so median would be the 5th observation So, Median = 14 Also 14 occurs 3 times . So, Mode = 14 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGbbGaaeOCaiaabkhacaqGHbGaaeOB aiaabEgacaqGLbGaaeiiaiaabMgacaqGUbGaaeiiaiaabggacaqGZb Gaae4yaiaabwgacaqGUbGaaeizaiaabMgacaqGUbGaae4zaiaabcca caqGVbGaaeOCaiaabsgacaqGLbGaaeOCaiaabccacaqG0bGaae4Bai aabccacaqGNbGaaeyzaiaabshacaqG6aaabaGaaeymaiaabkdacaqG SaGaaeymaiaabkdacaqGSaGaaeymaiaabodacaqGSaGaaeymaiaabo dacaqGSaGaaeymaiaabsdacaqGSaGaaeymaiaabsdacaqGSaGaaeym aiaabsdacaqGXaGaaeOnaiaabYcacaqGXaGaaeyoaaqaaiaab2eaca qGVbGaaeizaiaabwgacaqGGaGaae4BaiaabAgacaqGGaGaae4zaiaa bMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeizaiaabggacaqG0bGaae yyaiaabccacaqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabggacaqG 0bGaaeiiaiaabAhacaqGHbGaaeiBaiaabwhacaqGLbGaaeiiaiaab+ gacaqGMbGaaeiiaiaab+gacaqGIbGaae4CaiaabwgacaqGYbGaaeOD aiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabEhacaqGOb GaaeyAaiaabogacaqGObGaaeiiaiaab+gacaqGJbGaae4yaiaabwha caqGYbGaae4CaiaabccacaqGMbGaae4BaiaabkhaaeaacaqG0bGaae iAaiaabwgacaqGGaGaaeyBaiaab+gacaqGZbGaaeiDaiaabccacaqG UbGaaeyDaiaab2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabA gacaqGGaGaaeiDaiaabMgacaqGTbGaaeyzaiaabccacaqGHbGaaeOB aiaabsgacaqGGaGaaeyBaiaabwgacaqGKbGaaeyAaiaabggacaqGUb GaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqGObGaaeyzaiaabcca caqGTbGaaeyAaiaabsgacaqGKbGaaeiBaiaabwgacaqGGaGaae4Bai aabkgacaqGZbGaaeyzaiaabkhacaqG2bGaaeyyaiaabshacaqGPbGa ae4Baiaab6gacaqGUaaabaGaaeivaiaabIgacaqGLbGaaeOCaiaabw gacaqGGaGaaeyyaiaabkhacaqGLbGaaeiiaiaabMdacaqGGaGaaeOD aiaabggacaqGSbGaaeyDaiaabwgacaqGZbGaaeilaiaabccacaqGZb Gaae4BaiaabccacaqGTbGaaeyzaiaabsgacaqGPbGaaeyyaiaab6ga caqGGaGaae4Daiaab+gacaqG1bGaaeiBaiaabsgacaqGGaGaaeOyai aabwgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabwdacaqG0bGa aeiAaiaabccacaqGVbGaaeOyaiaabohacaqGLbGaaeOCaiaabAhaca qGHbGaaeiDaiaabMgacaqGVbGaaeOBaaqaaiaabofacaqGVbGaaeil aiaabccacaqGnbGaaeyzaiaabsgacaqGPbGaaeyyaiaab6gacqGH9a qpdaqjEaqaaiaabgdacaqG0aaaaaqaaiaabgeacaqGSbGaae4Caiaa b+gacaqGGaGaaeymaiaabsdacaqGGaGaae4BaiaabogacaqGJbGaae yDaiaabkhacaqGZbGaaeiiaiaabodacaqGGaGaaeiDaiaabMgacaqG TbGaaeyzaiaabohacaqGUaaabaGaae4uaiaab+gacaqGSaGaaeiiam aaL4babaGaaeytaiaab+gacaqGKbGaaeyzaiabg2da9iaabgdacaqG 0aaaaaaaaa@349E@

T e l l w h e t h e r t h e s t a t e m e n t i s t r u e o r f a l s e : ( i ) T h e m o d e i s a l w a y s o n e o f t h e n u m b e r s i n a d a t a . ( i i ) T h e m e a n c a n b e o n e o f t h e n u m b e r s i n a d a t a . ( i i i ) T h e m e d i a n i s a l w a y s o n e o f t h e n u m b e r s i n a d a t a . ( i v ) A d a t a a l w a y s h a s a m o d e . ( v ) T h e d a t a 6 , 4 , 3 , 8 , 9 , 1 2 , 1 3 , 9 h a s m e a n 9 . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaaieqacaWFubGaa8xzaiaa=XgacaWFSbGa a8hiaiaa=DhacaWFObGaa8xzaiaa=rhacaWFObGaa8xzaiaa=jhaca WFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=nhacaWF0bGaa8xyaiaa =rhacaWFLbGaa8xBaiaa=vgacaWFUbGaa8hDaiaa=bcacaWFPbGaa8 3Caiaa=bcacaWF0bGaa8NCaiaa=vhacaWFLbGaa8hiaiaa=9gacaWF YbGaa8hiaiaa=zgacaWFHbGaa8hBaiaa=nhacaWFLbGaa8Noaaqaai aa=HcacaWFPbGaa8xkaiaa=bcacaWFubGaa8hAaiaa=vgacaWFGaGa a8xBaiaa=9gacaWFKbGaa8xzaiaa=bcacaWFPbGaa83Caiaa=bcaca WFHbGaa8hBaiaa=DhacaWFHbGaa8xEaiaa=nhacaWFGaGaa83Baiaa =5gacaWFLbGaa8hiaiaa=9gacaWFMbGaa8hiaiaa=rhacaWFObGaa8 xzaiaa=bcacaWFUbGaa8xDaiaa=1gacaWFIbGaa8xzaiaa=jhacaWF ZbGaa8hiaiaa=LgacaWFUbGaa8hiaiaa=fgacaWFGaGaa8hzaiaa=f gacaWF0bGaa8xyaiaa=5caaeaadaqadaqaaiaa=LgacaWFPbaacaGL OaGaayzkaaGaa8hvaiaa=HgacaWFLbGaa8hiaiaa=1gacaWFLbGaa8 xyaiaa=5gacaWFGaGaa83yaiaa=fgacaWFUbGaa8hiaiaa=jgacaWF LbGaa8hiaiaa=9gacaWFUbGaa8xzaiaa=bcacaWFVbGaa8Nzaiaa=b cacaWF0bGaa8hAaiaa=vgacaWFGaGaa8NBaiaa=vhacaWFTbGaa8Ny aiaa=vgacaWFYbGaa83Caiaa=bcacaWFPbGaa8NBaiaa=bcacaWFHb Gaa8hiaiaa=rgacaWFHbGaa8hDaiaa=fgacaWFUaaabaWaaeWaaeaa caWFPbGaa8xAaiaa=LgaaiaawIcacaGLPaaacaWFubGaa8hAaiaa=v gacaWFGaGaa8xBaiaa=vgacaWFKbGaa8xAaiaa=fgacaWFUbGaa8hi aiaa=LgacaWFZbGaa8hiaiaa=fgacaWFSbGaa83Daiaa=fgacaWF5b Gaa83Caiaa=bcacaWFVbGaa8NBaiaa=vgacaWFGaGaa83Baiaa=zga caWFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=5gacaWF1bGaa8xBai aa=jgacaWFLbGaa8NCaiaa=nhacaWFGaGaa8xAaiaa=5gacaWFGaGa a8xyaiaa=bcacaWFKbGaa8xyaiaa=rhacaWFHbGaa8Nlaaqaamaabm aabaGaa8xAaiaa=zhaaiaawIcacaGLPaaacaWFGaGaa8xqaiaa=bca caWFKbGaa8xyaiaa=rhacaWFHbGaa8hiaiaa=fgacaWFSbGaa83Dai aa=fgacaWF5bGaa83Caiaa=bcacaWFObGaa8xyaiaa=nhacaWFGaGa a8xyaiaa=bcacaWFTbGaa83Baiaa=rgacaWFLbGaa8Nlaaqaamaabm aabaGaa8NDaaGaayjkaiaawMcaaiaa=bcacaWFubGaa8hAaiaa=vga caWFGaGaa8hzaiaa=fgacaWF0bGaa8xyaiaa=bcacaWF2aGaa8hlai aa=bcacaWF0aGaa8hlaiaa=bcacaWFZaGaa8hlaiaa=bcacaWF4aGa a8hlaiaa=bcacaWF5aGaa8hlaiaa=bcacaWFXaGaa8Nmaiaa=Xcaca WFGaGaa8xmaiaa=ndacaWFSaGaa8hiaiaa=LdacaWFGaGaa8hAaiaa =fgacaWFZbGaa8hiaiaa=1gacaWFLbGaa8xyaiaa=5gacaWFGaGaa8 xoaiaa=5caaaaa@25A6@

(i) True , because mode is the value that occurs most of         the time and it belongs to the given data . (ii) False , Mean may or may not belong to the data . (iii) True , because median is the middle observation         of given data . (iv) False The given data is: 6,4,3,8,9,12,13,9 . So, Mean = 6 + 4 + 3 + 8 + 9 + 12 + 13 + 9 8                       = 64 8                       = 8 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaWaauIh aeaacaqGubGaaeOCaiaabwhacaqGLbaaaiaabYcacaqGGaGaaeOyai aabwgacaqGJbGaaeyyaiaabwhacaqGZbGaaeyzaiaabccacaqGTbGa ae4BaiaabsgacaqGLbGaaeiiaiaabMgacaqGZbGaaeiiaiaabshaca qGObGaaeyzaiaabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzaiaa bccacaqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaab+gacaqGJbGaae 4yaiaabwhacaqGYbGaae4CaiaabccacaqGTbGaae4BaiaabohacaqG 0bGaaeiiaiaab+gacaqGMbaabaGaaGjbVlaaysW7caaMe8UaaGjbVl aabshacaqGObGaaeyzaiaabccacaqG0bGaaeyAaiaab2gacaqGLbGa aeiiaiaabggacaqGUbGaaeizaiaabccacaqGPbGaaeiDaiaabccaca qGIbGaaeyzaiaabYgacaqGVbGaaeOBaiaabEgacaqGZbGaaeiiaiaa bshacaqGVbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGNbGaae yAaiaabAhacaqGLbGaaeOBaiaabccacaqGKbGaaeyyaiaabshacaqG HbGaaeOlaaqaaiaabIcacaqGPbGaaeyAaiaabMcacaqGGaWaauIhae aacaqGgbGaaeyyaiaabYgacaqGZbGaaeyzaaaacaqGSaGaaeiiaiaa b2eacaqGLbGaaeyyaiaab6gacaqGGaGaaeyBaiaabggacaqG5bGaae iiaiaab+gacaqGYbGaaeiiaiaab2gacaqGHbGaaeyEaiaabccacaqG UbGaae4BaiaabshacaqGGaGaaeOyaiaabwgacaqGSbGaae4Baiaab6 gacaqGNbGaaeiiaiaabshacaqGVbGaaeiiaiaabshacaqGObGaaeyz aiaabccacaqGKbGaaeyyaiaabshacaqGHbGaaeOlaaqaaiaabIcaca qGPbGaaeyAaiaabMgacaqGPaGaaeiiamaaL4babaGaaeivaiaabkha caqG1bGaaeyzaaaacaqGSaGaaeiiaiaabkgacaqGLbGaae4yaiaabg gacaqG1bGaae4CaiaabwgacaqGGaGaaeyBaiaabwgacaqGKbGaaeyA aiaabggacaqGUbGaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqGOb GaaeyzaiaabccacaqGTbGaaeyAaiaabsgacaqGKbGaaeiBaiaabwga caqGGaGaae4BaiaabkgacaqGZbGaaeyzaiaabkhacaqG2bGaaeyyai aabshacaqGPbGaae4Baiaab6gaaeaacaaMe8UaaGjbVlaaysW7caaM e8Uaae4BaiaabAgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6 gacaqGGaGaaeizaiaabggacaqG0bGaaeyyaiaab6caaeaacaqGOaGa aeyAaiaabAhacaqGPaGaaeiiamaaL4babaGaaeOraiaabggacaqGSb Gaae4CaiaabwgaaaaabaGaaeivaiaabIgacaqGLbGaaeiiaiaabEga caqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabsgacaqGHbGaaeiDai aabggacaqGGaGaaeyAaiaabohacaqG6aGaaeiiaiaabAdacaqGSaGa aeinaiaabYcacaqGZaGaaeilaiaabIdacaqGSaGaaeyoaiaabYcaca qGXaGaaeOmaiaabYcacaqGXaGaae4maiaabYcacaqG5aGaaeOlaaqa aiaabofacaqGVbGaaeilaiaabccacaqGnbGaaeyzaiaabggacaqGUb Gaeyypa0ZaaSaaaeaacaaI2aGaey4kaSIaaGinaiabgUcaRiaaioda cqGHRaWkcaaI4aGaey4kaSIaaGyoaiabgUcaRiaaigdacaaIYaGaey 4kaSIaaGymaiaaiodacqGHRaWkcaaI5aaabaGaaGioaaaaaeaacaaM e8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaays W7caaMe8UaaGjbVlabg2da9maalaaabaGaaGOnaiaaisdaaeaacaaI 4aaaaaqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaays W7caaMe8UaaGjbVlaaysW7caaMe8Uaeyypa0JaaGioaaaaaa@65DF@

Q.16 Use the bar graph (Fig below) to a answer the following question (a) Which is the most popular pet? (b) How many children have dog as a pet?

(a) The bar showing cats is the tallest, so cat is the most popular pet. (b) From the bar graph, the number of children having dog as a pet is 8.

( i ) About how many books were sold in 1989? 1990? 1992? ( ii ) Inwhich year were about 475 books sold?         About 225 books sold? ( iii ) In which years were fewer than 25 0 books sold ? ( iv ) Can you explain how you would estimate the number         of books sold in 1989?

(i) In 1989, 175 books were sold in 1990, 475 books were sold . In 1992, 225 books were sold . (ii) From the graph, it can be concluded that 475 books were sold in the year 1990 and 225 books were sold in the year 1992 . (iii) From the graph, it can be concluded that in the years 1989 and 1992, the number of books sold were less than 250 . (iv) From the graph, it can be concluded that the number of books sold in the year 1989 is about 1 and 3 4 th   part of 1 cm The scale is taken as 1 cm = 100 books So, 100+ 3 4 × 100 = 100 + 75 = 175 Therefore, about 175 books were sold in the year 1989 . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeys aiaab6gacaqGGaGaaeymaiaabMdacaqG4aGaaeyoaiaabYcacaqGGa GaaeymaiaabEdacaqG1aGaaeiiaiaabkgacaqGVbGaae4BaiaabUga caqGZbGaaeiiaiaabEhacaqGLbGaaeOCaiaabwgacaqGGaGaae4Cai aab+gacaqGSbGaaeizaiaabccacaqGPbGaaeOBaiaabccacaqGXaGa aeyoaiaabMdacaqGWaGaaeilaiaabccacaqG0aGaae4naiaabwdaca qGGaGaaeOyaiaab+gacaqGVbGaae4AaiaabohacaqGGaGaae4Daiaa bwgacaqGYbGaaeyzaiaabccacaqGZbGaae4BaiaabYgacaqGKbGaae OlaaqaaiaabMeacaqGUbGaaeiiaiaabgdacaqG5aGaaeyoaiaabkda caqGSaGaaeiiaiaabkdacaqGYaGaaeynaiaabccacaqGIbGaae4Bai aab+gacaqGRbGaae4CaiaabccacaqG3bGaaeyzaiaabkhacaqGLbGa aeiiaiaabohacaqGVbGaaeiBaiaabsgacaqGUaaabaGaaeikaiaabM gacaqGPbGaaeykaiaabccacaqGgbGaaeOCaiaab+gacaqGTbGaaeii aiaabshacaqGObGaaeyzaiaabccacaqGNbGaaeOCaiaabggacaqGWb GaaeiAaiaabYcacaqGGaGaaeyAaiaabshacaqGGaGaae4yaiaabgga caqGUbGaaeiiaiaabkgacaqGLbGaaeiiaiaabogacaqGVbGaaeOBai aabogacaqGSbGaaeyDaiaabsgacaqGLbGaaeizaiaabccacaqG0bGa aeiAaiaabggacaqG0bGaaeiiaiaabsdacaqG3aGaaeynaiaabccaca qGIbGaae4Baiaab+gacaqGRbGaae4CaiaabccacaqG3bGaaeyzaiaa bkhacaqGLbaabaGaae4Caiaab+gacaqGSbGaaeizaiaabccacaqGPb GaaeOBaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyEaiaabwga caqGHbGaaeOCaiaabccacaqGXaGaaeyoaiaabMdacaqGWaGaaeiiai aabggacaqGUbGaaeizaiaabccacaqGYaGaaeOmaiaabwdacaqGGaGa aeOyaiaab+gacaqGVbGaae4AaiaabohacaqGGaGaae4Daiaabwgaca qGYbGaaeyzaiaabccacaqGZbGaae4BaiaabYgacaqGKbGaaeiiaiaa bMgacaqGUbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG5bGaae yzaiaabggacaqGYbGaaeiiaiaabgdacaqG5aGaaeyoaiaabkdacaqG UaaabaGaaeikaiaabMgacaqGPbGaaeyAaiaabMcacaqGGaGaaeOrai aabkhacaqGVbGaaeyBaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGa ae4zaiaabkhacaqGHbGaaeiCaiaabIgacaqGSaGaaeiiaiaabMgaca qG0bGaaeiiaiaabogacaqGHbGaaeOBaiaabccacaqGIbGaaeyzaiaa bccacaqGJbGaae4Baiaab6gacaqGJbGaaeiBaiaabwhacaqGKbGaae yzaiaabsgacaqGGaGaaeiDaiaabIgacaqGHbGaaeiDaiaabccacaqG PbGaaeOBaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyEaiaabw gacaqGHbGaaeOCaiaabohaaeaacaqGXaGaaeyoaiaabIdacaqG5aGa aeiiaiaabggacaqGUbGaaeizaiaabccacaqGXaGaaeyoaiaabMdaca qGYaGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOBaiaa bwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaab+gacaqGMbGaae iiaiaabkgacaqGVbGaae4BaiaabUgacaqGZbGaaeiiaiaabohacaqG VbGaaeiBaiaabsgacaqGGaGaae4DaiaabwgacaqGYbGaaeyzaiaabc cacaqGSbGaaeyzaiaabohacaqGZbGaaeiiaiaabshacaqGObGaaeyy aiaab6gacaqGGaGaaeOmaiaabwdacaqGWaGaaeOlaaqaaiaabIcaca qGPbGaaeODaiaabMcacaqGGaGaaeOraiaabkhacaqGVbGaaeyBaiaa bccacaqG0bGaaeiAaiaabwgacaqGGaGaae4zaiaabkhacaqGHbGaae iCaiaabIgacaqGSaGaaeiiaiaabMgacaqG0bGaaeiiaiaabogacaqG HbGaaeOBaiaabccacaqGIbGaaeyzaiaabccacaqGJbGaae4Baiaab6 gacaqGJbGaaeiBaiaabwhacaqGKbGaaeyzaiaabsgacaqGGaGaaeiD aiaabIgacaqGHbGaaeiDaiaabccacaqG0bGaaeiAaiaabwgacaqGGa GaaeOBaiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaab+ga caqGMbaabaGaaeOyaiaab+gacaqGVbGaae4AaiaabohacaqGGaGaae 4Caiaab+gacaqGSbGaaeizaiaabccacaqGPbGaaeOBaiaabccacaqG 0bGaaeiAaiaabwgacaqGGaGaaeyEaiaabwgacaqGHbGaaeOCaiaabc cacaqGXaGaaeyoaiaabIdacaqG5aGaaeiiaiaabMgacaqGZbGaaeii aiaabggacaqGIbGaae4BaiaabwhacaqG0bGaaeiiaiaabgdacaqGGa Gaaeyyaiaab6gacaqGKbGaaeiiamaalaaabaGaaG4maaqaaiaaisda aaGaaeiDaiaabIgacaaMe8UaaeiCaiaabggacaqGYbGaaeiDaiaabc cacaqGVbGaaeOzaiaabccacaqGXaGaaeiiaiaabogacaqGTbaabaGa aeivaiaabIgacaqGLbGaaeiiaiaabohacaqGJbGaaeyyaiaabYgaca qGLbGaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqGHbGaae4Aaiaa bwgacaqGUbGaaeiiaiaabggacaqGZbGaaeiiaiaabgdacaqGGaGaae 4yaiaab2gacaqGGaGaaeypaiaabccacaqGXaGaaeimaiaabcdacaqG GaGaaeOyaiaab+gacaqGVbGaae4AaiaabohaaeaacaqGtbGaae4Bai aabYcacaqGGaGaaeymaiaabcdacaqGWaGaae4kamaalaaabaGaaG4m aaqaaiaaisdaaaGaey41aqRaaGymaiaaicdacaaIWaGaeyypa0JaaG ymaiaaicdacaaIWaGaey4kaSIaaG4naiaaiwdacqGH9aqpcaaIXaGa aG4naiaaiwdaaeaacaqGubGaaeiAaiaabwgacaqGYbGaaeyzaiaabA gacaqGVbGaaeOCaiaabwgacaqGSaGaaeiiaiaabggacaqGIbGaae4B aiaabwhacaqG0bGaaeiiaiaabgdacaqG3aGaaeynaiaabccacaqGIb Gaae4Baiaab+gacaqGRbGaae4CaiaabccacaqG3bGaaeyzaiaabkha caqGLbGaaeiiaiaabohacaqGVbGaaeiBaiaabsgacaqGGaGaaeyAai aab6gacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabMhacaqGLbGa aeyyaiaabkhacaqGGaGaaeymaiaabMdacaqG4aGaaeyoaiaab6caaa aa@18E2@

Q.18 Number of children in six different classes are given below. Represent the data on a bar graph.

( a ) H o w w o u l d y o u c h o o s e a s c a l e? ( b ) A n s w e r t h e f o l l o w i n g q u e s t i o n s : ( i ) W h i c h c l a s s h a s t h e m a x i m u m n u m b e r o f c h i l d r e n ? a n d t h e m i n i m u m ? ( i i ) F i n d t h e r a t i o o f s t u d e n t s o f c l a s s s i x t h t o t h e s t u d e n t s o f c l a s s e i g h t .

(a)   Choose a scale as 1 unit = 10 children, because we can show a clear difference between the number of students of class 7th and class 9th . (b) (i) The bar representing the number of children of class fifth is tallest, there are maximum number of children in class fifth and the bar representing the number of children of class tenth is smallest, there are minimum number of children in class tenth . (ii) The number of student in class sixth is 120 and the number of student in class eight is 100 . So, the ratio ratio is = 120 100 = 20 × 6 20 × 5 = 6 5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyyaiaabMcacaaMe8Uaae4q aiaabIgacaqGVbGaae4BaiaabohacaqGLbGaaeiiaiaabggacaqGGa Gaae4CaiaabogacaqGHbGaaeiBaiaabwgacaqGGaGaaeyyaiaaboha caqGGaGaaeymaiaabccacaqG1bGaaeOBaiaabMgacaqG0bGaaeiiai aab2dacaqGGaGaaeymaiaabcdacaqGGaGaae4yaiaabIgacaqGPbGa aeiBaiaabsgacaqGYbGaaeyzaiaab6gacaqGSaGaaeiiaiaabkgaca qGLbGaae4yaiaabggacaqG1bGaae4CaiaabwgacaqGGaGaae4Daiaa bwgacaqGGaGaae4yaiaabggacaqGUbaabaGaae4CaiaabIgacaqGVb Gaae4DaiaabccacaqGHbGaaeiiaiaabogacaqGSbGaaeyzaiaabgga caqGYbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCai aabwgacaqGUbGaae4yaiaabwgacaqGGaGaaeOyaiaabwgacaqG0bGa ae4DaiaabwgacaqGLbGaaeOBaiaabccacaqG0bGaaeiAaiaabwgaca qGGaGaaeOBaiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabohacaqG0bGaaeyDaiaabsgacaqGLbGaae OBaiaabshacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaaqaaiaaboga caqGSbGaaeyyaiaabohacaqGZbGaaeiiaiaabEdacaqG0bGaaeiAai aabccacaqGHbGaaeOBaiaabsgacaqGGaGaae4yaiaabYgacaqGHbGa ae4CaiaabohacaqGGaGaaeyoaiaabshacaqGObGaaeOlaaqaaiaabI cacaqGIbGaaeykaaqaaiaabIcacaqGPbGaaeykaaqaaiaabsfacaqG ObGaaeyzaiaabccacaqGIbGaaeyyaiaabkhacaqGGaGaaeOCaiaabw gacaqGWbGaaeOCaiaabwgacaqGZbGaaeyzaiaab6gacaqG0bGaaeyA aiaab6gacaqGNbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGUb GaaeyDaiaab2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabAga caqGGaGaae4yaiaabIgacaqGPbGaaeiBaiaabsgacaqGYbGaaeyzai aab6gacaqGGaGaae4BaiaabAgacaqGGaGaae4yaiaabYgacaqGHbGa ae4CaiaabohacaqGGaGaaeOzaiaabMgacaqGMbGaaeiDaiaabIgaca qGGaGaaeyAaiaabohaaeaacaqG0bGaaeyyaiaabYgacaqGSbGaaeyz aiaabohacaqG0bGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGYb GaaeyzaiaabccacaqGHbGaaeOCaiaabwgacaqGGaGaaeyBaiaabgga caqG4bGaaeyAaiaab2gacaqG1bGaaeyBaiaabccacaqGUbGaaeyDai aab2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGa ae4yaiaabIgacaqGPbGaaeiBaiaabsgacaqGYbGaaeyzaiaab6gaca qGGaGaaeyAaiaab6gacaqGGaGaae4yaiaabYgacaqGHbGaae4Caiaa bohacaqGGaGaaeOzaiaabMgacaqGMbGaaeiDaiaabIgaaeaacaqGHb GaaeOBaiaabsgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkga caqGHbGaaeOCaiaabccacaqGYbGaaeyzaiaabchacaqGYbGaaeyzai aabohacaqGLbGaaeOBaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGa aeiDaiaabIgacaqGLbGaaeiiaiaab6gacaqG1bGaaeyBaiaabkgaca qGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqGJbGaaeiAaiaa bMgacaqGSbGaaeizaiaabkhacaqGLbGaaeOBaiaabccacaqGVbGaae OzaiaabccacaqGJbGaaeiBaiaabggacaqGZbGaae4CaiaabccacaqG 0bGaaeyzaiaab6gacaqG0bGaaeiAaiaabccacaqGPbGaae4Caaqaai aabohacaqGTbGaaeyyaiaabYgacaqGSbGaaeyzaiaabohacaqG0bGa aeilaiaabccacaqG0bGaaeiAaiaabwgacaqGYbGaaeyzaiaabccaca qGHbGaaeOCaiaabwgacaqGGaGaaeyBaiaabMgacaqGUbGaaeyAaiaa b2gacaqG1bGaaeyBaiaabccacaqGUbGaaeyDaiaab2gacaqGIbGaae yzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGaae4yaiaabIgacaqG PbGaaeiBaiaabsgacaqGYbGaaeyzaiaab6gacaqGGaGaaeyAaiaab6 gacaqGGaGaae4yaiaabYgacaqGHbGaae4CaiaabohacaqGGaGaaeiD aiaabwgacaqGUbGaaeiDaiaabIgacaqGUaaabaGaaeikaiaabMgaca qGPbGaaeykaaqaaiaabsfacaqGObGaaeyzaiaabccacaqGUbGaaeyD aiaab2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGa Gaae4CaiaabshacaqG1bGaaeizaiaabwgacaqGUbGaaeiDaiaabcca caqGPbGaaeOBaiaabccacaqGJbGaaeiBaiaabggacaqGZbGaae4Cai aabccacaqGZbGaaeyAaiaabIhacaqG0bGaaeiAaiaabccacaqGPbGa ae4CaiaabccacaqGXaGaaeOmaiaabcdacaqGGaGaaeyyaiaab6gaca qGKbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGUbGaaeyDaiaa b2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaaaba Gaae4CaiaabshacaqG1bGaaeizaiaabwgacaqGUbGaaeiDaiaabcca caqGPbGaaeOBaiaabccacaqGJbGaaeiBaiaabggacaqGZbGaae4Cai aabccacaqGLbGaaeyAaiaabEgacaqGObGaaeiDaiaabccacaqGPbGa ae4CaiaabccacaqGXaGaaeimaiaabcdacaqGUaaabaGaae4uaiaab+ gacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGaaeyy aiaabshacaqGPbGaae4BaiaabccacaqGYbGaaeyyaiaabshacaqGPb Gaae4BaiaabccacaqGPbGaae4CaiaabccacaqG9aGaaeiiamaalaaa baGaaGymaiaaikdacaaIWaaabaGaaGymaiaaicdacaaIWaaaaiabg2 da9maalaaabaGaaGOmaiaaicdacqGHxdaTcaaI2aaabaGaaGOmaiaa icdacqGHxdaTcaaI1aaaaiabg2da9maaL4babaWaaSaaaeaacaaI2a aabaGaaGynaaaaaaaaaaa@0D0B@

Q.19 The performance of student sin 1 st Term and 2 nd Term is given. Draw a double bar graph choosing appropriate s c a l e a n d a n s w e r t h e f o l l o w i n g :

(i) In which subject, has the child improved h is performance the most? (ii) in which subject is the improvement the least? (iii) Has the performance gone down in any subject?

(i) There was a maximum increase in the marks obtained in Maths . Thus, the child improved his performance the most in Maths . (ii) From the graph, we observe that the improvment was least in S . Science . (iii) From the graph, we observe that the performance in Hindi has grown down . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeiv aiaabIgacaqGLbGaaeOCaiaabwgacaqGGaGaae4DaiaabggacaqGZb GaaeiiaiaabggacaqGGaGaaeyBaiaabggacaqG4bGaaeyAaiaab2ga caqG1bGaaeyBaiaabccacaqGPbGaaeOBaiaabogacaqGYbGaaeyzai aabggacaqGZbGaaeyzaiaabccacaqGPbGaaeOBaiaabccacaqG0bGa aeiAaiaabwgacaqGGaGaaeyBaiaabggacaqGYbGaae4Aaiaabohaca qGGaGaae4BaiaabkgacaqG0bGaaeyyaiaabMgacaqGUbGaaeyzaiaa bsgacaqGGaGaaeyAaiaab6gaaeaacaqGnbGaaeyyaiaabshacaqGOb Gaae4Caiaab6cacaqGGaGaaeivaiaabIgacaqG1bGaae4CaiaabYca caqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqGObGaaeyAai aabYgacaqGKbGaaeiiaiaabMgacaqGTbGaaeiCaiaabkhacaqGVbGa aeODaiaabwgacaqGKbGaaeiiaiaabIgacaqGPbGaae4Caiaabccaca qGWbGaaeyzaiaabkhacaqGMbGaae4BaiaabkhacaqGTbGaaeyyaiaa b6gacaqGJbGaaeyzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae yBaiaab+gacaqGZbGaaeiDaaqaaiaabMgacaqGUbGaaeiiaiaab2ea caqGHbGaaeiDaiaabIgacaqGZbGaaeOlaaqaaiaabIcacaqGPbGaae yAaiaabMcacaqGGaGaiaiMbAeacGaGygOCaiacaIzGVbGaiaiMb2ga cGaGygiiaiacaIzG0bGaiaiMbIgacGaGygyzaiacaIzGGaGaiaiMbE gacGaGygOCaiacaIzGHbGaiaiMbchacGaGygiAaiacaIzGSaGaiaiM bccacGaGyg4DaiacaIzGLbGaiaiMbccacGaGyg4BaiacaIzGIbGaia iMbohacGaGygyzaiacaIzGYbGaiaiMbAhacGaGygyzaiacaIzGGaGa iaiMbshacGaGygiAaiacaIzGHbGaiaiMbshacaqGGaGaaeiDaiaabI gacaqGLbGaaeiiaiaabMgacaqGTbGaaeiCaiaabkhacaqGVbGaaeOD aiaab2gacaqGLbGaaeOBaiaabshacaqGGaGaae4DaiaabggacaqGZb GaaeiiaiaabYgacaqGLbGaaeyyaiaabohacaqG0baabaGaaeyAaiaa b6gacaqGGaGaae4uaiaab6cacaqGGaGaae4uaiaabogacaqGPbGaae yzaiaab6gacaqGJbGaaeyzaiaab6caaeaacaqGOaGaaeyAaiaabMga caqGPbGaaeykaiaabccacaqGgbGaaeOCaiaab+gacaqGTbGaaeiiai aabshacaqGObGaaeyzaiaabccacaqGNbGaaeOCaiaabggacaqGWbGa aeiAaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4Baiaabkgaca qGZbGaaeyzaiaabkhacaqG2bGaaeyzaiaabccacaqG0bGaaeiAaiaa bggacaqG0bGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGaae yzaiaabkhacaqGMbGaae4BaiaabkhacaqGTbGaaeyyaiaab6gacaqG JbGaaeyzaiaabccacaqGPbGaaeOBaiaabccacaqGibGaaeyAaiaab6 gacaqGKbGaaeyAaaqaaiaabIgacaqGHbGaae4CaiaabccacaqGNbGa aeOCaiaab+gacaqG3bGaaeOBaiaabccacaqGKbGaae4BaiaabEhaca qGUbGaaeOlaaaaaa@4E32@

Q.20 Consider this data collected from a survey of a colony.

( i ) D r a w a d o u b l e b a r g r a p h c h o o s i n g a n a p p r o p r i a t e s c a l e . W h a t d o y o u i n f e r f r o m t h e g r a p h ? ( i i ) W h i c h s p o r t i s m o s t p o p u l a r ? ( i i i ) W h i c h i s m o r e p r e f e rr e d w a t c h i n g o r p a r t i c i p a t i n g s p o r t ?

(ii) From the bar graph, the bar representing the number of people who like watching and participating in cricket is the tallest among all the bars .Thus, cricket is the most popular sport . (iii) The bars showing watchig spot are longer than the bars showing participating in spot . Thus, watching different types of sports is more preffered than participating in the sports . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMgacaqGPaGaaeii aiaabAeacaqGYbGaae4Baiaab2gacaqGGaGaaeiDaiaabIgacaqGLb GaaeiiaiaabkgacaqGHbGaaeOCaiaabccacaqGNbGaaeOCaiaabgga caqGWbGaaeiAaiaabYcacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiai aabkgacaqGHbGaaeOCaiaabccacaqGYbGaaeyzaiaabchacaqGYbGa aeyzaiaabohacaqGLbGaaeOBaiaabshacaqGPbGaaeOBaiaabEgaca qGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaab6gacaqG1bGaaeyBaiaa bkgacaqGLbGaaeOCaaqaaiaab+gacaqGMbGaaeiiaiaabchacaqGLb Gaae4BaiaabchacaqGSbGaaeyzaiaabccacaqGGaGaae4DaiaabIga caqGVbGaaeiiaiaabYgacaqGPbGaae4AaiaabwgacaqGGaGaae4Dai aabggacaqG0bGaae4yaiaabIgacaqGPbGaaeOBaiaabEgacaqGGaGa aeyyaiaab6gacaqGKbGaaeiiaiaabchacaqGHbGaaeOCaiaabshaca qGPbGaae4yaiaabMgacaqGWbGaaeyyaiaabshacaqGPbGaaeOBaiaa bEgacaqGGaGaaeyAaiaab6gacaqGGaGaae4yaiaabkhacaqGPbGaae 4yaiaabUgacaqGLbGaaeiDaaqaaiaabMgacaqGZbGaaeiiaiaabsha caqGObGaaeyzaiaabccacaqG0bGaaeyyaiaabYgacaqGSbGaaeyzai aabohacaqG0bGaaeiiaiaabggacaqGTbGaae4Baiaab6gacaqGNbGa aeiiaiaabggacaqGSbGaaeiBaiaabccacaqG0bGaaeiAaiaabwgaca qGGaGaaeOyaiaabggacaqGYbGaae4Caiaab6cacaqGubGaaeiAaiaa bwhacaqGZbGaaeilaiaabccacaqGJbGaaeOCaiaabMgacaqGJbGaae 4AaiaabwgacaqG0bGaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqG ObGaaeyzaiaabccacaqGTbGaae4BaiaabohacaqG0baabaGaaeiCai aab+gacaqGWbGaaeyDaiaabYgacaqGHbGaaeOCaiaabccacaqGZbGa aeiCaiaab+gacaqGYbGaaeiDaiaab6caaeaacaqGOaGaaeyAaiaabM gacaqGPbGaaeykaiaabccacaqGubGaaeiAaiaabwgacaqGGaGaaeOy aiaabggacaqGYbGaae4CaiaabccacaqGZbGaaeiAaiaab+gacaqG3b GaaeyAaiaab6gacaqGNbGaaeiiaiaabEhacaqGHbGaaeiDaiaaboga caqGObGaaeyAaiaabEgacaqGGaGaae4CaiaabchacaqGVbGaaeiDai aabccacaqGHbGaaeOCaiaabwgacaqGGaGaaeiBaiaab+gacaqGUbGa ae4zaiaabwgacaqGYbGaaeiiaiaabshacaqGObGaaeyyaiaab6gaca qGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkgacaqGHbGaaeOCaiaa bohaaeaacaqGZbGaaeiAaiaab+gacaqG3bGaaeyAaiaab6gacaqGNb GaaeiiaiaabchacaqGHbGaaeOCaiaabshacaqGPbGaae4yaiaabMga caqGWbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaaeyAai aab6gacaqGGaGaae4CaiaabchacaqGVbGaaeiDaiaab6cacaqGGaGa aeivaiaabIgacaqG1bGaae4CaiaabYcacaqGGaGaae4Daiaabggaca qG0bGaae4yaiaabIgacaqGPbGaaeOBaiaabEgacaqGGaGaaeizaiaa bMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGaae iiaiaabshacaqG5bGaaeiCaiaabwgacaqGZbaabaGaae4BaiaabAga caqGGaGaae4CaiaabchacaqGVbGaaeOCaiaabshacaqGZbGaaeiiai aabMgacaqGZbGaaeiiaiaab2gacaqGVbGaaeOCaiaabwgacaqGGaGa aeiCaiaabkhacaqGLbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgaca qGKbGaaeiiaiaabshacaqGObGaaeyyaiaab6gacaqGGaGaaeiCaiaa bggacaqGYbGaaeiDaiaabMgacaqGJbGaaeyAaiaabchacaqGHbGaae iDaiaabMgacaqGUbGaae4zaiaabccacaqGPbGaaeOBaiaabccacaqG 0bGaaeiAaiaabwgacaqGGaGaae4CaiaabchacaqGVbGaaeOCaiaabs hacaqGZbGaaeOlaaaaaa@811D@

Q.21 Take the data giving the mininum and the maximum temperature of various cities given in the beginning of this chapter.

P l o t a d o u b l e b a r g r a p h u s i n g t h e d a t a a n d a n s w e r t h e f o l l o w i n g : ( i ) W h i c h c i t y h a s t h e l a r g e s t d i f f e r e n c e i n t h e m i n i m u m a n d m a x i m u m t e m p e r a t u r e o n t h e g i v e n d a t e ? ( i i ) W h i c h i s t h e h o t t e s t c i t y a n d w h i c h i s t h e c o l d e s t c i t y ? ( i i i ) N a m e t w o c i t i e s w h e r e m a x i m u m t e m p e r a t u r e o f o n e w a s l e s s t h a n t h e m i n i m u m t e m p e r a t u r e o f t h e o t h e r . ( i v ) N a m e t h e c i t y w h i c h h a s t h e l e a s t d i f f e r e n c e b e t w e e n i t s m i n i m u m a n d t h e m a x i m u m t e m p e r a t u r e .

A double bar graph for given data is shown as below:

(i) From the graph, we observe that Jammu has the largest         difference in its minimum and maximum temprature on the         given date . (ii) From graph, we observe that Jammu is the hottest city           and Banglore is the coldest city . (iii) Banglore and Jaipur, Banglore and Ahemdabad .           For Banglore, the maximum temprature was 28 ° C , while           temprature of both cities was 29 ° C . (iv) From graph, we observe that the city which has the           least difference between its minimum and maximum           temprature is Mumbai . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeOr aiaabkhacaqGVbGaaeyBaiaabccacaqG0bGaaeiAaiaabwgacaqGGa Gaae4zaiaabkhacaqGHbGaaeiCaiaabIgacaqGSaGaaeiiaiaabEha caqGLbGaaeiiaiaab+gacaqGIbGaae4CaiaabwgacaqGYbGaaeODai aabwgacaqGGaGaaeiDaiaabIgacaqGHbGaaeiDaiaabccacaqGkbGa aeyyaiaab2gacaqGTbGaaeyDaiaabccacaqGObGaaeyyaiaabohaca qGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabYgacaqGHbGaaeOCaiaa bEgacaqGLbGaae4CaiaabshaaeaacaaMe8UaaGjbVlaaysW7caaMe8 UaaeizaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6ga caqGJbGaaeyzaiaabccacaqGPbGaaeOBaiaabccacaqGPbGaaeiDai aabohacaqGGaGaaeyBaiaabMgacaqGUbGaaeyAaiaab2gacaqG1bGa aeyBaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaaeyBaiaabggaca qG4bGaaeyAaiaab2gacaqG1bGaaeyBaiaabccacaqG0bGaaeyzaiaa b2gacaqGWbGaaeOCaiaabggacaqG0bGaaeyDaiaabkhacaqGLbGaae iiaiaab+gacaqGUbGaaeiiaiaabshacaqGObGaaeyzaiaabccaaeaa caaMe8UaaGjbVlaaysW7caaMe8Uaae4zaiaabMgacaqG2bGaaeyzai aab6gacaqGGaGaaeizaiaabggacaqG0bGaaeyzaiaab6caaeaacaqG OaGaaeyAaiaabMgacaqGPaGaaeiiaiaabAeacaqGYbGaae4Baiaab2 gacaqGGaGaae4zaiaabkhacaqGHbGaaeiCaiaabIgacaqGSaGaaeii aiaabEhacaqGLbGaaeiiaiaab+gacaqGIbGaae4CaiaabwgacaqGYb GaaeODaiaabwgacaqGGaGaaeiDaiaabIgacaqGHbGaaeiDaiaabcca caqGkbGaaeyyaiaab2gacaqGTbGaaeyDaiaabccacaqGPbGaae4Cai aabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiAaiaab+gacaqG0bGa aeiDaiaabwgacaqGZbGaaeiDaiaabccacaqGJbGaaeyAaiaabshaca qG5baabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caqGHbGaaeOB aiaabsgacaqGGaGaaeOqaiaabggacaqGUbGaae4zaiaabYgacaqGVb GaaeOCaiaabwgacaqGGaGaaeyAaiaabohacaqGGaGaaeiDaiaabIga caqGLbGaaeiiaiaabogacaqGVbGaaeiBaiaabsgacaqGLbGaae4Cai aabshacaqGGaGaae4yaiaabMgacaqG0bGaaeyEaiaab6caaeaacaqG OaGaaeyAaiaabMgacaqGPbGaaeykaiaabccacaqGcbGaaeyyaiaab6 gacaqGNbGaaeiBaiaab+gacaqGYbGaaeyzaiaabccacaqGHbGaaeOB aiaabsgacaqGGaGaaeOsaiaabggacaqGPbGaaeiCaiaabwhacaqGYb GaaeilaiaabccacaqGcbGaaeyyaiaab6gacaqGNbGaaeiBaiaab+ga caqGYbGaaeyzaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaaeyqai aabIgacaqGLbGaaeyBaiaabsgacaqGHbGaaeOyaiaabggacaqGKbGa aeOlaaqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaeOraiaab+ gacaqGYbGaaeiiaiaabkeacaqGHbGaaeOBaiaabEgacaqGSbGaae4B aiaabkhacaqGLbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGGa GaaeyBaiaabggacaqG4bGaaeyAaiaab2gacaqG1bGaaeyBaiaabcca caqG0bGaaeyzaiaab2gacaqGWbGaaeOCaiaabggacaqG0bGaaeyDai aabkhacaqGLbGaaeiiaiaabEhacaqGHbGaae4CaiaabccacaqGYaGa aeioaiabgclaWkaadoeacaqGSaGaaeiiaiaabEhacaqGObGaaeyAai aabYgacaqGLbaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caqG 0bGaaeyzaiaab2gacaqGWbGaaeOCaiaabggacaqG0bGaaeyDaiaabk hacaqGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabkgacaqGVbGaaeiD aiaabIgacaqGGaGaae4yaiaabMgacaqG0bGaaeyAaiaabwgacaqGZb GaaeiiaiaabEhacaqGHbGaae4CaiaabccacaqGYaGaaeyoaiabgcla WkaadoeacaGGUaaabaGaaeikaiaabMgacaqG2bGaaeykaiaabccaca qGgbGaaeOCaiaab+gacaqGTbGaaeiiaiaabEgacaqGYbGaaeyyaiaa bchacaqGObGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqGVbGaae OyaiaabohacaqGLbGaaeOCaiaabAhacaqGLbGaaeiiaiaabshacaqG ObGaaeyyaiaabshacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabo gacaqGPbGaaeiDaiaabMhacaqGGaGaae4DaiaabIgacaqGPbGaae4y aiaabIgacaqGGaGaaeiAaiaabggacaqGZbGaaeiiaiaabshacaqGOb GaaeyzaaqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaeiBaiaa bwgacaqGHbGaae4CaiaabshacaqGGaGaaeizaiaabMgacaqGMbGaae OzaiaabwgacaqGYbGaaeyzaiaab6gacaqGJbGaaeyzaiaabccacaqG IbGaaeyzaiaabshacaqG3bGaaeyzaiaabwgacaqGUbGaaeiiaiaabM gacaqG0bGaae4CaiaabccacaqGTbGaaeyAaiaab6gacaqGPbGaaeyB aiaabwhacaqGTbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGTb GaaeyyaiaabIhacaqGPbGaaeyBaiaabwhacaqGTbaabaGaaGjbVlaa ysW7caaMe8UaaGjbVlaaysW7caqG0bGaaeyzaiaab2gacaqGWbGaae OCaiaabggacaqG0bGaaeyDaiaabkhacaqGLbGaaeiiaiaabMgacaqG ZbGaaeiiaiaab2eacaqG1bGaaeyBaiaabkgacaqGHbGaaeyAaiaab6 caaaaa@113C@

Q.22 Tell whether the following is cretain to happen, impossible, can happed but not certain. (i) You are older today than yesterday. (ii) Atossed 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) Tomorrowl will be a coludy day.

(i) Certain (ii) Can happen but not certain (iii) Impossible (iv) Can happen but not certain (v) Can happen but not certain

Q.23 There are 6 marbles in a box with numbers from 1 to 6 marked on each o0f them. (i) What is the probability of drawing a marble with number 2? (ii) What is the probability of drawing a marble with umber 5?

(i)   Since,   Probability = Number of favourable outcomes Number of possible outcomes         So,P(apperance of 2) = 1 6 (ii) P(apperance of 2) = 1 6 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaaMe8Uaae4u aiaabMgacaqGUbGaae4yaiaabwgacaqGSaGaaGjbVpaaL4babaGaae iuaiaabkhacaqGVbGaaeOyaiaabggacaqGIbGaaeyAaiaabYgacaqG PbGaaeiDaiaabMhacaqGGaGaeyypa0JaaeiiamaalaaabaGaaeOtai aabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaab+gacaqGMbGa aeiiaiaabAgacaqGHbGaaeODaiaab+gacaqG1bGaaeOCaiaabggaca qGIbGaaeiBaiaabwgacaqGGaGaae4BaiaabwhacaqG0bGaae4yaiaa b+gacaqGTbGaaeyzaiaabohaaeaacaqGobGaaeyDaiaab2gacaqGIb GaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGaaeiCaiaab+ga caqGZbGaae4CaiaabMgacaqGIbGaaeiBaiaabwgacaqGGaGaae4Bai aabwhacaqG0bGaae4yaiaab+gacaqGTbGaaeyzaiaabohaaaaaaaqa aiaaysW7caaMe8UaaGjbVlaaysW7caqGtbGaae4BaiaabYcacaqGqb GaaeikaiaabggacaqGWbGaaeiCaiaabwgacaqGYbGaaeyyaiaab6ga caqGJbGaaeyzaiaabccacaqGVbGaaeOzaiaabccacaqGYaGaaeykai aabccacqGH9aqpdaqjEaqaamaalaaabaGaaGymaaqaaiaaiAdaaaaa aaqaaiaabIcacaqGPbGaaeyAaiaabMcacaqGGaGaaeiuaiaabIcaca qGHbGaaeiCaiaabchacaqGLbGaaeOCaiaabggacaqGUbGaae4yaiaa bwgacaqGGaGaae4BaiaabAgacaqGGaGaaeOmaiaabMcacaqGGaGaey ypa0ZaauIhaeaadaWcaaqaaiaaigdaaeaacaaI2aaaaaaaaaaa@B6EA@

Q.24 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 namely − Head and Tail . One team can opt either Head or Tail . Since, Probability = Number of favourable outcomes Number of favourable outcomes So, Probability(our team start first) = 1 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGbbGaaeiiaiaabogacaqGVbGaaeyA aiaab6gacaqGGaGaaeiAaiaabggacaqGZbGaaeiiaiaabshacaqG3b Gaae4BaiaabccacaqGMbGaaeyyaiaabogacaqGLbGaae4Caiaabcca caqGUbGaaeyyaiaab2gacaqGLbGaaeiBaiaabMhacqGHsislcaqGib GaaeyzaiaabggacaqGKbGaaeiiaiaabggacaqGUbGaaeizaiaabcca caqGubGaaeyyaiaabMgacaqGSbGaaeOlaaqaaiaab+eacaqGUbGaae yzaiaabccacaqG0bGaaeyzaiaabggacaqGTbGaaeiiaiaabogacaqG HbGaaeOBaiaabccacaqGVbGaaeiCaiaabshacaqGGaGaaeyzaiaabM gacaqG0bGaaeiAaiaabwgacaqGYbGaaeiiaiaabIeacaqGLbGaaeyy aiaabsgacaqGGaGaae4BaiaabkhacaqGGaGaaeivaiaabggacaqGPb GaaeiBaiaab6caaeaacaqGtbGaaeyAaiaab6gacaqGJbGaaeyzaiaa bYcacaqGGaWaauIhaeaacaqGqbGaaeOCaiaab+gacaqGIbGaaeyyai aabkgacaqGPbGaaeiBaiaabMgacaqG0bGaaeyEaiabg2da9maalaaa baGaaeOtaiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaab+ gacaqGMbGaaeiiaiaabAgacaqGHbGaaeODaiaab+gacaqG1bGaaeOC aiaabggacaqGIbGaaeiBaiaabwgacaqGGaGaae4BaiaabwhacaqG0b Gaae4yaiaab+gacaqGTbGaaeyzaiaabohaaeaacaqGobGaaeyDaiaa b2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGaae OzaiaabggacaqG2bGaae4BaiaabwhacaqGYbGaaeyyaiaabkgacaqG SbGaaeyzaiaabccacaqGVbGaaeyDaiaabshacaqGJbGaae4Baiaab2 gacaqGLbGaae4CaaaaaaaabaGaae4uaiaab+gacaqGSaGaaeiiaiaa bcfacaqGYbGaae4BaiaabkgacaqGHbGaaeOyaiaabMgacaqGSbGaae yAaiaabshacaqG5bGaaeikaiaab+gacaqG1bGaaeOCaiaabccacaqG 0bGaaeyzaiaabggacaqGTbGaaeiiaiaabohacaqG0bGaaeyyaiaabk hacaqG0bGaaeiiaiaabAgacaqGPbGaaeOCaiaabohacaqG0bGaaeyk aiaabccacqGH9aqpdaqjEaqaaiaabccadaWcaaqaaiaaigdaaeaaca aIYaaaaaaaaaaa@EB22@

Q.25 A box contains pairs of socks of two colours (black and white) I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that it will make a pair?

While closing the eyes, one can draw either a black sock or a white sock . Therefore, there are two possible cases . Since, Probability= Number of favourable outcomes Number of favourable outcomes So, P(a pair of white socks will be formed) = 1 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGxbGaaeiAaiaabMgacaqGSbGaaeyz aiaabccacaqGJbGaaeiBaiaab+gacaqGZbGaaeyAaiaab6gacaqGNb GaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGLbGaaeyEaiaabwga caqGZbGaaeilaiaabccacaqGVbGaaeOBaiaabwgacaqGGaGaae4yai aabggacaqGUbGaaeiiaiaabsgacaqGYbGaaeyyaiaabEhacaqGGaGa aeyzaiaabMgacaqG0bGaaeiAaiaabwgacaqGYbGaaeiiaiaabggaca qGGaGaaeOyaiaabYgacaqGHbGaae4yaiaabUgacaqGGaGaae4Caiaa b+gacaqGJbGaae4Aaaqaaiaab+gacaqGYbGaaeiiaiaabggacaqGGa Gaae4DaiaabIgacaqGPbGaaeiDaiaabwgacaqGGaGaae4Caiaab+ga caqGJbGaae4Aaiaab6caaeaacaqGubGaaeiAaiaabwgacaqGYbGaae yzaiaabAgacaqGVbGaaeOCaiaabwgacaqGSaGaaeiiaiaabshacaqG ObGaaeyzaiaabkhacaqGLbGaaeiiaiaabggacaqGYbGaaeyzaiaabc cacaqG0bGaae4Daiaab+gacaqGGaGaaeiCaiaab+gacaqGZbGaae4C aiaabMgacaqGIbGaaeiBaiaabwgacaqGGaGaae4yaiaabggacaqGZb GaaeyzaiaabohacaqGUaaabaGaae4uaiaabMgacaqGUbGaae4yaiaa bwgacaqGSaGaaeiiamaaL4babaGaaeiuaiaabkhacaqGVbGaaeOyai aabggacaqGIbGaaeyAaiaabYgacaqGPbGaaeiDaiaabMhacaqG9aWa aSaaaeaacaqGobGaaeyDaiaab2gacaqGIbGaaeyzaiaabkhacaqGGa Gaae4BaiaabAgacaqGGaGaaeOzaiaabggacaqG2bGaae4Baiaabwha caqGYbGaaeyyaiaabkgacaqGSbGaaeyzaiaabccacaqGVbGaaeyDai aabshacaqGJbGaae4Baiaab2gacaqGLbGaae4Caaqaaiaab6eacaqG 1bGaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabc cacaqGMbGaaeyyaiaabAhacaqGVbGaaeyDaiaabkhacaqGHbGaaeOy aiaabYgacaqGLbGaaeiiaiaab+gacaqG1bGaaeiDaiaabogacaqGVb GaaeyBaiaabwgacaqGZbaaaaaaaeaacaqGtbGaae4BaiaabYcacaqG GaGaaeiuaiaabIcacaqGHbGaaeiiaiaabchacaqGHbGaaeyAaiaabk hacaqGGaGaae4BaiaabAgacaqGGaGaae4DaiaabIgacaqGPbGaaeiD aiaabwgacaqGGaGaae4Caiaab+gacaqGJbGaae4AaiaabohacaqGGa Gaae4DaiaabMgacaqGSbGaaeiBaiaabccacaqGIbGaaeyzaiaabcca caqGMbGaae4BaiaabkhacaqGTbGaaeyzaiaabsgacaqGPaGaaeiiai aab2dadaqjEaqaamaalaaabaGaaGymaaqaaiaaikdaaaaaaaaaaa@0BF3@

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Faqs (frequently asked questions), 1. how many exercises are there in chapter 3 data handling of the ncert solutions for class 7 mathematics.

This chapter has four exercises. Exercise 3.1 has five short and four long answer questions. Exercise 3.2 has two short and three long answer questions. Exercise 3.3 has three short and three long answer questions and Exercise 3.4 has two short and two long answer questions.

2. What are the key formulas in Chapter 3 of the NCERT textbook for Class 7 Mathematics?

Calculation of arithmetic mean, mean, median, and mode, probability, and other formulas are described in these solutions. These are the foundations for learning data handling techniques. These formulas, as well as the key definitions, can be found at the end of the chapter for a quick review.

3. Why is Chapter 3 of the NCERT Solutions for Class 7 Mathematics so important?

The concepts covered in NCERT Solutions for Class 7 Mathematics Chapter 3 are excellent for honing  study skills because the examples and exercises in the chapter are based on experiential learning. Students can access hundreds of practise problems on mean, median, mode, bar graphs, and probability, questions. Besides this they can use the problem-solving techniques explained in the examples to stay ahead in the competition.

4. Is it mandatory for me to practise all of the questions in NCERT Solutions Class 7 Mathematics Data Handling?

To improve their data handling skills, students should complete all of the examples and practise questions. The sample examples and exercise questions will ensure that students have a strong foundation to deal with higher mathematical concepts and even Mathematics for competitive exams in the future.

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NCERT Solutions Class 7 Maths Chapter 3 Data Handling

Data is the fuel in this world for exploring and interpreting a lot of opportunities. The NCERT Solutions for Class 7 maths chapter 3 data handling introduces the concept of data and its importance. Starting with the collection of data to how it can be stored or displayed and how the central tendency of the data collected can be ascertained is explained in this chapter. NCERT solutions class 7 maths chapter 3 data handling explains how graphs, specifically bar graphs, can be used in different ways for the effective display of data. In addition , this chapter introduces the concept of probability and its theorem to find out the possibility of the occurrence of an event.

Class 7 maths NCERT Solutions chapter 3 data handling will provide the building block for understanding and handling data. In a world that is driven by data making it of utmost importance, this chapter provides the key foundational concepts on this topic. The importance of finding the central tendency of a data set is a part of statistical science which is required in almost all the branches of science, including the day to day life. Also, how the Average, Median, and Mode are relevant to different data sets are discussed with relevant examples in chapter 3. The chapter and exercises can be found out in the links given below and also find some of these in the exercises given below.

• NCERT Solutions Class 7 Maths Chapter 3 Ex 3.1
• NCERT Solutions Class 7 Maths Chapter 3 Ex 3.2
• NCERT Solutions Class 7 Maths Chapter 3 Ex 3.3
• NCERT Solutions Class 7 Maths Chapter 3 Ex 3.4

NCERT Solutions for Class 7 Maths Chapter 3 PDF

The NCERT solutions class 7 maths chapter has exercises pertaining to data analysis, calculations of central tendency, how to arrange data in tabular form, and questions on probability. All these exercise questions are available for free pdf download and can be easily accessed through the links given below.

☛ Download Class 7 Maths NCERT Solutions Chapter 3 Data Handling

NCERT Class 7 Maths Chapter 3   Download PDF

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling

The NCERT solutions provide a good combination of questions that will allow the students to explore a wide range of concepts through the problems. Consistent practice will improve their understanding of the ideas. These questions are also examined and produced by professionals, making them extremely important. Students can have a look at the components of the NCERT solutions Class 7 maths chapter 3 data handling below.

• Class 7 Maths Chapter 3 Ex 3.1 - 9 Questions
• Class 7 Maths Chapter 3 Ex 3.2 - 5 Questions
• Class 7 Maths Chapter 3 Ex 3.3 - 6 Questions
• Class 7 Maths Chapter 3 Ex 3.4 - 3 Questions

☛ Download Class 7 Maths Chapter 3 NCERT Book

Topics Covered: The various topics covered under data handling are arranging data in tabular format, finding the central tendency of the dataset for which explanation is given for Mean, Median, and Mode . Also, plotting of the dataset as histograms and how graphs are to be plotted as well as read. Other topics included in these Class 7 maths NCERT solutions chapter 3 are the explanation of probability theory and its basic attributes.

Total Questions: The NCERT Class 7 maths chapter 3 data handling chapter 3 spans 23 questions in order to make students understand the theories and formulas in data handling. Students will find 18 of these easy and 5 as long-form questions.

List of Formulas in NCERT Solutions Class 7 Maths Chapter 3

The NCERT solutions Class 7 maths chapter 3 covers lots of important concepts crucial for understanding higher grade maths. The concept behind the calculation of mean, median, and mode is explained precisely and the students will be able to recall them easily. Like ‘mean’ refers to the average value, ‘median’ to the mid-point value, and ‘mode’ as the most common value. Similarly, other formulas discussed in NCERT Solutions for Class 7 maths chapter 3 are given below.

• Arithmetic Mean: Sum of all the given observations at a given point / Total number of observations
• Probability Formula: Total number of successful events / Total number of outcomes

Important Questions for Class 7 Maths NCERT Solutions Chapter 3

Ncert solutions for class 7 maths video chapter 3, faqs on ncert solutions class 7 maths chapter 3, what are the important formulas in ncert solutions class 7 maths chapter 3.

The important formulas covered in the NCERT solutions class 7 maths chapter 3 are the calculation of arithmetic mean, the mean, median and mode, probability, etc. These are the fundamentals in giving a good start in learning the data handling techniques. The students can find these formulas as well as the important definitions at the last of the chapter for a quick revision.

Why Should I Practice NCERT Solutions Class 7 Maths data handling chapter 3?

The NCERT solutions class 7 chapter 3 has been prepared by expert scholars and researchers in their field, making sure that all the important aspects related to data handling like analysis and reading of data, as well as its representation, are covered in a language that is easily understood by everyone. Also, the CBSE board recommends studying from the NCERT solutions, which makes it very certain that the students can expect the examination questions from this book itself.

Why are Class 7 Maths NCERT Solutions Chapter 3 Important?

The concepts discussed in the NCERT Solutions Class 7 Maths Chapter 3 are excellent for honing the exam study abilities as the examples and the exercises presented in the chapter are taken from real-world scenarios making it more useful practically. The students can have access to hundreds of practice problems on mean, median, mode, bar graphs, and probability, which will help them prepare for exams, giving them a good head start as they also have access to the problem-solving techniques which are explained in the examples.

How CBSE Students can utilize NCERT Solutions Class 7 Maths Chapter 3 effectively?

The CBSE board recommends studying from the NCERT solutions, which means that the students should follow the pattern of problem-solving as discussed in the examples provided and should grasp the techniques which are mentioned in those examples to arrive at the answer logically and strategically.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 7 Maths data handling?

It would be beneficial to the students if they ensure that they solve all the examples and the exercise questions to level up their data handling skills. The sample examples and exercise questions provided in the NCERT Solutions Class 7 Maths Chapter 3 will ensure that the students have their foundation knowledge strong to deal with higher maths and even the maths for competitive exams ahead.

How Many Questions are there in Class 7 Maths NCERT Solutions Chapter 3 data handling?

The class 7 maths chapter 3 data handling chapter 3 has 23 questions. 14 of which are related to mean, median, mode. The six of them are related to the depiction and analysis of data through bar graphs, while the remaining 3 are of probability calculation.

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NCERT Solutions and Notes for Class 7 Maths Chapter 3: Data Handling Notes and Solutions (Free PDF)

• Updated on  
• Mar 1, 2024

In the NCERT Class 6 Maths Chapter 9 of Data Handling, we have briefly learned how to collect data, tabulate it and then represent it in the form of bar graphs. This form of collection, interpretation and representation of data helps us in better understanding of the data and to bring out useful results from them. In NCERT Class 7 Maths Chapter 3, Data Handling, we will study a step ahead of the previous data collection and handling. In this chapter, students will come across more types of data and their presentations. Let us now have a look at the NCERT Class 7 Maths Chapter 3 Data Handling Notes and Solutions (PDF).

Download NCERT Class 7 Maths Chapter 3 Data Handling Notes and Solutions PDF

Explore all the Chapters of Class Mathematics:-

Table of Contents

• 1.1 Topic 1: Collecting and Organising Data
• 1.2 Topic 2: Arithmetic Mean (AM) 
• 1.3 Topic 3: Range
• 1.4 Topic 4: Mode
• 1.5 Topic 5: Median
• 1.6 Topic 6: Chance
• 1.7 Topic 7: Probability
• 2.1 Exercise 3.1 Solutions
• 2.2 Exercise 3.2 Solutions
• 2.3 Exercise 3.3 Solutions
• 2.4 Exercise 3.4 Solutions

NCERT Class 7 Maths Chapter 3 Notes – PDF Available

Check the topic-wise notes for NCERT Maths Class 7 – Chapter 3 below. You can also download the PDF of the notes and take a printout to study later when you need quick revision before going to the exam hall. 

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Topic 1: Collecting and Organising Data

Let us see the term that is used often in data collection and organisation below. 

• Average: An average is a number that represents or shows the central tendency of a group of observations or data.

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Topic 2: Arithmetic Mean (AM) 

The most common representative value of a group of data is the arithmetic mean or the mean.

The average or Arithmetic Mean (A.M.) or simply mean is calculated by:

Topic 3: Range

The difference between the highest and the lowest observation gives us an idea of the spread of the observations.  We call the result of this difference the range of the observation.

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Topic 4: Mode

It must be noted that the Mean is not the only measure of central tendency or the only form of

representative value. For different requirements from data, other measures of central tendencies are used.

• Mode: The mode of a set of observations is the observation that occurs most often.

For example, in the data – 1, 1, 1, 2, 2, 2, 2, 3, 4, 4, “2” occurs the most number of times. So, 2 is the mode of the given observations. 

We have seen how to determine the mode of a short set of data. This form of determining the mode of the given data is not feasible with the large set of data. Now, let us see how we can determine the mode for large data.  

• Mode for Larger Data: In case of larger data, we tabulate the data. Tabulation can begin by putting tally marks and finding the frequency.

Topic 5: Median

Until now, we have seen that in some situations, the arithmetic mean is an appropriate measure of central tendency whereas, in some other situations, mode is the appropriate measure of central tendency.

The median helps to determine the middle of the data when the data set is arranged either in increasing or decreasing values. Thus we say that the median refers to the value which lies in the middle of the data (when arranged in an increasing or decreasing order) with half of the observations above it and the other half below it. 

Thus, in a given data, arranged in ascending or descending order, the median gives us the middle observation.

For n number of values:

Median = (n/2) th term, if n is even

And, Median = ½ × (n + 1) th term, if n is odd

Topic 6: Chance

There are situations in our life, that are certain to happen, some that are impossible and some that may or may not happen. The situation that may or may not happen has a chance of happening.

Topic 7: Probability

Probability is the possibility of an event to happen. Events that have many possibilities can have a probability between 0 and 1. Those which have no chance of happening have probability 0 and those that are bound to happen have probability 1. 

The formula for probability is:

NCERT Solutions of Class 7 Maths Chapter 3: Data Handling- Free PDF Download 

Below we have provided solutions for NCERT Class 7 Maths Chapter 3: Data Handling. Go through for answers to some important questions. 

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Exercise 3.1 Solutions

Q 1. 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

• Which number is the highest? 
• Which number is the lowest?
• What is the range of the data? 
• Find the arithmetic mean.

Solutions. Let us tabulate the given data below. 

• From the table, it is clear that the highest number is 9.
• The lowest number is 1.
• The range of data can be determined by subtracting the lowest value from the highest value. 

Range = Highest Value – Lowest Value

∴ Range = 9 – 1 = 8

•  We know that:

∴ Arithmetic Mean = 100/20 = 5

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Q 2. A cricketer scores the following runs in eight innings:

58, 76, 40, 35, 46, 45, 0, 100.

Find the mean score.

Solutions. The mean scores can be calculated by:

So, the mean score of the cricketer is 50. 

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Exercise 3.2 Solutions

Q 1. The scores in 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?

Solutions. We know that the most frequent value in a set of observations is the mode of the given data set. From the given test scores, it is clear that “20” is the mode of the given data of test scores.

Now, to determine the median, let us arrange the given test scores in increasing order below.

5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25

Now, the median of this organised data is the middle value of the data.

We have scores of 15 students, i.e. the total number of test scores is 15, which is odd. 

Median for odd number of observations = ½ × (n + 1) th term = ½ × (15 + 1) th term

The median of the given test scores = 8 th  term

∴ the median of the given test scores is 20. 

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Q 2. 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 the same?

Solutions. The mean of the given scores can be calculated as:

Thus, mean = 39 .

The mode of a given data is the most frequent value in the data set. By observing the given set of runs, it is clear that the mode of the given data is 15 .

Now, to determine the median, let us arrange the given data in increasing order of their values.

6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120

We have runs scored by 11 cricketers, which is an odd number, i.e. n = 11.

Median for odd number of observations = ½ × (n + 1) th term = ½ × (11 + 1) th term

The median of the given test scores = 6 th  term

∴ median = 15 . 

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Exercise 3.3 Solutions

Q 1. Consider this data collected from a survey of a colony.

• Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
• Which sport is most popular?
• Which is more preferred, watching or participating in sports?

Solutions. The answers to each question are given below.

• The double bar graph for the given data is given below. 
• From observing the bar graph above, it is clear that most people like to watch and participate in cricket. So, cricket is the most popular sport.
• Again, from the double bar graph above, it is clear that people prefer watching sports over participating.

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Exercise 3.4 Solutions

Q 1. Tell whether the following is certain to happen, impossible, can happen but not certain.

• You are older today than yesterday. 
• A tossed coin will land heads up.
• A dice when tossed shall land up with 8 on top.
• The next traffic light seen will be green. 
• Tomorrow will be a cloudy day.

Solutions:  The answers are given below. 

• It is a fact that you are older today than yesterday. So, this is certain to happen.
• Given the two sides of a coin, there are 50-50% chances of landing a heads or a tails. So, the event of landing of a tossed coin into heads can happen but is not certain.
• There are only 6 faces on a dice, numbered from 1 to 6. So, landing a dice with an 8 on top is impossible. 
• Again, just like (b), this event can happen but is not certain.
• This event can happen but not certain. 

Data Handling – Chapter 3 – Introduction – Class 7  

Q 2. There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.

• What is the probability of drawing a marble with number 2?
• What is the probability of drawing a marble with the number 5?

Solutions: The answers are given below. 

• Here the favourable outcome is – drawing a marble with the number 2 while the total number of outcomes is 6 (as there are 6 marbles numbered from 1 to 6). We know that probability can be calculated by:

∴ Probability of drawing a marble with the number 2 on it:

The probability is ⅙. 

• As there is only 1 marble numbered 5, the probability of drawing a marble with the number 5 is:

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Explore Notes of All subjects of CBSE Class 7:-

Check out Maths Class 6 Notes and Exercise Solutions for other chapters below. 

Ans: The most common representative value of a group of data is the arithmetic mean or the mean. The average or Arithmetic Mean (A.M.) or simply mean is calculated by dividing the sum of all observations by the total number of observations. 

Ans: The difference between the highest and the lowest observation gives us an idea of the spread of the observations.  We call the result of this difference the range of the observation.

Ans: The mode of a set of observations is the observation that occurs most often. For example, in the data – 1, 1, 1, 2, 2, 2, 2, 3, 4, 4, “2” occurs the most number of times. So, 2 is the mode of the given observations.

This was all about NCERT Class 7 Maths Chapter 3, Data Handling in which we studied the different properties of integers. Download the NCERT Class 7 Maths Chapter 3 Notes and Solutions PDF to ace your exam preparations. Follow the CBSE Class 7 Maths Solutions and Notes for more such chapter notes and important questions and answers for preparation for CBSE Class 7 Maths. 

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• RD Sharma Solutions
• Chapter 24 Data Handling Iii Construction Of Bar Graphs

RD Sharma Solutions for Class 7 Maths Chapter 24 Data Handling - III (Constructions of Bar Graphs)

The PDF of RD Sharma Solutions for Class 7 Chapter 24 Data Handling – III (Constructions of Bar Graphs) is provided here. The subject experts at BYJU’S outline the concepts in a clear and precise manner based on the grasping abilities of students. The solutions to all questions in RD Sharma Solutions are given here in a detailed and step-by-step way to help students understand more effectively.

By solving this chapter, the students will understand how to draw a bar graph and a double bar graph for the given data. RD Sharma Solutions for Class 7 are given here, which include the answers to all the questions present in this chapter. Let us have a look at some of the concepts that are discussed in this chapter.

• Definition and meaning of bar graphs
• Steps to construct bar graphs
• RD Sharma Solutions Class 7 Maths Chapter 1 Integers
• RD Sharma Solutions Class 7 Maths Chapter 2 Fractions
• RD Sharma Solutions Class 7 Maths Chapter 3 Decimals
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers
• RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers
• RD Sharma Solutions Class 7 Maths Chapter 6 Exponents
• RD Sharma Solutions Class 7 Maths Chapter 7 Algebraic Expressions
• RD Sharma Solutions Class 7 Maths Chapter 8 Linear Equations in One Variable
• RD Sharma Solutions Class 7 Maths Chapter 9 Ratio and Proportion
• RD Sharma Solutions Class 7 Maths Chapter 10 Unitary Method
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage
• RD Sharma Solutions Class 7 Maths Chapter 12 Profit and Loss
• RD Sharma Solutions Class 7 Maths Chapter 13 Simple Interest
• RD Sharma Solutions Class 7 Maths Chapter 14 Lines and Angles
• RD Sharma Solutions Class 7 Maths Chapter 15 Properties of Triangles
• RD Sharma Solutions Class 7 Maths Chapter 16 Congruence
• RD Sharma Solutions Class 7 Maths Chapter 17 Constructions
• RD Sharma Solutions Class 7 Maths Chapter 18 Symmetry
• RD Sharma Solutions Class 7 Maths Chapter 19 Visualising Solid Shapes
• RD Sharma Solutions Class 7 Maths Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures)
• RD Sharma Solutions Class 7 Maths Chapter 21 Mensuration II (Area of Circle)
• RD Sharma Solutions Class 7 Maths Chapter 22 Data Handling I (Collection and Organisation of Data)
• RD Sharma Solutions Class 7 Maths Chapter 23 Data Handling II (Central Values)
• RD Sharma Solutions Class 7 Maths Chapter 24 Data Handling III (Constructions of Bar Graphs)
• RD Sharma Solutions Class 7 Maths Chapter 25 Data Handling IV (Probability)

RD Sharma Solutions for Class 7 Maths Chapter 24 Data Handling – III (Constructions of Bar graphs)

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Access answers to Maths RD Sharma Solutions for Class 7 Chapter 24 – Data Handling – III (Constructions of Bar graphs)

Exercise 24.1 Page No: 24.5

1. Two hundred students of Classes VI and VII were asked to name their favourite colours so as to decide upon what should be the colour of their school house. The results are shown in the following table.

Represent the given data on a bar graph.

(i) Which is the most preferred colour and which is the least?

(ii) How many colours are there in all?

Steps of constructing bar graph:

1. Mark the horizontal axis OX as Name of the Colour and the vertical axis OY as Number of Students.

2. Along the horizontal axis OX, choose bars of uniform (equal) width, with a uniform gap between them.

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 small division to represent 10 students.

(i) The most preferred colour is blue, and the least preferred is green.

(ii) In all, there are 5 colours.

2. Following data gives total marks (out of 600) obtained by six children of a particular class.

Represent the data by a bar graph

1. Mark the horizontal axis OX as Name of the Students and the vertical axis OY as Marks Obtained.

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 small division to represent 100 marks.

3. Number of children in six different classes is given below. Represent the data on a bar graph.

(i) How do you choose the scale.

(ii) Which class has the maximum number of children?

(iii) Which class has the minimum number of children?

1. Mark the horizontal axis OX as Class and the vertical axis OY as Number of Children.

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 40 children.

(i) We choose 1 big to represent 40 children.

(ii)The maximum numbers of students are in class V.

(iii) The minimum number of students are in class X.

4. The performance of students in 1st term and 2nd term is as given below. Draw a double bar graph choosing appropriate scale, and answer the following:

(i) In which subject, has the children improved their performance the most?

(ii) Has the performance gone down in any subject?

1. Mark the horizontal axis OX as Subject and the vertical axis is OY as Marks.

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 10 marks.

(i) In Maths, the students showed their greatest improvement.

(ii) The students performed worst in Hindi

5. Consider the following data gathered from a survey of a colony:

Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?

(i) Which sport is most popular?

(ii) What is more preferred watching or participating in sports?

Steps of constructing graph:

1. Mark the horizontal axis OX as Favourite Sports and the vertical axis OY as Number of People.

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 2 big divisions to represent 400 people.

(i) Cricket is the most popular sport.

(ii)Watching is preferred over participation.

6. The production of saleable steel in some of the steel plants of our country during 1999 is given below:

Construct a bar graph to represent the above data on a graph paper by using the scale 1 big division = 20 thousand tonnes.

1. Mark the horizontal axis OX as Name of the Steel Plant and the vertical axis OY as Production (in thousand tonnes).

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 20 thousand tonnes.

7. The following data gives the number (in thousands) of applicants registered with an Employment Exchange during, 1995-2000:

Construct a bar graph to represent the above data.

1. Mark the horizontal axis OX as Years and the vertical axis OY as Number of Applicants Registered (in thousands).

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 4 thousand applicants.

8. The following table gives the route length (in thousand kilometres) of the Indian Railways in some of the years:

Represent the above data with the help of a bar graph.

1. Mark the horizontal axis OX as Years and the vertical axis OY as Route Length (in thousand kilometres).

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 10000 Km.

9. The following data gives the amount of loans (in crores of rupees) disbursed by a bank during some years:

(i) Represent the above data with the help of a bar graph.

(ii) With the help of the bar graph, indicate the year in which amount of loan is not increased over that of the preceding year.

1. Mark the horizontal axis OX as Years and the vertical axis OY as Loan (in crores of rupees).

2. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 10 crores of rupees.

(ii) In 1995, the loan amount was not increased over that of the preceding year.

10. The following table shows the interest paid by a company (in lakhs):

Draw the bar graph to represent the above information.

1. Mark the horizontal axis OX as Years and the vertical axis OY as Interest (in lakhs of rupees).

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big divisions to represent 5 lakhs rupees.

11. The following data shows the average age of men in various countries in a certain year:

Represent the above information by a bar graph.

1. Mark the horizontal axis OX as Countries and the vertical axis OY as Average Age of men (in years).

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 10 year.

12. The following data gives the production of food grains (in thousand tonnes) for some years:

1. Mark the horizontal axis OX as Years and the vertical axis OY as Production of food grains (in thousand tonnes).

13. The following data gives the amount of manure (in thousand tonnes) manufactured by a company during some years:

(i) Represent the above data with the help of bar graph.

(ii) Indicate with the help of the bar graph the year in which the amount of manure manufactured by the company was maximum.

(iii) Choose the correct alternative:

The consecutive years during which there was maximum decrease in manure production are:

(a) 1994 and 1995 (b) 1992 and 1993

(c) 1996 and 1997 (d) 1995 and 1996

1. Mark the horizontal axis OX as Years and the vertical axis OY as Manure (in thousand tonnes).

3. Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 big division to represent 5 thousand tonnes.

(ii) In the year 1994, the amount of manure manufactured by the company was maximum.

(iii) 1996 and 1997.

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NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF Download

The NCERT Solutions for Class 7 Maths Chapter 3 Data Handling covers all the exercises included in the textbook. By referring to the Chapter 3 Data Handling Maths questions, students can easily understand each and every step so that they can build their problem-solving skills as well as comprehension skills which is a must to score good marks.

The NCERT Solutions for Class 7 Maths Chapter 3 Data Handling are well structured and systematic so that students can get an overall viewpoint of the important questions. By referring to the important questions of Chapter 3 Data Handling, students can improve their confidence level which can be implemented in the Maths final exam. 

NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF

If students are looking for accurate and comprehensive answers then they can prefer utilising the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF. Students generally prefer utilising the NCERT Solutions after solving the Chapter 3 Data Handling Maths questions so that they can match their answers and accordingly they can work upon. Students can refer to the NCERT Solutions of Chapter 3 Data Handling from the Selfstudys from their comfort zone. 

How to Download the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling?

To download the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling, students need to follow the given steps, those steps are: 

• Visit the Selfstudys website. 

• Bring the arrow towards NCERT Books & Solutions. 

• A drop down menu will appear, select NCERT Solutions from the list. 

• A new page will appear, select Class 7 from the list of classes. 
• Click Maths from the list of classes. 
• Again a new page will appear, select Chapter 3 Data Handling from the list of classes. 

Features of NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF

It is advisable for students to know everything about the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF; those important features are discussed below: 

• Diagrams and Illustrations are Included: In the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling, diagrams and illustrations are included like wherever there is a need so that students can visualise the concepts well. 
• Answers are Included: Inside the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling revision, answers of all the NCERT questions are included so that students can understand the steps easily. 
• Provides Practise Questions: Students can easily practise questions from the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling theory as it is already available in it. By practising questions of Class 7 Maths, students can easily improve their problem solving skills. 
• Easy to Understand the Language: The Class 7 Maths language used in the NCERT Solutions are very easy to understand so that students can easily understand the questions and answers well. 
• All Topics are Covered: In the NCERT Class 7 Maths Solutions, all the topics of Chapter 3 Data Handling are covered so that students don’t miss out on any of the topics. 
• Step Wise Solutions are Given: All the questions of Chapter 3 Data Handling of Class 7 Maths are explained in a stepwise manner so that students can easily understand each and every step. 

What Are the Advantages of NCERT Solutions for Class 7 Maths Chapter 3 Data Handling?

The NCERT Solutions for Class 7 Maths Chapter 3 Data Handling provides several benefits to students; some of the benefits are discussed below: 

• Covered in a Comprehensive Way: Topics in the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling revision are covered in a comprehensive way so that students can understand all the chapters in a proper way. 
• Reduces Number of Mistakes: By practising more and more questions from the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling theory, students can learn to reduce the number of mistakes then and there. 
• Understands Concepts Better: By referring to the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF, students can understand the concepts in a better way.
• Make More Confidence: Solving maximum number of Class 7 Maths questions from the NCERT Solutions can help students to improve their problem solving skills and accordingly they can improve their confidence level. 
• Can Maintain Accuracy: Accuracy is considered to be closeness of the answers; students can easily maintain the accuracy by solving questions of Chapter 3 Data Handling from the NCERT Class 7 Maths Solutions. 
• Improves Problem Solving Skills: It is the ability to identify the problem and to analyse the answers so that students can score well; students can maintain their accuracy level by solving Chapter 3 Data Handling questions from the NCERT Class 7 Maths Solutions. 

How to Utilise NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF in an Effective Way? 

Here are some tips for students to follow so that they can utilise the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF in an effective way; those tips are discussed below: 

• Read the Chapter: Before attempting the questions from the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling, students need to read the chapter in a proper way so that they can understand the concepts and formulas in a better way. 
• Solve the Problems On Your Own: Students are advised to solve the problems of NCERT Solutions for Class 7 Maths Chapter 3 Data Handling theory so that they can know their strengths and weaknesses.  
• Check the Answers: After attempting questions from the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF, students need to check the answers. This will help students to understand the mistakes made while solving questions of Maths Chapter 3 Data Handling so that they can understand the correct method. 
• Understand the Logic Behind the Each Step: While referring to the NCERT Class 7 Maths Solutions, students should make sure that they have understood the logic behind each and every step. 
• Practise Questions on a Regular Basis: Students need to solve more questions of Chapter 3 Data Handling Maths from Class 7 NCERT Solutions on a regular basis so that they can retain it in their memory. 
• Revise Regularly: It is advisable for students to revise the Chapter 3 Data Handling topics and concepts on a regular basis so that students can solve questions from the NCERT Class 7 Maths Solutions. 

Who Can Benefit from NCERT Solutions for Class 7 Maths Chapter 3 Data Handling?

The NCERT Solutions for Class 7 Maths Chapter 3 Data Handling can benefit wide range of students; those are: 

• Class 7 Students: The NCERT Solutions for Class 7 Maths Chapter 3 Data Handling revision are designed in such a way that students can understand the topics in a better way and accordingly they can solve the problems on their own. 
• Competitive Exam Aspirants: Students who are preparing for competitive exam aspirants can go through the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling theory so that they can build a strong foundation. 
• Teachers and Tutors: Teachers and tutors can also utilise the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF so that they can easily understand the concepts without any complexity. The answers of Class 7 Maths questions are explained in a step by step approach so that students can understand it better. 
• Students Who Need Additional Practice: Students who want extra practice of Class 7 Maths Chapter 3 Data Handling as well as a strong foundation can use the NCERT Solutions so that they can build their own strategy. 
• Students Struggling With The Chapter: Students who find Chapter 3 Data Handling of Class 7 Maths difficult can utilise the NCERT Solutions so that they can eliminate the difficulty. 
• Students Who Need Self Study: The NCERT Class 7 Maths Solutions are considered to be the best study resources as it helps students in studying on their own. 

When Is the Best Time to Use NCERT Solutions for Class 7 Maths Chapter 3 Data Handling?

The best time to use the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling really depends on the student’s preferences and student’s way of learning; some of the scenarios are discussed below: 

• After Completing the Chapter: Once students have the basic understanding of the Chapter 3 Data Handling in the Class 7 Maths syllabus, they can start practising questions from the NCERT Solutions.  
• After Attempting the Questions: After attempting the Chapter 3 Data Handling questions on their own, students can refer to the NCERT Class 7 Maths Solutions on their own so that they can know the correct method of solving. 
• While Preparing: The NCERT Solutions for Class 7 Maths Chapter 3 Data Handling revision can be utilised while preparing so that students can strengthen their understanding level.
• To Revise: Students can prefer utilising the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling theory so that they can quickly revise the concepts. 
• To Complete Homework: If students want to complete the Maths Chapter 3 Data Handling from the Class 7, then they can prefer utilising the NCERT Solutions. 
• To Improve Problem Solving Skills: If in case, a student's problem solving skills is weak then he or she can prefer utilising the NCERT Solutions for Class 7 Maths Chapter 3 Data Handling PDF. 

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Case Study Questions for Class 7 Maths

• Last modified on: 7 months ago
• Reading Time: 7 Minutes

Table of Contents

Here in this article, we are providing case study questions for class 7 maths.

Maths Class 7 Chapter List

Latest chapter list (2023-24).

There is total 13 chapters.

Chapter 1 Integers Case Study Questions Chapter 2 Fractions and Decimals Case Study Questions Chapter 3 Data Handling Case Study Questions Chapter 4 Simple Equations Case Study Questions Chapter 5 Lines and Angles Case Study Questions Chapter 6 The Triangles and its Properties Case Study Questions Chapter 7 Comparing Quantities Case Study Questions Chapter 8 Rational Numbers Case Study Questions Chapter 9 Perimeter and Area Case Study Questions Chapter 10 Algebraic Expressions Case Study Questions Chapter 11 Exponents and Powers Case Study Questions Chapter 12 Symmetry Case Study Questions Chapter 13 Visualising Solid Shapes Case Study Questions

Old Chapter List

Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangles 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

Deleted Chapter:

• Chapter 7 Congruence of Triangles
• Chapter 10 Practical Geometry

Tips for Answering Case Study Questions for Class 7 Maths in Exam

1. Comprehensive Reading for Context: Prioritize a thorough understanding of the provided case study. Absorb the contextual details and data meticulously to establish a strong foundation for your solution.

2. Relevance Identification: Pinpoint pertinent mathematical concepts applicable to the case study. By doing so, you can streamline your thinking process and apply appropriate methods with precision.

3. Deconstruction of the Problem: Break down the complex problem into manageable components or steps. This approach enhances clarity and facilitates organized problem-solving.

4. Highlighting Key Data: Emphasize critical information and data supplied within the case study. This practice aids quick referencing during the problem-solving process.

5. Application of Formulas: Leverage pertinent mathematical formulas, theorems, and principles to solve the case study. Accuracy in formula selection and unit usage is paramount.

6. Transparent Workflow Display: Document your solution with transparency, showcasing intermediate calculations and steps taken. This not only helps track progress but also offers insight into your analytical process.

7. Variable Labeling and Definition: For introduced variables or unknowns, offer clear labels and definitions. This eliminates ambiguity and reinforces a structured solution approach.

8. Step Explanation: Accompany each step with an explanatory note. This reinforces your grasp of concepts and demonstrates effective application.

9. Realistic Application: When the case study pertains to real-world scenarios, infuse practical reasoning and logic into your solution. This ensures alignment with real-life implications.

10. Thorough Answer Review: Post-solving, meticulously review your answer for accuracy and coherence. Assess its compatibility with the case study’s context.

11. Solution Recap: Before submission, revisit your solution to guarantee comprehensive coverage of the problem and a well-organized response.

12. Previous Case Study Practice: Boost your confidence by practicing with past case study questions from exams or textbooks. This familiarity enhances your readiness for the question format.

13. Efficient Time Management: Strategically allocate time for each case study question based on its complexity and the overall exam duration.

14. Maintain Composure and Confidence: Approach questions with poise and self-assurance. Your preparation equips you to conquer the challenges presented.

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Data Handling Ncert Solution for Class 7 Maths Chapter 3 Exercise 3.2

Data Handling Class 7 Ex. 3.2;

New Ncert Solution for Class 7 Maths Chapter 3 Data Handling Free Solution;

Exercise 3.2

Question 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?

Solution 1:- Given scores-data: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Taking arrangement of the given data in increasing order 5, 9, 10, 12, 15, 16 , 19, 20, 20, 20, 20, 23, 24, 25, 25

The-mode (a set of observations), is the observation that occurs, most-often. Since, 20 occurs 4-times (highest) So, Mode = 20 n = 15 (odd)    [ where, n = number of observation.]

the median, gives us the middle observation.

∴ Median = (n+1)/2 th term = (15+1)/2 = 8th term = 20 So, median = 20 and mode = 20 Finally, Mode and median are same as 20.

Question 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?

Solution 2:- Given data: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15

Taking arrangment of the given data in increasing order, we get

6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120,

Here, 15 occurs 3 times (highest) ∴ Mode = 15 n = 11 (odd) ∴ Median = (11+1)/2 th term = 6th term = 15 Thus mean = 39, mode = 15 and median = 15

No, the mean, mode and median are not given the same data.

Question 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? Solution 3:- Given data: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47

Arranging in increasing order, we get 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50,

(i) Here, 38 and 43 occur 3 times (highest) Thus mode = 38 and 43 Since, n = 15(odd)  [where, n = number of observation data.] Median = (n+1)/2 th term =(15+1)/2 th term = 8th term = 40 Thus mode 38 and 43 and median = 40

(ii) Yes, the given data has two modes i.e. 38 and 43.

Question 4 :- Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14 Solution 4:- 13, 16, 12, 14, 19, 12, 14, 13, 14 Arranging the given data in increasing order, we get

12, 12, 13, 13, 14, 14, 14, 16, 19

Here, 14 occur 3 times (highest) Thus, mode = 14 n = 9(odd)      [where, n = number of observation.] ∴ Median = (n+1)/2 th  term = (9+1)/2 th  term = 5th term = 14 Hence, mode = 14 and median = 14.

Question 5 :- Tell whether the statement is true or false. (i) The mode is always one of the number 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 5:- (i) True (ii) False (iii) True (iv) False.

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