case study 1 class 9 maths

Class 9 Maths Case Study Questions of Chapter 1 Real Numbers

• Post author: studyrate
• Post published:
• Post category: class 9th
• Post comments: 0 Comments

Case study Questions in Class 9 Mathematics Chapter 1  are very important to solve for your exam. Class 9 Maths Chapter 1 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 9 Maths Case Study Questions  Chapter 1 Real Numbers

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 9 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Real Numbers Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 1 Real Numbers

Case Study/Passage-Based Questions

Case Study 1: A Mathematics Exhibition is being conducted in your school and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.

Observe the following factor tree and answer the following:

1. What will be the value of x?

Answer: b) 13915

2. What will be the value of y?

Answer: c) 11

3. What will be the value of z?

Answer: b) 23

4. According to the Fundamental Theorem of Arithmetic 13915 is a

a) Composite number

b) Prime number

c) Neither prime nor composite

d) Even number

Answer: a) Composite number

5. The prime factorization of 13915 is

a) 5 × 11 3  × 13 2

b) 5 × 11 3  × 23 2

c) 5 × 11 2  × 23

d) 5 × 11 2  × 13 2

Answer: c) 5 × 112 × 23

Case Study 2: Srikanth has made a project on real numbers, where he finely explained the applicability of exponential laws and divisibility conditions on real numbers. He also included some assessment questions at the end of his project as listed below. Answer them.

(i) For what value of n, 4 n  ends in 0?

(a) 10 (b) when n is even (c) when n is odd (d) no value of n

Answer: (d) no value of n3

(ii) If a is a positive rational number and n is a positive integer greater than 1, then for what value of n, an is a rational number?

(a) when n is any even integer (b) when n is any odd integer (c) for all n > 1 (d) only when n=0

Answer: (c) for all n > 1

(iii) If x and y are two odd positive integers, then which of the following is true?

(a) x 2 +y 2  is even (b) x 2 +y 2  is not divisible by 4 (c) x 2 +y 2   is odd (d) both (a) and (b)

Answer: (d) both (a) and (b)

(iv) The statement ‘One of every three consecutive positive integers is divisible by 3’ is

(a) always true (b) always false (c) sometimes true (d) None of these

Answer:(a) always true

(v) If n is any odd integer, then n 2 – 1 is divisible by

(a) 22 (b) 55 (c) 88 (d) 8

Answer: (d) 8

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Mathematics Chapter 1 Real Numbers with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Maths Real Numbers Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

You Might Also Like

Class 9 maths case study questions of chapter 15 probability pdf download, mcq questions of class 9 social science history chapter 8 clothing a social history with answers, mcq questions of class 9 maths chapter 13 surface areas and volumes with answers, leave a reply cancel reply.

Save my name, email, and website in this browser for the next time I comment.

myCBSEguide

• Mathematics
• CBSE Class 9 Mathematics...

CBSE Class 9 Mathematics Case Study Questions

Table of Contents

myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!

Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

• Now he told Raju to draw another line CD as in the figure
• The teacher told Ajay to mark  ∠ AOD  as 2z
• Suraj was told to mark  ∠ AOC as 4y
• Clive Made and angle  ∠ COE = 60°
• Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

• 2y + z = 90°
• 2y + z = 180°
• 4y + 2z = 120°
• (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

• (a) 31.6 m²
• (c) 513.3 m³
• (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

Mycbseguide: blessing in disguise.

Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

Test Generator

Create question paper PDF and online tests with your own name & logo in minutes.

Question Bank, Mock Tests, Exam Papers, NCERT Solutions, Sample Papers, Notes

Related Posts

• Competency Based Learning in CBSE Schools
• Class 11 Physical Education Case Study Questions
• Class 11 Sociology Case Study Questions
• Class 12 Applied Mathematics Case Study Questions
• Class 11 Applied Mathematics Case Study Questions
• Class 11 Mathematics Case Study Questions
• Class 11 Biology Case Study Questions
• Class 12 Physical Education Case Study Questions

14 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

MATHS PAAGAL HAI

All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.

Where is search ? bar

maths is love

Can I have more questions without downloading the app.

I love math

Hello l am Devanshu chahal and l am an entorpinior. I am started my card bord business and remanded all the existing things this all possible by math now my business is 120 crore and my business profit is 25 crore in a month. l find the worker team because my business is going well Thanks

Leave a Comment

Save my name, email, and website in this browser for the next time I comment.

• Andhra Pradesh
• Chhattisgarh
• West Bengal
• Madhya Pradesh
• Maharashtra
• Jammu & Kashmir
• NCERT Books 2022-23
• NCERT Solutions
• NCERT Notes
• NCERT Exemplar Books
• NCERT Exemplar Solution
• States UT Book
• School Kits & Lab Manual
• NCERT Books 2021-22
• NCERT Books 2020-21
• NCERT Book 2019-2020
• NCERT Book 2015-2016
• RD Sharma Solution
• TS Grewal Solution
• DK Goel Solution
• TR Jain Solution
• Selina Solution
• Frank Solution
• ML Aggarwal Solution
• Lakhmir Singh and Manjit Kaur Solution
• I.E.Irodov solutions
• ICSE - Goyal Brothers Park
• ICSE - Dorothy M. Noronhe
• Sandeep Garg Textbook Solution
• Micheal Vaz Solution
• S.S. Krotov Solution
• Evergreen Science
• KC Sinha Solution
• ICSE - ISC Jayanti Sengupta, Oxford
• ICSE Focus on History
• ICSE GeoGraphy Voyage
• ICSE Hindi Solution
• ICSE Treasure Trove Solution
• Thomas & Finney Solution
• SL Loney Solution
• SB Mathur Solution
• P Bahadur Solution
• Narendra Awasthi Solution
• MS Chauhan Solution
• LA Sena Solution
• Integral Calculus Amit Agarwal Solution
• IA Maron Solution
• Hall & Knight Solution
• Errorless Solution
• Pradeep's KL Gogia Solution
• OP Tandon Solutions
• Sample Papers
• Previous Year Question Paper
• Value Based Questions
• CBSE Syllabus
• CBSE MCQs PDF
• Assertion & Reason
• New Revision Notes
• Revision Notes
• HOTS Question
• Marks Wise Question
• Toppers Answer Sheets
• Exam Paper Aalysis
• Concept Map
• CBSE Text Book
• Additional Practice Questions
• Vocational Book
• CBSE - Concept
• KVS NCERT CBSE Worksheets
• Formula Class Wise
• Formula Chapter Wise
• JEE Crash Course
• JEE Previous Year Paper
• Important Info
• JEE Mock Test
• JEE Sample Papers
• SRM-JEEE Mock Test
• VITEEE Mock Test
• BITSAT Mock Test
• Manipal Engineering Mock Test
• AP EAMCET Previous Year Paper
• COMEDK Previous Year Paper
• GUJCET Previous Year Paper
• KCET Previous Year Paper
• KEAM Previous Year Paper
• Manipal Previous Year Paper
• MHT CET Previous Year Paper
• WBJEE Previous Year Paper
• AMU Previous Year Paper
• TS EAMCET Previous Year Paper
• SRM-JEEE Previous Year Paper
• VITEEE Previous Year Paper
• BITSAT Previous Year Paper
• UPSEE Previous Year Paper
• CGPET Previous Year Paper
• CUSAT Previous Year Paper
• AEEE Previous Year Paper
• Crash Course
• Previous Year Paper
• NCERT Based Short Notes
• NCERT Based Tests
• NEET Sample Paper
• Previous Year Papers
• Quantitative Aptitude
• Numerical Aptitude Data Interpretation
• General Knowledge
• Mathematics
• Agriculture
• Accountancy
• Business Studies
• Political science
• Enviromental Studies
• Mass Media Communication
• Teaching Aptitude
• NAVODAYA VIDYALAYA
• SAINIK SCHOOL (AISSEE)
• Mechanical Engineering
• Electrical Engineering
• Electronics & Communication Engineering
• Civil Engineering
• Computer Science Engineering
• CBSE Board News
• Scholarship Olympiad
• School Admissions
• Entrance Exams
• All Board Updates
• Miscellaneous
• State Wise Books
• Engineering Exam

CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

SHARING IS CARING If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.

Each question has five sub-questions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation.

CBSE Class 9 All Students can also Download here Class 9 Other Study Materials in PDF Format.

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 10 Maths
• CBSE Syllabus 2023-24
• Social Media Channels
• Login Customize Your Notification Preferences
• CBSE Class 9th 2023-24 : Science Practical Syllabus; Download PDF 19 April, 2023, 4:52 pm
• CBSE Class 9 Maths Practice Book 2023 (Released By CBSE) 23 March, 2023, 6:16 pm
• CBSE Class 9 Science Practice Book 2023 (Released By CBSE) 23 March, 2023, 5:56 pm
• CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF 10 February, 2023, 6:20 pm
• CBSE Class 9th Maths 2023 : Important Assertion Reason Question with Solution Download Pdf 9 February, 2023, 12:16 pm
• CBSE Class 9th Exam 2023 : Social Science Most Important Short Notes; Download PDF 16 January, 2023, 4:29 pm
• CBSE Class 9th Mathematics 2023 : Most Important Short Notes with Solutions 27 December, 2022, 6:05 pm
• CBSE Class 9th English 2023 : Chapter-wise Competency-Based Test Items with Answer; Download PDF 21 December, 2022, 5:16 pm
• CBSE Class 9th Science 2023 : Chapter-wise Competency-Based Test Items with Answers; Download PDF 20 December, 2022, 5:37 pm

• Second click on the toggle icon

Provide prime members with unlimited access to all study materials in PDF format.

Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding.

Provide prime users with access to exclusive PDF study materials that are not available to regular users.

(Education-Your Door To The Future)

CBSE Class 9 Maths Most Important Case Study Based Questions With Solution

Cbse class 9 mathematics case study questions.

In this post I have provided CBSE Class 9 Maths Case Study Based Questions With Solution. These questions are very important for those students who are preparing for their final class 9 maths exam.

All these questions provided in this article are with solution which will help students for solving the problems. Dear students need to practice all these questions carefully with the help of given solutions.

As you know CBSE Class 9 Maths exam will have a set of cased study based questions in the form of MCQs. CBSE Class 9 Maths Question Bank given in this article can be very helpful in understanding the new format of questions for new session.

All Of You Can Also Read

Case studies in class 9 mathematics.

The Central Board of Secondary Education (CBSE) has included case study based questions in the Class 9 Mathematics paper in current session. According to new pattern CBSE Class 9 Mathematics students will have to solve case based questions. This is a departure from the usual theoretical conceptual questions that are asked in Class 9 Maths exam in this year.

Each question provided in this post has five sub-questions, each followed by four options and one correct answer. All CBSE Class 9th Maths Students can easily download these questions in PDF form with the help of given download Links and refer for exam preparation.

There is many more free study materials are available at Maths And Physics With Pandey Sir website. For many more books and free study material all of you can visit at this website.

Given Below Are CBSE Class 9th Maths Case Based Questions With Their Respective Download Links.

CBSE Case Study Questions for Class 9 Maths - Pdf PDF Download

Cbse case study questions for class  9 maths.

CBSE Case Study Questions for Class 9 Maths are a type of assessment where students are given a real-world scenario or situation and they need to apply mathematical concepts to solve the problem. These types of questions help students to develop their problem-solving skills and apply their knowledge of mathematics to real-life situations.

Chapter Wise Case Based Questions for Class 9 Maths

The CBSE Class 9 Case Based Questions can be accessed from Chapetrwise Links provided below:

Chapter-wise case-based questions for Class 9 Maths are a set of questions based on specific chapters or topics covered in the maths textbook. These questions are designed to help students apply their understanding of mathematical concepts to real-world situations and events.

Chapter 1: Number System

• Case Based Questions: Number System

Chapter 2: Polynomial

• Case Based Questions: Polynomial

Chapter 3: Coordinate Geometry

• Case Based Questions: Coordinate Geometry

Chapter 4: Linear Equations

• Case Based Questions: Linear Equations - 1
• Case Based Questions: Linear Equations -2

Chapter 5: Introduction to Euclid’s Geometry

• Case Based Questions: Lines and Angles

Chapter 7: Triangles

• Case Based Questions: Triangles

Chapter 8: Quadrilaterals

• Case Based Questions: Quadrilaterals - 1
• Case Based Questions: Quadrilaterals - 2

Chapter 9: Areas of Parallelograms

• Case Based Questions: Circles

Chapter 11: Constructions

• Case Based Questions: Constructions

Chapter 12: Heron’s Formula

• Case Based Questions: Heron’s Formula

Chapter 13: Surface Areas and Volumes

• Case Based Questions: Surface Areas and Volumes

Chapter 14: Statistics

• Case Based Questions: Statistics

Chapter 15: Probability

• Case Based Questions: Probability

Weightage of Case Based Questions in Class 9 Maths

Why are Case Study Questions important in Maths Class  9?

• Enhance critical thinking:  Case study questions require students to analyze a real-life scenario and think critically to identify the problem and come up with possible solutions. This enhances their critical thinking and problem-solving skills.
• Apply theoretical concepts:  Case study questions allow students to apply theoretical concepts that they have learned in the classroom to real-life situations. This helps them to understand the practical application of the concepts and reinforces their learning.
• Develop decision-making skills:  Case study questions challenge students to make decisions based on the information provided in the scenario. This helps them to develop their decision-making skills and learn how to make informed decisions.
• Improve communication skills:  Case study questions often require students to present their findings and recommendations in written or oral form. This helps them to improve their communication skills and learn how to present their ideas effectively.
• Enhance teamwork skills:  Case study questions can also be done in groups, which helps students to develop teamwork skills and learn how to work collaboratively to solve problems.

In summary, case study questions are important in Class 9 because they enhance critical thinking, apply theoretical concepts, develop decision-making skills, improve communication skills, and enhance teamwork skills. They provide a practical and engaging way for students to learn and apply their knowledge and skills to real-life situations.

Class 9 Maths Curriculum at Glance

The Class 9 Maths curriculum in India covers a wide range of topics and concepts. Here is a brief overview of the Maths curriculum at a glance:

• Number Systems:  Students learn about the real number system, irrational numbers, rational numbers, decimal representation of rational numbers, and their properties.
• Algebra:  The Algebra section includes topics such as polynomials, linear equations in two variables, quadratic equations, and their solutions.
• Coordinate Geometry:  Students learn about the coordinate plane, distance formula, section formula, and slope of a line.
• Geometry:  This section includes topics such as Euclid’s geometry, lines and angles, triangles, and circles.
• Trigonometry: Students learn about trigonometric ratios, trigonometric identities, and their applications.
• Mensuration: This section includes topics such as area, volume, surface area, and their applications.
• Statistics and Probability:  Students learn about measures of central tendency, graphical representation of data, and probability.

The Class 9 Maths curriculum is designed to provide a strong foundation in mathematics and prepare students for higher education in the field. The curriculum is structured to develop critical thinking, problem-solving, and analytical skills, and to promote the application of mathematical concepts in real-life situations. The curriculum is also designed to help students prepare for competitive exams and develop a strong mathematical base for future academic and professional pursuits.

Students can also access Case Based Questions of all subjects of CBSE Class 9

• Case Based Questions for Class 9 Science
• Case Based Questions for Class 9 Social Science
• Case Based Questions for Class 9 English
• Case Based Questions for Class 9 Hindi
• Case Based Questions for Class 9 Sanskrit

Frequently Asked Questions (FAQs) on Case Based Questions for Class 9 Maths

What is case-based questions.

Case-Based Questions (CBQs) are open-ended problem solving tasks that require students to draw upon their knowledge of Maths concepts and processes to solve a novel problem. CBQs are often used as formative or summative assessments, as they can provide insights into how students reason through and apply mathematical principles in real-world problems.

What are case-based questions in Maths?

Case-based questions in Maths are problem-solving tasks that require students to apply their mathematical knowledge and skills to real-world situations or scenarios.

What are some common types of case-based questions in class 9 Maths?

Common types of case-based questions in class 9 Maths include word problems, real-world scenarios, and mathematical modeling tasks.

Top Courses for Class 9

Faqs on cbse case study questions for class 9 maths - pdf, past year papers, mock tests for examination, cbse case study questions for class 9 maths - pdf, important questions, practice quizzes, sample paper, viva questions, semester notes, shortcuts and tricks, extra questions, previous year questions with solutions, objective type questions, video lectures, study material.

CBSE Case Study Questions for Class 9 Maths - Pdf Free PDF Download

Importance of cbse case study questions for class 9 maths - pdf, cbse case study questions for class 9 maths - pdf notes, cbse case study questions for class 9 maths - pdf class 9, study cbse case study questions for class 9 maths - pdf on the app, welcome back, create your account for free.

Forgot Password

Unattempted tests, change country.

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Maths
• Chapter 1: Number Systems
• Exercise 1.1

NCERT Solutions for class 9 Maths Chapter 1 - Number Systems Exercise 1.1

NCERT Solutions Class 9 Maths Chapter 1 Number Systems Exercise 1.1 are provided here. Our subject experts have prepared the NCERT Maths solutions for Class 9 chapter-wise so that it helps students to solve problems easily while using it as a reference. They also focus on creating solutions for these exercises in such a way that it is easy to understand for the students.

The first exercise in Number Systems Exercise 1.1 is the introduction. They provide a detailed and stepwise explanation of each answer to the questions given in the exercises in the NCERT textbook for Class 9. The NCERT Solutions are always prepared by following NCERT guidelines so that it covers the whole syllabus accordingly. These are very helpful in scoring well in CBSE examinations.

NCERT Solutions for Class 9 Maths Chapter 1- Number Systems Exercise 1.1

carouselExampleControls111

Previous Next

Exercise 1.2 Solutions 4 Questions (3 long and 1 short)

Exercise 1.3 Solutions 9 Questions (9 long)

Exercise 1.4 Solutions 2 Questions (2 long)

Exercise 1.5 Solutions 5 Questions (4 long and 1 short)

Exercise 1.6 Solutions 3 Questions (3 long)

Access Answers to Maths NCERT Class 9 Chapter 1 – Number Systems Exercise 1.1

1. Is zero a rational number? Can you write it in the form p/q where p and q are integers and q ≠ 0?

We know that a number is said to be rational if it can be written in the form p/q , where p and q are integers and q ≠ 0.

Taking the case of ‘0’,

Zero can be written in the form 0/1, 0/2, 0/3 … as well as , 0/1, 0/2, 0/3 ..

Since it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be positive or negative number.

Hence, 0 is a rational number.

2. Find six rational numbers between 3 and 4.

There are infinite rational numbers between 3 and 4.

As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6+1 = 7 (or any number greater than 6)

i.e., 3 × (7/7) = 21/7

and, 4 × (7/7) = 28/7. The numbers between 21/7 and 28/7 will be rational and will fall between 3 and 4.

Hence, 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 are the 6 rational numbers between 3 and 4.

3. Find five rational numbers between 3/5 and 4/5.

There are infinite rational numbers between 3/5 and 4/5.

To find out 5 rational numbers between 3/5 and 4/5, we will multiply both the numbers 3/5 and 4/5

with 5+1=6 (or any number greater than 5)

i.e., (3/5) × (6/6) = 18/30

and, (4/5) × (6/6) = 24/30

The numbers between18/30 and 24/30 will be rational and will fall between 3/5 and 4/5.

Hence, 19/30, 20/30, 21/30, 22/30, 23/30 are the 5 rational numbers between 3/5 and 4/5

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

Natural numbers- Numbers starting from 1 to infinity (without fractions or decimals)

i.e., Natural numbers= 1,2,3,4…

Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)

i.e., Whole numbers= 0,1,2,3…

Or, we can say that whole numbers have all the elements of natural numbers and zero.

Every natural number is a whole number; however, every whole number is not a natural number.

(ii) Every integer is a whole number.

Integers- Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal numbers.

i.e., integers= {…-4,-3,-2,-1,0,1,2,3,4…}

i.e., Whole numbers= 0,1,2,3….

Hence, we can say that integers include whole numbers as well as negative numbers.

Every whole number is an integer; however, every integer is not a whole number.

(iii) Every rational number is a whole number.

Rational numbers- All numbers in the form p/q, where p and q are integers and q≠0.

i.e., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7…

All whole numbers are rational; however, all rational numbers are not whole numbers.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.1 is the first exercise of Chapter 1 of Class 9 Maths. This exercise explains how to find rational numbers between two given numbers.

Key Features of NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems Exercise 1.1

• These NCERT Solutions help you solve and revise all questions of Exercise 1.1.
• After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
• It follows NCERT guidelines which help in preparing the students accordingly.
• It contains all the important questions from the examination point of view.

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Counselling

Talk to our experts

1800-120-456-456

• CBSE Class 9 Maths Important Questions for Chapter 1 - Number System

CBSE Class 9 Maths Important Questions Chapter 1 - Number System Free PDF Download

Chapter 1 of Mathematics Class 9 deals with an introduction to various other topics. Those who are planning to pursue a career in mathematics should prepare well for this chapter. Mathematics is the subject to deal with practical life calculations and Class 9 Maths Chapter 1 Important Questions will help set a good base for the students.

Based on these crucial questions, students can prepare for mathematics finals without any hassle. Class 9 is the base to prepare well for 10th boards. Hence students need to master their concepts and utilise their time efficiently. According to CBSE’s basic guidelines, these Important Questions for Class 9 Maths Chapter 1 Number System are prepared.

As a result, students do not need to be concerned or go elsewhere for answers to such inquiries. The first chapter of Mathematics 9th standard is Number System, which deals with whole numbers, rational and irrational numbers, and integers. Read about the topics included in the Crucial Questions for Class 9 Mathematics number system and comprehensive curriculum created by specialists in detail. Vedantu offers students a free PDF download option for all updated CBSE textbook solutions. Topics such as Science, Math, and English will become easier to study if you have access to NCERT Class 9 Science, Math, and other answers that are only accessible on Vedantu.

Download CBSE Class 9 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 9 Maths Important Questions for other chapters:

Important Topics Covered in Class 9 Maths Chapter 1

Introduction to number system

Irrational Number

Real Number and Their Decimal Expansion

Representation of Real Number on Number Line

Operations on Real Number

Laws of Exponents for Real Number

Study Important Questions for Class 9 Maths Chapter 1 – Number Systems

1 Marks Questions

1. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is irrational number.

Ans: We know that the square root of every positive integer will not yield an integer.

We know that \[\sqrt{4}\] is $2$, which is an integer. But, $\sqrt{7}$ or $\sqrt{10}$ will give an irrational number.

Therefore, we conclude that the square root of every positive integer is not an irrational number.

2. Write three numbers whose decimal expansions are non-terminating non-recurring.

Ans: The three numbers that have their expansions as non-terminating on recurring decimals are given below.

0.04004000400004.... 

0.07007000700007....

0.13001300013000013.... 

3. Find three different irrational numbers between the rational numbers $\frac{\text{5}}{\text{11}}$ and $\frac{\text{9}}{\text{11}}$ .

Ans: Let us convert $\frac{5}{11}$ and $\frac{9}{11}$ into decimal form, to get

$\frac{5}{7}=0.714285....and\frac{9}{11}=0.818181....$

Three irrational numbers that lie between $0.714285....$ and $0.818181....$ are:

0.73073007300073...

0.74074007400074.... 

0.76076007600076.... 

4. Which of the following rational numbers have terminating decimal representation?

 $(i)\frac{3}{5}$             

$(ii)\frac{2}{13} $

$(iii)\frac{40}{27}   $    

$(iv)\frac{23}{7}$

Ans: $(i)\frac{3}{5}$

5. How many rational numbers can be found between two distinct rational numbers?

(iv) Infinite

Ans: (iv) Infinite

6. The value of $\left( \text{2+}\sqrt{\text{3}} \right)\left( \text{2-}\sqrt{\text{3}} \right)$ in

(i) $\text{1}$

(ii) $\text{-1}$

(iii) $\text{2}$

(iv) none of these

Ans: (i) $1$

7. ${{\left( \text{27} \right)}^{\text{-2/3}}}$ is equal to

(i) $\text{9}$

(ii) $\text{1/9}$

(iii) $\text{3}$

Ans: (ii) $1/9$

8. Every natural number is

(i) not an integer

(ii) always a whole number

(iii) an irrational number

(iv) not a fraction

Ans: (ii) always a whole number

9. Select the correct statement from the following

(i) $\frac{\text{7}}{\text{9}}\text{}\frac{\text{4}}{\text{5}}$

(ii) $\frac{\text{2}}{\text{6}}\text{}\frac{\text{3}}{\text{9}}$

(iii) $\frac{\text{-2}}{\text{3}}\text{}\frac{\text{-4}}{\text{5}}$

(iv)$\frac{\text{-5}}{\text{7}}\text{}\frac{\text{-3}}{\text{4}}$

Ans: (iii) $\frac{-2}{3}>\frac{-4}{5}$

10. $\text{7}\text{.}\overline{\text{2}}$ is equal to

(i) $\frac{\text{68}}{\text{9}}$

(ii) $\frac{\text{64}}{\text{9}}$

(iii) $\frac{\text{65}}{\text{9}}$

(iv) $\frac{\text{63}}{\text{9}}$

Ans: (iii) $\frac{65}{9}$

11. $\text{0}\text{.83458456}......$ is

(i) an irrational number

(ii) rational number

(iii) a natural number

(iv) a whole number

Ans: (i) an irrational number

12. A terminating decimal is

(i) a natural number

(ii) a rational number

(iii) a whole number

(iv) an integer.

Ans: (ii) a rational number

13. The $\frac{\text{p}}{\text{q}}$ form of the number $\text{0}\text{.8}$ is

(i) $\frac{\text{8}}{\text{10}}$

(ii) $\frac{\text{8}}{\text{100}}$

(iii) $\frac{\text{1}}{\text{8}}$

(iv) $\text{1}$

Ans: (i) $\frac{8}{10}$

14. The value of $\sqrt[\text{3}]{\text{1000}}$ is

Ans: (ii) $10$

15. The sum of rational and an irrational number

(i) may be natural

(ii) may be irrational

(iii) is always irrational

(iv) is always rational

Ans: (iii) is always rational

16. The rational number not lying between $\frac{\text{3}}{\text{5}}$ and $\frac{\text{2}}{\text{3}}$ is

(i) $\frac{\text{49}}{\text{75}}$

(ii) $\frac{\text{50}}{\text{75}}$

(iii) \[\frac{\text{47}}{\text{75}}\]

(iv) $\frac{\text{46}}{\text{75}}$

Ans: (B) $\frac{50}{75}$

17. $\text{0}\text{.12}\overline{\text{3}}$ is equal to

(i) $\frac{\text{122}}{\text{90}}$

(ii) $\frac{\text{122}}{\text{100}}$

(iii) $\frac{\text{122}}{\text{99}}$

(iv) None of these

Ans: (a) $\frac{122}{990}$

18. The number ${{\left( \text{1+}\sqrt{\text{3}} \right)}^{\text{2}}}$ is

(a) natural number

(b) irrational number

(c) rational number

(d) integer

Ans: (b) irrational number

19. The simplest form of $\sqrt{\text{600}}$ is

(i) $\text{10}\sqrt{\text{60}}$

(ii) $\text{100}\sqrt{\text{6}}$

(iii) $\text{20}\sqrt{\text{3}}$

(iv) $\text{10}\sqrt{\text{6}}$

Ans: (D) $10\sqrt{6}$

20. The value of $\text{0}\text{.}\overline{\text{23}}\text{+0}\text{.}\overline{\text{22}}$ is

(i) $\text{0}\text{.4}\overline{\text{5}}$

(ii) $\text{0}\text{.4}\overline{\text{4}}$

(iii) $\text{0}\text{.}\overline{\text{45}}$

(iv) $\text{0}\text{.}\overline{\text{44}}$

Ans: (A) $0.\overline{23}=0.232323....$

$0.\overline{22}=0.222222....$

$0.\overline{23}+0.\overline{22}=0.454545....$

$=0.\overline{45}$

21. The value of ${{\text{2}}^{\frac{\text{1}}{\text{3}}}}\text{ }\!\!\times\!\!\text{ }{{\text{2}}^{\text{-}\frac{\text{4}}{\text{3}}}}$ is

(i) $\text{2}$

(ii) $\frac{\text{1}}{\text{2}}$

Ans: (B) ${{2}^{\frac{1}{3}}}\times {{2}^{-\frac{4}{3}}}={{2}^{\frac{1}{3}-\frac{4}{3}}}={{2}^{\frac{1-4}{3}}}={{2}^{-\frac{3}{3}}}$

22. \[\text{16}\sqrt{\text{13}}\text{ }\!\!\div\!\!\text{ 9}\sqrt{\text{52}}\] is equal to

(i) $\frac{\text{3}}{\text{9}}$

(ii) $\frac{\text{9}}{\text{8}}$

(iii) \[\frac{\text{8}}{\text{9}}\]

Ans: $16\sqrt{13}\div 9\sqrt{52}$

$\frac{16\sqrt{13}}{9\sqrt{52}}=\frac{16}{9}\sqrt{\frac{13}{52}}=\frac{8}{9}$

23. $\sqrt{\text{8}}$ is an

(i) natural number

(iii) integer

(iv) irrational number

Ans: (D) $\sqrt{8}$ is an irrational number

$\therefore \sqrt{4\times 2}=2\sqrt{2}$

2 Marks Questions

1. Is zero a rational number? Can you write it in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$?

Ans: Consider the definition of a rational number. A rational number is the one that can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.

Zero can be written as $\frac{0}{1},\frac{0}{2},\frac{0}{3},\frac{0}{4},\frac{0}{5}......$

So, we arrive at the conclusion that $0$ can be written in the form $\frac{p}{q}$, where $q$is any integer.

Therefore, zero is a rational number.

2. Find six rational numbers between $3$ and $4$.

Ans: We know that there are infinite rational numbers between any two numbers.

A rational number is the one that can be written in the form of $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.

We know that the numbers $3.1,3.2,3.3,3.4,3.5$ and $3.6$ all lie between $3$ and $4$.

We need to rewrite the numbers $3.1,3.2,3.3,3.4,3.5$ and $3.6$ in $\frac{p}{q}$ form to get the rational numbers between $3$ and $4$.

So, after converting we get $\frac{32}{10},\frac{32}{10},\frac{33}{10},\frac{34}{10},\frac{35}{10},$ and $\frac{36}{10},$ into lowest fractions.

On converting the fractions into lowest fractions, we get $\frac{16}{5},\frac{17}{5},\frac{7}{2}$ and $\frac{18}{5}$.

Therefore, six rational numbers between $3$ and $4$are \[\frac{31}{10},\frac{16}{5},\frac{33}{10},\frac{17}{5},\frac{7}{2}\] and $\frac{18}{5}$.

3. Find five rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.

We know that the numbers $\frac{3}{5}$ and $\frac{4}{5}$ can also be written as $0.6\text{ and }0.8$.

We can conclude that the numbers$0.61,0.62,0.63,0.64$ and $0.65$ in \[\frac{p}{q}\] form to get the rational numbers between $3\text{ and }4$.

So, after converting, we get $\frac{61}{100},\frac{62}{100},\frac{63}{100},\frac{64}{100}\text{ and }\frac{65}{100}$.

We can further convert the rational numbers $\frac{62}{100},\frac{64}{100}\text{ and }\frac{65}{100}$ into lowest fractions.

On converting the fractions, we get $\frac{31}{50},\frac{16}{25}\text{ and }\frac{13}{20}$.

Therefore, six rational numbers between $3\text{ and }4$ are $\frac{61}{100},\frac{31}{50},\frac{63}{100},\frac{16}{50}\text{ and }\frac{13}{50}$.

4. Show how $\sqrt{5}$ can be represented on the number line.

Ans: According to Pythagoras theorem, we can conclude that

${{\left( \sqrt{5} \right)}^{2}}={{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}$.

We need to draw a line segment $AB\text{ of }1$unit on the number line. Then draw a straight line segment $BC\text{ of }2$ units. Then join the points $C$ and $A$, to form a line segment $BC$.

Then draw the arc $ACD$, to get the number $\sqrt{5}$ on the number line.

5. You know that $\frac{1}{7}=0.142857....$. Can you predict what the decimal expansion of $\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ are, without actually doing the long division? If so, how?

(Hint: Study the remainder while finding the value of $\frac{1}{7}$ carefully.)

Ans: We are given that $\frac{1}{7}=0.\overline{142857}$ or $\frac{1}{7}=0.142857....$

We need to find the value of \[\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7}\text{ and }\frac{6}{7}\], without performing long division.

We know that \[\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7}\text{ and }\frac{6}{7}\] can be rewritten as

$2\times \frac{1}{7},3\times \frac{1}{7},4\times \frac{1}{7},5\times \frac{1}{7}\text{ and }6\times \frac{1}{7}$.

On substituting value of $\frac{1}{7}$ as $0.142857....$, we get 

$2\times \frac{1}{7}=2\times 0.142857....=0.285714.... $

$3\times \frac{1}{7}=3\times 0.142857....=0.428571.... $ 

$4\times \frac{1}{7}=4\times 0.142857....=0.571428....$ 

$5\times \frac{1}{7}=5\times 0.142857....=0.714285.... $ 

 $6\times \frac{1}{7}=6\times 0.142857....=0.857142.... $ 

Therefore, we conclude that, we can predict the values of \[\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7}\text{ and }\frac{6}{7}\], without performing long division, to get 

\[\frac{2}{7}=0.\overline{285714},\frac{3}{7}=0.\overline{428571},\frac{4}{7}=0.\overline{571428},\frac{5}{7}=0.\overline{714285},\frac{6}{7}=0.\overline{857142}\]

6. Express $0.99999....$in the form $\frac{p}{q}$. Are you surprised by your answer? Discuss why the answer makes sense with your teacher and classmates.

Ans: Let $x=0.99999....\text{ }......(a)$

We need to multiply both sides by $10$ to get 

$10x=9.9999....\text{ }......(b)$

We need to subtract $(a)\text{ from }(b)$, to get

10x=9.99999.... 

x=0.99999.... 

_____________

We can also write $9x=9\text{ as }x=\frac{9}{9}\text{ or }x=1$.

Therefore, on converting $0.99999....$ in the $\frac{p}{q}$ form, we get the answer as $1$.

Yes, at a glance we are surprised at our answer. But the answer makes sense when we observe that $0.99999....$ goes on forever. So there is no gap between $1$  and $0.9999....$ and hence they are equal.

7. Visualize $3.765$ on the number line using successive magnification.

Ans: We know that the number $3.765$ will lie between $3.764\text{ and }3.766$.

We know that the number $3.764$and $3.766$ will lie between $3.76\text{ and }3.77$.

We know that the number $3.76\text{ and }3.77$. will lie between $3.7\text{ and }3.8$.

We know that the number  $3.7\text{ and }3.8$ will lie between $3\text{ and }~4$.

Therefore, we can conclude that we need to use the successive magnification, after locating numbers $3\text{ and }~4$ on the number line

(Image will be uploaded soon)

8. Visualize $4.\overline{26}$ on the number line, upto $4$decimal places.

Ans: We know that the number $4.\overline{26}$ can also be written as$4.262....$ .

We know that the number $4.262....$ will lie between $4.261\text{ and }4.263$.

We know that the number  $4.261\text{ and }4.263$ will lie between $4.26\text{ and }4.27$.

We know that the number $4.26\text{ and }4.27$ will lie between $4.2\text{ and }4.3$.

We know that the number $4.2\text{ and }4.3$ will lie between $4\text{ and }5$.

Therefore, we can conclude that we need to use the successive magnification, after locating numbers $4\text{ and }5$ on the number line.

9. Recall, $\pi $is defined as the ratio of the circumference (say $c$) of a circle of its diameter (say $d$). That is, $\pi =\frac{c}{d}$. This seems to contradict the fact that $\pi $ is irrational. How you resolve this contradiction?

Ans: We know that when we measure the length of the line or a figure by using a scaleneory device, we do not get an exact measurement. In fact, we get an approximate rational value. So, we are not able to realize that either the circumference ($c$) or diameter ($d$) of a circle is irrational.

Therefore, we can conclude that as such there is not any contradiction regarding the value of $\pi $ and we realize that the value of $\pi $ is irrational.

10. Represent $9.3$ on the number line.

Ans: Mark the distance $9.3$ units from a fixed point $A$ on a given line to obtain a point $B$ such that $AB=9.3$ units. From $B$ mark a distance of $1$ unit and call the new point as $C$. Find themid-point of $AC$ and call that point as $O$. Draw a semi-circle with centre $O$ and radius $OC=5.15$units. Draw a line perpendicular to $AC$ passing through $B$ cutting the semi-circle at $D$.

Then $BD=\sqrt{9.3}$.

11. Find (i) ${{64}^{\frac{1}{5}}}$ (ii) ${{32}^{\frac{1}{5}}}$ (iii) ${{125}^{\frac{1}{3}}}$

(i) ${{64}^{\frac{1}{2~}}}$

We know that${{a}^{\frac{1}{n}}}=\sqrt[n]{a},\text{ where }a>0$

We conclude that  ${{64}^{\frac{1}{2~}}}$can also be written as $\sqrt[2]{64}=\sqrt[2]{8\times 8}$

$\sqrt[2]{64}=\sqrt[2]{8\times 8}$$=8$

Therefore, the value of  ${{64}^{\frac{1}{2~}}}$will be $8$.

(ii) ${{32}^{\frac{1}{5}}}$

We conclude that  ${{32}^{\frac{1}{5}}}$can also be written as $\sqrt[5]{32}=\sqrt[5]{2\times 2\times 2\times 2\times 2}$

$\sqrt[5]{32}=\sqrt[5]{2\times 2\times 2\times 2\times 2}=2$

Therefore, the value of  ${{32}^{\frac{1}{5}}}$will be $2$.

(iii) ${{125}^{\frac{1}{3}}}$

We conclude that ${{125}^{\frac{1}{3}}}$can also be written as $\sqrt[3]{125}=\sqrt[3]{5\times 5\times 5}$

$\sqrt[3]{125}=\sqrt[3]{5\times 5\times 5}=5$

Therefore, the value of ${{125}^{\frac{1}{3}}}$will be $5$.

12. Simplify $\sqrt[3]{2}\times \sqrt[4]{3}$

Ans: $\sqrt[3]{2}\times \sqrt[4]{3}$

${{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{4}}}$

The LCM of $3\text{ and }4\text{ is }12$

$\therefore {{2}^{\frac{1}{3}}}={{2}^{\frac{4}{12}}}={{\left( {{2}^{4}} \right)}^{\frac{1}{12}}}={{16}^{\frac{1}{12}}} $

${{3}^{\frac{1}{4}}}={{3}^{\frac{3}{12}}}={{\left( {{3}^{3}} \right)}^{\frac{1}{12}}}={{27}^{\frac{1}{12}}} $

${{2}^{\frac{1}{3}}}\times {{3}^{\frac{1}{4}}}={{16}^{\frac{1}{12}}}\times {{27}^{\frac{1}{12}}}={{\left( 16\times 27 \right)}^{\frac{1}{12}}} $

$={{\left( 432 \right)}^{\frac{1}{12}}} $

13. Find the two rational numbers between$\frac{1}{2}$ and $\frac{1}{3}$.

Ans: First rational number between $\frac{1}{2}$ and $\frac{1}{3}$

$=\frac{1}{2}\left[ \frac{1}{2}+\frac{1}{3} \right]\Rightarrow \frac{1}{2}\left[ \frac{3+2}{6} \right]\Rightarrow \frac{5}{12} $

$ =\frac{1}{2},\frac{5}{12}\text{ and }\frac{1}{3} $

Second rational number between $\frac{1}{2}$ and $\frac{1}{3}$

$=\frac{1}{2}\left[ \frac{1}{2}+\frac{5}{12} \right]\Rightarrow \frac{1}{2}\left[ \frac{6+5}{12} \right]\Rightarrow \frac{11}{24}$

$=\frac{5}{12}\text{ and }\frac{11}{24}$ are two rational numbers between $\frac{1}{2}$ and $\frac{1}{3}$.

14. Find two rational numbers between $2$ and $3$.

Ans: Irrational numbers between $2$ and $3$ is $\sqrt{2\times 3}=\sqrt{6}$

Irrational number between $2$ and $3$ is $\sqrt{6}$.

$\sqrt{2\times \sqrt{6}}={{2}^{\frac{1}{2}}}\times {{6}^{\frac{1}{4}}}={{2}^{2\times \frac{1}{4}}}\times {{6}^{\frac{1}{4}}} $

$ ={{\left( {{2}^{2}} \right)}^{\frac{1}{4}}}\times {{6}^{\frac{1}{4}}}={{4}^{\frac{1}{4}}}\times {{6}^{\frac{1}{4}}}={{\left( 24 \right)}^{\frac{1}{4}}}=\sqrt[4]{24} $

$\sqrt{6}\text{ and }\sqrt{24}$ are two rational numbers between $2$ and $3$.

15. Multiply $\left( 3-\sqrt{5} \right)$ by $\left( 6+\sqrt{2} \right)$.

Ans: $\left( 3-\sqrt{5} \right)$$\left( 6+\sqrt{2} \right)$

$  =3\left( 6-\sqrt{2} \right)-\sqrt{5}\left( 6+\sqrt{2} \right) $

$=18+3\sqrt{2}-6\sqrt{5}-\sqrt{5}\times \sqrt{2} $

$  =18+3\sqrt{2}-6\sqrt{5}-\sqrt{10} $

16. Evaluate (i) $\sqrt[3]{125}$ (ii) $\sqrt[4]{1250}$

(i) $\sqrt[3]{125}$$={{\left( 5\times 5\times 5 \right)}^{\frac{1}{3}}}={{\left( {{5}^{3}} \right)}^{\frac{1}{3}}}=5$

(ii) $\sqrt[4]{1250}$$\begin{align}

$={{\left( 2\times 5\times 5\times 5\times 5 \right)}^{\frac{1}{4}}}={{\left( 2\times {{5}^{4}} \right)}^{\frac{1}{4}}} $ 

$={{2}^{\frac{1}{4}}}\times {{\left( {{5}^{4}} \right)}^{\frac{1}{4}}}=5\times \sqrt[4]{2} $

17. Find rationalizing factor of $\sqrt{300}$.

Ans: $\sqrt{300}=\sqrt{2\times 2\times 3\times 5\times 5}$

$  =\sqrt{{{2}^{2}}\times 3\times {{5}^{2}}} $

$ =2\times 5\sqrt{3}=10\sqrt{3} $

Rationalizing factor is $\sqrt{3}$

18. Rationalizing the denominator $\frac{1}{\sqrt{5}+\sqrt{2}}$ and subtract it from $\sqrt{5}-\sqrt{2}$.

Ans: $\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{\sqrt{5}-\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$

Difference between $\left( \sqrt{5}-\sqrt{2} \right)\text{ and }\left( \sqrt{5}-\frac{\sqrt{2}}{3} \right)$

$=\sqrt{5}-\sqrt{2}-\left( \frac{\sqrt{5}-\sqrt{2}}{3} \right) $

$ =\sqrt{5}-\sqrt{2}-\frac{\sqrt{5}}{3}+\frac{\sqrt{2}}{3} $

$=\left( \sqrt{5}-\frac{\sqrt{5}}{3} \right)-\left( \sqrt{2}-\frac{\sqrt{2}}{3} \right) $

$=\frac{2\sqrt{5}}{3}-\frac{2\sqrt{2}}{3}=\frac{2}{3}\left( \sqrt{5}-\sqrt{2} \right) $

19. Show that $\sqrt{7}-3$ is irrational.

Ans: Suppose $\sqrt{7}-3$ is rational

Let $\sqrt{7}-3=x$ ($x$ is a rational number)

$\sqrt{7}=x+3$

$x$ is a rational number $3$ is also a rational number

$\therefore x+3$ is a rational number

But is $\sqrt{7}$ irrational number which is contradiction

$\therefore \sqrt{7}-3$ is an irrational number.

20. Find two rational numbers between $7$ and $5$.

Ans: First rational number $=\frac{1}{2}\left[ 7+5 \right]=\frac{12}{2}=6$

Second rational number $=\frac{1}{2}\left[ 7+6 \right]=\frac{1}{2}\times 13=\frac{13}{2}$

Two rational numbers between $7\text{ and }5\text{ are }6\text{ and }\frac{13}{2}$.

21. Show that $5+\sqrt{2}$ is not a rational number.

Ans: Let $5+\sqrt{2}$ is a rational number.

Say $5+\sqrt{2=x}$ i.e., $\sqrt{2}=x-5$

$x$ is a rational number $5$ is also rational number

$\therefore x-5$ is also a rational number.

But $\sqrt{2}$ is irrational number which is a contradiction

$\therefore 5+\sqrt{2}$ is an irrational number.

22. Simplify ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}$.

Ans: ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}+2\sqrt{5}\times \sqrt{2}=5+2+2\sqrt{10}=7+2\sqrt{2}$

23. Evaluate $\frac{{{11}^{\frac{5}{2}}}}{{{11}^{\frac{3}{2}}}}$.

Ans: $\frac{{{11}^{\frac{5}{2}}}}{{{11}^{\frac{3}{2}}}}={{11}^{\frac{5}{2}-\frac{3}{2}}}\left[ \because \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \right]$

$ ={{11}^{\frac{5-3}{2}}}={{11}^{\frac{2}{2}}} $

$=11 $ 

24. Find four rational numbers between $\frac{3}{7}$ and $\frac{4}{7}$.

$\frac{3}{7}\times \frac{10}{10}=\frac{30}{70}\text{ and }\frac{4}{7}\times \frac{10}{10}=\frac{40}{70}$

Take any four rational numbers between $\frac{30}{70}\text{ and }\frac{40}{70}$ i.e., rational numbers between  $\frac{3}{7}$ and $\frac{4}{7}$ are $\frac{31}{70},\frac{32}{70},\frac{33}{70},\frac{34}{70},\frac{35}{70}$

25. Write the following in decimal form (i) $\frac{36}{100}$ (ii) $\frac{2}{11}$

(i) $\frac{36}{100}=0.36$

(ii) $\frac{2}{11}=0.\overline{18}$

26. Express $2.417\overline{8}$ in the form $\frac{a}{b}$

Ans: $x=2.4\overline{178}$

$10x=24.\overline{178}$$......(1)$$[\text{Multiplying both sides by }10]$

$10x=24.178178178.... $

$1000\times 10x=1000\times 24.178178178....$Multiplying both sides by 1000

$10,000x=24178.178178.... $

$ 10000x=24178.\overline{178}\text{ }......(2) $

Subtracting $(1)\text{ from }(2)$

$10,000x-x=24178.\overline{178}-24.\overline{178} $

$9990x=24154 $

$x=\frac{24154}{9990} $

$ 2.4\overline{178}=\frac{24154}{9990}+\frac{12077}{4995} $

27. Multiply $\sqrt{3}$ by $\sqrt[3]{5}$.

Ans: $\sqrt{3}\text{ and }\sqrt[3]{5}$

Or ${{3}^{\frac{1}{2}}}\text{ and }{{5}^{\frac{1}{3}}}$

$LCM\text{ of }2\text{ and }3\text{ is }6 $

${{3}^{\frac{1}{2}}}={{3}^{\frac{1}{2}\times \frac{3}{3}}}={{\left( {{3}^{3}} \right)}^{\frac{1}{6}}}={{\left( 27 \right)}^{\frac{1}{6}}} $

 ${{5}^{\frac{1}{3}}}={{5}^{\frac{1}{3}\times \frac{2}{2}}}={{\left( {{5}^{2}} \right)}^{\frac{1}{6}}}={{\left( 25 \right)}^{\frac{1}{6}}} $

 $\sqrt{3}\times \sqrt[3]{5}={{\left( 27 \right)}^{\frac{1}{6}}}\times {{\left( 25 \right)}^{\frac{1}{6}}}={{\left( 27\times 25 \right)}^{\frac{1}{6}}} $

$ ={{675}^{\frac{1}{6}}}=\sqrt[6]{675} $

28. Find the value of $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{5}}$ if $\sqrt{5}=2.236$ and $\sqrt{10}=3.162$.

Ans: $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{10}+5}{5}=\frac{8.162}{5}=1.6324$

29. Convert $0.\overline{25}$ into rational number.

Ans: Let \[x=0.\overline{25}\]                            ......(i)

$x=0.252525....$

Multiply both sides by 100

$100x=25.252525....$

$100x=25.\overline{25}$                                                  ......(ii)

Subtract (i) from (ii)

$100x-x=25.\overline{25}-0.\overline{25} $

 $x=\frac{25}{99} $

30. Simplify $\left( 3\sqrt{3}+2\sqrt{2} \right)\left( 2\sqrt{3}+3\sqrt{2} \right)$.

Ans: By multiplying each terms in the given product we have,

 $ \left( 3\sqrt{3}+2\sqrt{2} \right)\left( 2\sqrt{3}+3\sqrt{2} \right) $

 $=3\sqrt{3}\left( 2\sqrt{3}+3\sqrt{2} \right)+2\sqrt{2}\left( 2\sqrt{3}+3\sqrt{2} \right) $

 $=18+9\sqrt{6}+4\sqrt{6}+12 $

 $ =30+\left( 9+4 \right)\sqrt{6} $

 $=30+13\sqrt{6} $

31. Simplify $\frac{{{9}^{\frac{3}{2}}}\times {{9}^{-\frac{4}{2}}}}{{{9}^{\frac{1}{2}}}}$.

Ans: By using the formulas of exponents with same base we get,

$\frac{{{9}^{\frac{3}{2}}}\times {{9}^{-\frac{4}{2}}}}{{{9}^{\frac{1}{2}}}}=\frac{{{9}^{\frac{3}{2}-\frac{4}{2}}}}{{{9}^{\frac{1}{2}}}}\left[ {{a}^{m}}.{{a}^{n}}={{a}^{m-n}} \right]$

$\frac{{{9}^{-\frac{1}{2}}}}{{{9}^{\frac{1}{2}}}}=\frac{1}{{{9}^{\frac{1}{2}+\frac{1}{2}}}}\left[ {{a}^{-m}}=\frac{1}{{{a}^{m}}} \right] $

$ =\frac{1}{{{9}^{\frac{2}{2}}}}=\frac{1}{9} $

3 Marks Questions

1. State whether the following statements are true or false. Give

reasons for your answers.

i. Every natural number is a whole number.

Separately, consider whole numbers and natural numbers.

We know that the whole number series is 0,1,2,3,4,5....

We know that the natural number series is 0,1,2,3,4,5....

As a result, every number in the natural number series may be found in the whole number series.

Therefore, we can safely conclude that any natural number is a whole number.

ii. Every integer is a whole number.

Ans: Separately, consider whole numbers and integers.

We know that integers are those numbers that can be written in the form of $\frac{p}{q}$ where q=1.

In the case of an integer series, we now have.... 4,3,2,1,0,1,2,3,4....

We can conclude that all whole number series numbers belong to the integer series.

However, the whole number series does not contain every number of integer series.

As a result, we can conclude that no integer is a whole number.

iii. Every rational number is a whole number.

Ans: Separately, consider whole numbers and rational numbers.

We know that integers are those numbers that can be written in the form of $\frac{p}{q}$ where $q\ne 0$.

We know that every number of whole number series can be written in the form of $\frac{p}{q}$ as $\frac{0}{1},\frac{1}{1},\frac{2}{1},\frac{3}{1},\frac{4}{1},\frac{5}{1}...$

We conclude that every number of the whole number series is a rational number. 

But, every rational number does not appear in the whole number series.

2. State whether the following statements are true or false. Justify your answers.

i. Every irrational number is a real number.

Ans: Separately, consider irrational numbers and real numbers.

We know that irrational numbers are the numbers that cannot be converted in the form $\frac{p}{q}$, where p and q are integers and $q\ne 0$.

A real number is made up of both rational and irrational numbers, as we all know.

As a result, we might conclude that any irrational number is, in fact, a real number.

ii. Every point on the number line is of the form $\sqrt{m}$, where m is a natural number.

Ans: Consider a number line. We know that we can express both negative and positive numbers on a number line.

We know that when we take the square root of any number, we cannot receive a negative value.

Therefore, we conclude that not every number point on the number line is of the form $\sqrt{m}$, where m is a natural number.

iii. Every real number is an irrational number.

As a result, we can deduce that any irrational number is actually a real number. However, not every real number is irrational.

Therefore, we conclude that every real number is not a rational number.

3. Express the following in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$.

i. $0.\overline{6}$

Let $x=0.\overline{6}$

$\Rightarrow x=0.6666$            ......(a)

Multiplying both sides by 10 we get

$10x=6.6666$             ......(b)

We need to subtract (a) from (b), to get

We can also write $9x=6$ as $x=\frac{6}{9}$ or $x=\frac{2}{3}$.

Therefore, on converting $0.\overline{6}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{2}{3}$.

ii. $0.4\overline{7}$

Ans: Let $x=0.4\overline{7}\Rightarrow x=0.47777$                     ......(a)

Multiplying both sides by 10 we get 

$10x=4.7777$                                            ......(b)

We can also write $9x=4.3$ as $x=\frac{4.3}{9}$ or $x=\frac{43}{90}$

Therefore, on converting $0.4\overline{7}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{43}{90}$.

iii. $0.\overline{001}$

Ans: Let $x=0.\overline{001}\Rightarrow x=0.001001$                     ......(a)

Multiplying both sides by 1000 we get 

$1000x=1.001001$                                         ......(b)

We can also write $999x=1$ as $x=\frac{1}{999}$

Therefore, on converting $0.\overline{001}$ in the $\frac{p}{q}$ form, we get the answer as $\frac{1}{999}$.

4. What can the maximum number of digits be in the recurring block of digits in the decimal expansion of $\frac{1}{17}$ ? Perform the division to check your answer.

Ans: The number of digits in the recurring block of $\frac{1}{17}$ must be determined.

To acquire the repeating block of $\frac{1}{17}$ we'll use long division.

We need to divide 1 by 17, to get 0.0588235294117647.... and we got the remainder as 1, which will continue to be 1 after carrying out 16 continuous divisions.

Therefore, we conclude that

\[\frac{1}{17}=0.0588235294117647\] or \[\frac{1}{17}=0.\overline{0588235294117647}\] which is a non-terminating decimal and recurring decimal.

5. Look at several examples of rational numbers in the form $\frac{p}{q}\left( q\ne 0 \right)$ where $p$ and $q$ are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property $q$ must satisfy?

Ans: Let us consider the examples of the form $\frac{p}{q}$ that are terminating decimals .

$ \frac{5}{2}=2.5 $ 

$ \frac{5}{4}=1.25 $ 

$ \frac{2}{5}=0.4 $

$  \frac{5}{16}=0.3125 $ 

It can be observed that the denominators of the above rational numbers have powers of 2,5 or both.

Therefore, we can conclude that property, which $q$ must satisfy in $\frac{p}{q}$ , so that the rational number $\frac{p}{q}$ is a terminating decimal is that q must have powers of 2,5 or both.

6. Classify the following numbers as rational or irrational:

i. $2-\sqrt{5}$

Ans: $2-\sqrt{5}$

We know that $\sqrt{5}=2.236....$, which is an irrational number.

$2-\sqrt{5}=2-2.236....$

$=-0.236...$, which is also an irrational number.

As a result, we can deduce that $2-\sqrt{5}$ is an irrational number.

ii. $\left( 3+\sqrt{23} \right)-\sqrt{23}$

Ans: $\left( 3+\sqrt{23} \right)-\sqrt{23}$

$\left( 3+\sqrt{23} \right)-\sqrt{23}=3+\sqrt{23}-\sqrt{23}=3$

As a result, we can deduce that $\left( 3+\sqrt{23} \right)-\sqrt{23}$ is a rational number.

iii. $\frac{2\sqrt{7}}{7\sqrt{7}}$

Ans: $\frac{2\sqrt{7}}{7\sqrt{7}}$

We can cancel $\sqrt{7}$ in the numerator and denominator to get $\frac{2\sqrt{7}}{7\sqrt{7}}=\frac{2}{7}$, because $\sqrt{7}$ is a common number in both the numerator and denominator.

iv. $\frac{1}{\sqrt{2}}$

Ans: $\frac{1}{\sqrt{2}}$

We know that $\sqrt{2}=1.4142...$, which is an irrational number.

$\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$

$=\frac{1.4142...}{2}=0.707...$ which is also an irrational number.

As a result, we can deduce that $\frac{1}{\sqrt{2}}$ is an irrational number.

Ans: $2\pi $

We know that $\pi =3.1415....,$ which is an irrational number.

We can conclude that $2\pi $ will also be an irrational number.

As a result, we can deduce that $2\pi $ is an irrational number.

7. Simplify each of the following expression:

i. $\left( 3+3\sqrt{3} \right)\left( 2+\sqrt{2} \right)$

$(3+3\sqrt{3})(2+\sqrt{2})$

Applying distributive law,

\[(3+3\sqrt{3})(2+\sqrt{2})=3(2+\sqrt{2})\sqrt{3}(2+\sqrt{2})\]

\[=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\]

ii. \[\left( 3+3\sqrt{3} \right)3-\sqrt{3}\]

Ans: $(3+3\sqrt{3})(3-\sqrt{3})$

$  (3+3\sqrt{3})(3-\sqrt{3})=(3-\sqrt{3})+\sqrt{3}(3-\sqrt{3}) $ 

$ =9-3\sqrt{3}+3\sqrt{3}-3 $

iii. ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}$

Ans: ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}$

Applying the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}+2\times \sqrt{5}\times \sqrt{2}+{{\left( \sqrt{2} \right)}^{2}} $

$ =5+2\sqrt{10}+2 $

$=7+2\sqrt{10}$

iv. $\left( 5+\sqrt{2} \right)\left( 5+\sqrt{2} \right)$

Ans: $\left( 5+\sqrt{2} \right)\left( 5+\sqrt{2} \right)$

Applying the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$

$ \left( 5+\sqrt{2} \right)\left( 5+\sqrt{2} \right)={{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}} $

8. Find 

i. ${{9}^{\frac{3}{2}}}$

Ans: We know that ${{a}^{\frac{1}{n}}}=\sqrt[n]{a},a>0$

As a result, we can deduce that ${{9}^{\frac{3}{2}}}$ can also be written as

$\sqrt[2]{{{\left( 9 \right)}^{3}}}=\sqrt[2]{9\times 9\times 9}=\sqrt[2]{3\times 3\times 3\times 3\times 3\times 3}$

$=3\times 3\times 3 $

Therefore, the value of ${{9}^{\frac{3}{2}}}$ will be $27$ .

ii. ${{32}^{\frac{2}{5}}}$

As a result, we can deduce that ${{32}^{\frac{2}{5}}}$ can also be written as

$ \sqrt[5]{{{\left( 32 \right)}^{2}}}=\sqrt[5]{\left( 2\times 2\times 2\times 2\times 2 \right)\left( 2\times 2\times 2\times 2\times 2 \right)} $

$=2\times 2 $ 

Therefore, the value of ${{32}^{\frac{2}{5}}}$ will be $4$.

iii. ${{16}^{\frac{3}{4}}}$

As a result, we can deduce that ${{16}^{\frac{3}{4}}}$ can also be written as 

$\sqrt[4]{{{\left( 16 \right)}^{3}}}=\sqrt[4]{\left( 2\times 2\times 2\times 2 \right)\left( 2\times 2\times 2\times 2 \right)\left( 2\times 2\times 2\times 2 \right)} $

$ =2\times 2\times 2 $

 & =8 $

Therefore, the value of ${{16}^{\frac{3}{4}}}$ will be $8$ .

iv. ${{125}^{-\frac{1}{3}}}$

Ans: We know that ${{a}^{-n}}=\frac{1}{{{a}^{n}}}$

As a result, we can deduce that ${{125}^{-\frac{1}{3}}}$ can also be written as $\frac{1}{{{125}^{\frac{1}{3}}}},or{{\left( \frac{1}{125} \right)}^{\frac{1}{3}}}$

We know that ${{a}^{\frac{1}{n}}}=\sqrt[n]{a},a>0$

$ \sqrt[3]{\frac{1}{125}}=\sqrt[3]{\left( \frac{1}{5}\times \frac{1}{5}\times \frac{1}{5} \right)} $

$=\frac{1}{5} $

Therefore, the value of ${{125}^{-\frac{1}{3}}}$ will be $\frac{1}{5}$.

9. Simplify

i. ${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}$

Ans: We know that ${{a}^{m}}.{{a}^{n}}={{a}^{\left( m+n \right)}}$

As a result, we can deduce that ${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}={{\left( 2 \right)}^{\frac{2}{3}+\frac{1}{5}}}$

${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}=\left( 2 \right)\frac{10+3}{15}={{\left( 2 \right)}^{\frac{13}{15}}}$

Therefore, the value of ${{2}^{\frac{2}{3}}}{{.2}^{\frac{1}{5}}}$ will be ${{\left( 2 \right)}^{\frac{13}{15}}}$.

ii. ${{\left( {{3}^{\frac{1}{3}}} \right)}^{7}}$

As a result, we can deduce that ${{\left( {{3}^{\frac{1}{3}}} \right)}^{7}}$ can also be written as ${{3}^{\frac{7}{3}}}$

iii. $\frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}$

Ans: We know that $\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{\left( m-n \right)}}$

As a result, we can deduce that  $\begin{align}

$ \frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}={{11}^{\frac{1}{2}}}-{{11}^{\frac{1}{4}}} $

$={{11}^{\frac{2-1}{4}}}={{11}^{\frac{1}{4}}} $

Therefore, the value of  $\frac{{{11}^{\frac{1}{2}}}}{{{11}^{\frac{1}{4}}}}$ will be ${{11}^{\frac{1}{4}}}$.

iv. ${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}$

Ans: We know that ${{a}^{m}}.{{b}^{m}}={{\left( a\times b \right)}^{m}}$

As a result, we can deduce that ${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}={{\left( 7\times 8 \right)}^{\frac{1}{2}}}.$

${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}={{\left( 7\times 8 \right)}^{\frac{1}{2}}}={{\left( 56 \right)}^{\frac{1}{2}}}.$

Therefore, the value of ${{7}^{\frac{1}{2}}}{{.8}^{\frac{1}{2}}}$ will be ${{\left( 56 \right)}^{\frac{1}{2}}}$.

10. Express $0.8888....$ in the form $\frac{p}{q}$.

Ans: Let us assume that the given decimal as,

$x=0.\overline{8}......\left( 1 \right)$

$10x=10\times 0.8888$ (Multiply both sides by 10)

$ 10x=8.8888 $

$10x=8.\overline{8}.....\left( 2 \right) $

$10x-x=8.\overline{8}-0.\overline{8}$ (Subtracting (1) from (2))

$x=\frac{8}{9} $

11.  Simplify by rationalizing denominator $\frac{7+3\sqrt{5}}{7-3\sqrt{5}}$.

Ans: We are given the fraction to rationalize. By rationalizing the denominator we get,

$\frac{7+3\sqrt{5}}{7-3\sqrt{5}}=\frac{7+3\sqrt{5}}{7-3\sqrt{5}}\times \frac{7+3\sqrt{5}}{7+3\sqrt{5}}$ 

$=\frac{{{\left( 7+3\sqrt{5} \right)}^{2}}}{{{7}^{2}}-{{\left( 3\sqrt{5} \right)}^{2}}} $

$ =\frac{{{7}^{2}}+{{\left( 3\sqrt{5} \right)}^{2}}+2\times 7\times 3\sqrt{5}}{49-{{3}^{2}}\times 5} $

$=\frac{49+9\times 5+42\sqrt{5}}{49-45} $

$=\frac{49+45+42\sqrt{5}}{4} $

$ =\frac{94+42\sqrt{5}}{4} $ 

$ =\frac{94}{4}+\frac{42}{4}\sqrt{5} $

$ =\frac{47}{2}+\frac{21}{2}\sqrt{5} $

12 . Simplify ${{\left\{ {{\left[ {{625}^{-}}^{\frac{1}{2}} \right]}^{-\frac{1}{4}}} \right\}}^{2}}$.

Ans:  Let us take the given expression to simplify and using the exponents formulas we get,

  \[{{\left\{ {{\left[ {{625}^{-}}^{\frac{1}{2}} \right]}^{-\frac{1}{4}}} \right\}}^{2}}\]

$ ={{\left\{ {{\left( \frac{1}{{{625}^{\frac{1}{2}}}} \right)}^{-\frac{1}{4}}} \right\}}^{2}} $

$={{\left\{ {{\left( \frac{1}{{{\left( {{25}^{2}} \right)}^{\frac{1}{2}}}} \right)}^{-\frac{1}{4}}} \right\}}^{2}}\left( {{a}^{-m}}=\frac{1}{{{a}^{m}}} \right) $

 $ =\left\{ {{\left( \frac{1}{25} \right)}^{-\frac{1}{4}\times 2}} \right\} $

$=\left( \frac{1}{{{25}^{-\frac{1}{2}}}} \right)=\frac{1}{{{\left( {{5}^{2}} \right)}^{-\frac{1}{2}}}}=\frac{1}{{{5}^{-1}}}=5 $

13. Visualize 3.76 on the number line using successive magnification.

14. Prove that $\frac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\frac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\frac{1}{1+{{x}^{a-c}}+{{x}^{b-c}}}=1$

Ans: We are asked to prove the expression,

$\frac{1}{1+{{x}^{b-a}}+{{x}^{c-a}}}+\frac{1}{1+{{x}^{a-b}}+{{x}^{c-b}}}+\frac{1}{1+{{x}^{a-c}}+{{x}^{b-c}}}=1$

Let us take the LHS of the given expression that is,

$LHS=\frac{1}{1+{{x}^{b}}.{{x}^{-a}}+{{x}^{c}}.{{x}^{-a}}}+\frac{1}{1+{{x}^{a}}.{{x}^{-b}}+{{x}^{c}}.{{x}^{-b}}}+\frac{1}{1+{{x}^{a}}.{{x}^{-c}}+{{x}^{b}}.{{x}^{-c}}} $

$=\frac{1}{{{x}^{-a}}.{{x}^{a}}+{{x}^{b}}.{{x}^{-a}}+{{x}^{c}}.{{x}^{-a}}}+\frac{1}{{{x}^{b}}.{{x}^{-b}}+{{x}^{a}}.{{x}^{-b}}+{{x}^{c}}.{{x}^{-b}}}+\frac{1}{{{x}^{c}}.{{x}^{-c}}+{{x}^{a}}.{{x}^{-c}}+{{x}^{b}}.{{x}^{-c}}} $

$ =\frac{1}{{{x}^{-a}}\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{1}{{{x}^{-b}}\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{1}{{{x}^{-c}}\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)} $

$=\frac{{{x}^{a}}}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{{{x}^{b}}}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}+\frac{{{x}^{c}}}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)} $ $ =\frac{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}{\left( {{x}^{a}}+{{x}^{b}}+{{x}^{c}} \right)}=1 $

15. Represent $\sqrt{3}$ on number line.

Ans: Consider a number line $\text{OD}$ such that the construction to form two triangles is done as shown below.

Take $OA=AB=1$ unit.

And $\angle A=90{}^\circ $

In $\Delta OAB$, by using the Pythagorean theorem we get,

$O{{B}^{2}}={{1}^{2}}+{{1}^{2}}$

$O{{B}^{2}}=2 $

$ OB=\sqrt{2}$ 

Now from triangle $\text{ }\!\!\Delta\!\!\text{ OBD}$, using the Pythagorean theorem we get,

$O{{D}^{2}}=O{{B}^{2}}+B{{D}^{2}} $

$ O{{D}^{2}}={{\left( \sqrt{2} \right)}^{2}}+{{\left( 1 \right)}^{1}} $

$O{{D}^{2}}=2+1=3 $

$OD=\sqrt{3} $

Now, if the point $\text{O}$ is $0$ units then the point $\text{D}$ represents $\sqrt{3}$units.

16. Simplify ${{\left( 3\sqrt{2}+2\sqrt{3} \right)}^{2}}{{\left( 3\sqrt{2}-2\sqrt{3} \right)}^{2}}$.

Ans: We are given the expression as,

${{\left( 3\sqrt{2}+2\sqrt{3} \right)}^{2}}{{\left( 3\sqrt{2}-2\sqrt{3} \right)}^{2}}$

Now, by regrouping the terms in the above expression we have,

$ =\left( 3\sqrt{2}+2\sqrt{3} \right)\left( 3\sqrt{2}+2\sqrt{3} \right)\left( 3\sqrt{2}-2\sqrt{3} \right)\left( 3\sqrt{2}-2\sqrt{3} \right) $

$ =\left( 3\sqrt{2}+2\sqrt{3} \right)\left( 3\sqrt{2}-2\sqrt{3} \right)\left( 3\sqrt{2}+2\sqrt{3} \right)\left( 3\sqrt{2}-2\sqrt{3} \right) $

$=\left[ {{\left( 3\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}} \right]\left[ {{\left( 3\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}} \right] $

$ =\left[ 9\times 2-4\times 3 \right]\left[ 9\times 2-4\times 3 \right] $

$ =\left[ 18-12 \right]\left[ 18-12 \right] $ 

$=6\times 6=36 $

17. Express $2.\overline{4178}$ in the form $\frac{p}{q}$.

Ans: Let $\frac{p}{q}=2.\overline{4178}$

$\frac{p}{q}=2.4178178178$

Multiply by 10

$10\frac{p}{q}=24.178178$

Multiply by 1000

$10000\frac{p}{q}=1000\times 24.178178 $ 

$1000\frac{p}{q}-\frac{p}{q}=24178.178178-14.178178 $

$9999\frac{p}{q}=24154 $

$\frac{p}{q}=\frac{24154}{9999} $

18. Simplify ${{\left( 27 \right)}^{-\frac{2}{3}}}\div {{9}^{\frac{1}{2}}}{{.3}^{-\frac{3}{2}}}$.

Ans: ${{\left( 27 \right)}^{-\frac{2}{3}}}\div {{9}^{\frac{1}{2}}}{{.3}^{-\frac{3}{2}}}$

$ =\frac{{{\left( 3\times 3\times 3 \right)}^{-\frac{2}{3}}}\times {{3}^{\frac{3}{2}}}}{{{\left( 3\times 3 \right)}^{\frac{1}{2}}}}\left[ {{a}^{-m}}=\frac{1}{{{a}^{m}}} \right] $

$ =\frac{{{\left( {{3}^{3}} \right)}^{-\frac{2}{3}}}\times {{3}^{\frac{3}{2}}}}{{{\left( {{3}^{2}} \right)}^{\frac{1}{2}}}} $

$=\frac{{{3}^{\frac{3}{2}-2}}}{3}=\frac{{{3}^{-\frac{1}{3}}}}{3} $

$=\frac{1}{{{3}^{\frac{4}{3}}}}=\frac{1}{\sqrt[3]{81}} $

19. Find three rational numbers between $2.\overline{2}$ and $2.\overline{3}.$

Ans: The irrational numbers are the numbers that do not end after the decimal point nor repeat its numbers in a sequence. 

Representing the given numbers in decimal form we have,

$ 2.\overline{2}=2.222222222...... $

$ 2.\overline{3}=2.333333333....... $ 

So any numbers between these two numbers that do not end nor repeat in any sequence gives the required irrational numbers.

Three rational numbers between $2.\overline{2}$ and $2.\overline{3}$ are $2.222341365....$, $2.28945187364....$ and $2.2321453269....$

20. Give an example of two irrational numbers whose

i. Sum is a rational number

Ans: The required two irrational numbers are $2+\sqrt{2}$ and $2-\sqrt{2}$

Sum $2+\sqrt{2}+2-\sqrt{2}=4$ which is a rational number.

ii. Product is a rational number

Ans: The required two irrational numbers are $3\sqrt{2}$ and $6\sqrt{2}$

Product $3\sqrt{2}\times 6\sqrt{2}=18\times 2=36$ which is rational.

iii. Quotient is a rational number

Ans: The required two irrational numbers are $2\sqrt{125}$ and $3\sqrt{5}$

Quotient $\frac{2\sqrt{125}}{3\sqrt{5}}=\frac{2}{3}\sqrt{\frac{125}{5}}=\frac{2}{3}\times 5=\frac{10}{3}$

21 . If $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, find the value of $\frac{5}{\sqrt{2}+\sqrt{3}}$.

Ans: First let us take the given expression and by rationalizing the denominator we get,

$\frac{5}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$ 

$\frac{5\left( \sqrt{2}-\sqrt{3} \right)}{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}} $

$=\frac{5\left( \sqrt{2}-\sqrt{3} \right)}{2-3} $

Now, substituting the required values of irrational numbers we get,

$=-5\left[ 1.414-1.732 \right] $

$  =-5\times -0.318 $

$  =1.59 $ 

22. Visualize 2.4646 on the number line using successive magnification.

23. Rationalizing the denominator of $\frac{1}{4+2\sqrt{3}}$.

Ans: First let us take the given expression and rationalizing the denominator by multiplying the numerator and denominator with its conjugate we get,

$\frac{1}{4+2\sqrt{3}}=\frac{1}{4+2\sqrt{3}}\times \frac{4-2\sqrt{3}}{4+2\sqrt{3}} $

$ =\frac{4-2\sqrt{3}}{{{\left( 4 \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}}} $

$=\frac{4-2\sqrt{3}}{16-{{\left( 2\sqrt{3} \right)}^{2}}} $ 

$=\frac{4-2\sqrt{3}}{16-12} $ 

$  =\frac{4-2\sqrt{3}}{4} $

$ =\frac{2\left( 2-\sqrt{3} \right)}{4} $

$ =\frac{2-\sqrt{3}}{2} $

24. Visualize the representation of $5.3\overline{7}$ on the number line up to 3 decimal places.

Ans: The representation of $5.3\overline{7}$ on the number line is given below:

25. Show that \[5\sqrt{2}\] is not a rational number.

Ans: Let us assume that \[5\sqrt{2}\] is a rational number.

Take \[x=5\sqrt{2}\] , with \[x\]being rational as well.

\[x=5\sqrt{2}\]

\[\Rightarrow \frac{x}{5}=\sqrt{2}\]

Let us compare the terms in LHS and RHS.

In LHS, we have\[\frac{x}{5}\] , with \[x\] and $5$ being rational numbers (Here \[x\] is rational, based on our assumption). So \[\frac{x}{5}\] is a rational number.

In RHS, we have$\sqrt{2}$, which is not a rational number, but an irrational number.

This is a contradiction, i.e. $LHS\ne RHS$.

So, we can conclude that  \[5\sqrt{2}\] is not a rational number.

26. Simplify \[3\sqrt[3]{250}+7\sqrt[3]{16}-4\sqrt[3]{54}\].

Ans: Let us first find the cube roots of given numbers to their simplest forms by using the prime factorization then we get,

\[3\sqrt[3]{250}+7\sqrt[3]{16}-4\sqrt[3]{54}=3\sqrt[3]{5\times 5\times 5\times 2}+7\sqrt[3]{2\times 2\times 2\times 2}-4\sqrt[3]{3\times 3\times 3\times 2}\]

\[=\left( 3\times 5\sqrt[3]{2} \right)+\left( 7\times 2\sqrt[3]{2} \right)-\left( 4\times 3\sqrt[3]{2} \right)\]

\[=\left( 15\sqrt[3]{2} \right)+\left( 14\sqrt[3]{2} \right)-\left( 12\sqrt[3]{2} \right)\]

\[=\left( 15+14-12 \right)\sqrt[3]{2}\]

\[=17\sqrt[3]{2}\]

Thus, we get \[3\sqrt[3]{250}+7\sqrt[3]{16}-4\sqrt[3]{54}=17\sqrt[3]{2}\]

27. Simplify \[3\sqrt{48}-\frac{5}{2}\sqrt{\frac{1}{3}}+4\sqrt{3}\].

Ans: Let us first find the square roots of given numbers to their simplest forms by using the prime factorization then we get,

\[3\sqrt{48}-\frac{5}{2}\sqrt{\frac{1}{3}}+4\sqrt{3}=\left( 3\sqrt{2\times 2\times 2\times 2\times 3} \right)-\left[ \frac{5}{2}\left( \sqrt{\frac{1}{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \right) \right]+\left( 4\sqrt{3} \right)\]

\[=\left( 3\times 2\times 2\sqrt{3} \right)-\left[ \frac{5}{2}\left( \frac{\sqrt{3}}{3} \right) \right]+\left( 4\sqrt{3} \right)\]

\[=\left( 12\sqrt{3} \right)-\left( \frac{5\sqrt{3}}{6} \right)+\left( 4\sqrt{3} \right)\]

\[=\left( 12-\frac{5}{6}+4 \right)\sqrt{3}\]

\[=\left( \frac{72-5+24}{6} \right)\sqrt{3}\]

\[=\frac{91}{6}\sqrt{3}\]

Thus, we get \[3\sqrt{48}-\frac{5}{2}\sqrt{\frac{1}{3}}+4\sqrt{3}=\frac{91}{6}\sqrt{3}\]

28. If $\frac{1}{7}=0.\overline{142857}$. Find the value of $\frac{2}{7},\frac{3}{7},\frac{4}{7}$

Ans: It is given that – $\frac{1}{7}=0.\overline{142857}$

(i) $\frac{2}{7}=2\times \frac{1}{7}$

$=2\times 0.\overline{142857}$

$=0.\overline{285714}$

$\Rightarrow \frac{2}{7}=0.\overline{285714}$

(ii) $\frac{3}{7}=3\times \frac{1}{7}$

$=3\times 0.\overline{142857}$

$=0.\overline{428571}$

$\Rightarrow \frac{3}{7}=0.\overline{428571}$

(iii) $\frac{4}{7}=4\times \frac{1}{7}$

$=4\times 0.\overline{142857}$

$=0.\overline{571428}$

$\Rightarrow \frac{4}{7}=0.\overline{571428}$

29. Find $6$ rational numbers between $\frac{6}{5}$ and $\frac{7}{5}$

Ans: It is possible to divide the interval between $\frac{6}{5}$ and $\frac{7}{5}$ into $10$ equal parts.

Then we will have – $\frac{6}{5},\frac{6.1}{5},\frac{6.2}{5},\frac{6.3}{5},\frac{6.4}{5},\frac{6.5}{5},\frac{6.6}{5},\frac{6.7}{5},\frac{6.8}{5},\frac{6.9}{5},\frac{7}{5}$

i.e. $\frac{60}{50},\frac{61}{50},\frac{62}{50},\frac{63}{50},\frac{64}{50},\frac{65}{50},\frac{66}{50},\frac{67}{50},\frac{68}{50},\frac{69}{50},\frac{70}{50}$

From these fractions, it is possible to choose $6$ rational numbers between $\frac{6}{5}$ and $\frac{7}{5}$

Thus , $6$ rational numbers between $\frac{6}{5}$ and $\frac{7}{5}$ are $\frac{61}{50},\frac{62}{50},\frac{63}{50},\frac{64}{50},\frac{65}{50},\frac{66}{50}$

30. Show how $\sqrt{4}$ can be represented on the number line.

Ans: Take $AB=OA=1\text{ }unit$ on a number line.

Also, $\angle A={{90}^{\circ }}$

In $\vartriangle OAB$, apply Pythagoras Theorem,

$\therefore O{{A}^{2}}+A{{B}^{2}}=O{{B}^{2}}$

$\Rightarrow O{{B}^{2}}={{1}^{2}}+{{1}^{2}}$

$\Rightarrow O{{B}^{2}}=1+1$

$\Rightarrow O{{B}^{2}}=2$

$\Rightarrow OB=\sqrt{2}$

Now, draw $OB=O{{A}_{1}}=\sqrt{2}$

And, ${{A}_{1}}{{B}_{1}}=1\text{ unit}$ with$\angle {{A}_{1}}={{90}^{\circ }}$

In \[\vartriangle O{{A}_{1}}{{B}_{1}}\], apply Pythagoras Theorem,

$\therefore O{{A}_{1}}^{2}+{{A}_{1}}{{B}_{1}}^{2}=O{{B}_{1}}^{2}$

$\Rightarrow O{{B}_{1}}^{2}={{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow O{{B}_{1}}^{2}=2+1$

$\Rightarrow O{{B}_{1}}^{2}=3$

$\Rightarrow O{{B}_{1}}=\sqrt{3}$

Now, draw $O{{B}_{1}}=O{{A}_{2}}=\sqrt{3}$

And, \[{{A}_{2}}{{B}_{2}}=1\text{ unit}\] with$\angle {{A}_{2}}={{90}^{\circ }}$

In \[\vartriangle O{{A}_{2}}{{B}_{2}}\], apply Pythagoras Theorem,

$\therefore O{{A}_{2}}^{2}+{{A}_{2}}{{B}_{2}}^{2}=O{{B}_{2}}^{2}$

$\Rightarrow O{{B}_{2}}^{2}={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow O{{B}_{2}}^{2}=3+1$

$\Rightarrow O{{B}_{2}}^{2}=4$

$\Rightarrow O{{B}_{2}}=\sqrt{4}$

Now, draw $O{{B}_{2}}=O{{A}_{3}}=\sqrt{4}$

Thus line segment $O{{A}_{3}}=\sqrt{4}$

Short Answer Questions (4 Marks)

1. Write the following in decimal form and say what kind of decimal expansion each has:

i. $\frac{36}{100}$

Ans: Performing long division of $36$ by $100$

$\begin{matrix} &{0.36}\\ 100&{\overline{)\;36\quad}}\\ &\underline{-0\quad}\\ &360\\ &\underline{-300\quad}\\ &\;\;600\\ &\underline{-600}\\ &\underline{\quad 0 \;\;} \end{matrix}$

Thus, $\frac{36}{100}=0.36$ - this is a terminating decimal.

ii. $\frac{1}{11}$

Ans: Performing long division of $1$ by $11$

$\begin{matrix} {} & 0.0909.. \\ 11 & \overline{)\text{ }1\text{ }} \\ {} & \underline{-0} \\ {} & 10 \\ {} & \underline{-0} \\ {} & 100 \\ {} & \underline{-99} \\ {} & 10 \\ {} & \underline{-0} \\ {} & 100 \\ {} & \underline{-99} \\ {} & 1 \\ \end{matrix}$

It can be seen that performing further division will produce a reminder of $1$ continuously.

Thus, $\frac{1}{11}=0.09090...$ i.e. $\frac{1}{11}=0.\overline{09}$, this is a non-terminating, but recurring decimal.

iii. $4\frac{1}{8}$

Ans: First convert the mixed fraction into an improper fraction –

$4\frac{1}{8}=\frac{(4\times 8)+1}{8}=\frac{33}{8}$

Performing long division of $33$ by $8$

$\begin{matrix} {} & 4.125 \\ 8 & \overline{\left){\text{ }33\text{ }}\right.} \\ {} & \underline{-32} \\ {} & 10 \\ {} & \underline{-8} \\ {} & 20 \\ {} & \underline{-16} \\ {} & 40 \\ {} & \underline{-40} \\ {} & 0 \\ \end{matrix}$

Thus, $4\frac{1}{8}=4.125$ - this is a terminating decimal.

iv. $\frac{3}{13}$

Ans: Performing long division of $3$ by $13$

$\begin{matrix} {} & 0.230769.. \\ 13 & \overline{\left){\text{ }3\text{ }}\right.} \\ {} & \underline{-0} \\ {} & 30 \\ {} & \underline{-26} \\ {} & 40 \\ {} & \underline{-39} \\ {} & 10 \\ {} & \underline{-0} \\ {} & 100 \\ {} & \underline{-91} \\ {} & 90 \\ {} & \underline{-78} \\ {} & 120 \\ {} & \underline{-117} \\ {} & 3 \\ \end{matrix}$

It can be seen that performing further division will produce a reminder of $3$ periodically, after every six divisions.

Thus, $\frac{3}{13}=0.230769...$ i.e. \[\frac{3}{13}=0.\overline{230769}\], this is a non-terminating, but recurring decimal.

v. $\frac{2}{11}$

Ans: Performing long division of $2$ by $11$

$\begin{matrix} {} & 0.1818.. \\ 11 & \overline{)\text{ 2 }} \\ {} & \underline{-0} \\ {} & 20 \\ {} & \underline{-11} \\ {} & 90 \\ {} & \underline{-88} \\ {} & 20 \\ {} & \underline{-11} \\ {} & 90 \\ {} & \underline{-88} \\ {} & 2 \\ \end{matrix}$

It can be seen that performing further division will produce a reminder of $2$followed by $9$ alternatively.

Thus, $\frac{2}{11}=0.181818...$ i.e. $\frac{2}{11}=0.\overline{18}$this is a non-terminating, but recurring decimal.

vi. $\frac{329}{400}$

Ans: Performing long division of $33$ by $8$

$\begin{matrix} {} & 0.8225 \\ 400 & \overline{)\text{ 329 }} \\ {} & \underline{-0} \\ {} & 3290 \\ {} & \underline{-3200} \\ {} & 900 \\ {} & \underline{-800} \\ {} & 1000 \\ {} & \underline{-800} \\ {} & 2000 \\ {} & \underline{-2000} \\ {} & 0 \\ \end{matrix}$

Thus, $\frac{329}{400}=0.8225$ - this is a terminating decimal.

2. Classify the following as rational or irrational:

i. $\sqrt{23}$

Ans: It is known that the root of $23$ will produce a non-terminating and non-recurring decimal number (it is not a perfect square value), also it cannot be represented as a fraction. Thus we can say that $\sqrt{23}$ is an irrational number.

ii. $\sqrt{225}$

Ans: It is known that $\sqrt{225}=15$, which is an integer.

Thus $\sqrt{225}$ is a rational number.

iii. $0.3796$

Ans: Here, $0.3796$ is a terminating decimal number, and also it can be expressed as a fraction.

i.e. $0.3796=\frac{3796}{10000}=\frac{949}{2500}$

Thus $0.3796$ is a rational number.

iv. $7.478478...$

Ans: Here, $7.478478...$ is a non-terminating, but recurring decimal number, and also it can be expressed as a fraction.

i.e. $7.478478...=7.\overline{487}$

Converting it into fraction 

If $x=7.478478...\text{          (1)}$

Then $1000x=7478.478478...\text{          (2)}$

Subtract equations $(2)-(1)$

$ 1000x=7478.478478... $

$  \underline{-\text{      }x=\text{      }7.478478...} $

$ \text{  }999x=7471 $

Now, $999x=7471$

$\Rightarrow x=\frac{7471}{999}$

i.e. $7.\overline{478}=\frac{7471}{999}$

Thus $7.478478...$ is a rational number.

v. $1.101001000100001...$

Ans: Here,  $1.101001000100001...$ is a non-terminating and non-recurring decimal number and also it cannot be represented as a fraction. Thus we can say that $1.101001000100001...$ is an irrational number.

3. Rationalize the denominator of the following:

(i) $\frac{1}{\sqrt{7}}$

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{7}}$ by $\sqrt{7}$

$\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$

Rationalizing the denominator of  $\frac{1}{\sqrt{7}}$ produces $\frac{\sqrt{7}}{7}$.

ii. $\frac{1}{\sqrt{7}-\sqrt{6}}$

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{7}-\sqrt{6}}$ by $\sqrt{7}+\sqrt{6}$

$\frac{1}{\sqrt{7}-\sqrt{6}}\times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{\left( \sqrt{7}-\sqrt{6} \right)\left( \sqrt{7}+\sqrt{6} \right)}$

Using the identity - \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

$=\frac{\sqrt{7}+\sqrt{6}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}}$

$=\frac{\sqrt{7}+\sqrt{6}}{7-6}$

$=\frac{\sqrt{7}+\sqrt{6}}{1}$

$\Rightarrow \frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6}$

Rationalizing the denominator of  $\frac{1}{\sqrt{7}-\sqrt{6}}$ produces $\sqrt{7}+\sqrt{6}$.

iii. $\frac{1}{\sqrt{5}+\sqrt{2}}$

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{5}+\sqrt{2}}$ by $\sqrt{5}-\sqrt{2}$

$\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)}$

$=\frac{\sqrt{5}-\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}$

$=\frac{\sqrt{5}-\sqrt{2}}{5-2}$

$=\frac{\sqrt{5}-\sqrt{2}}{3}$

$\Rightarrow \frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{3}$

Rationalizing the denominator of  $\frac{1}{\sqrt{5}+\sqrt{2}}$ produces $\frac{\sqrt{5}-\sqrt{2}}{3}$.

iv. $\frac{1}{\sqrt{7}-2}$

Ans: In order to rationalize the denominator, we multiply and divide $\frac{1}{\sqrt{7}-2}$ by $\sqrt{7}+2$

$\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{\left( \sqrt{7}-2 \right)\left( \sqrt{7}+2 \right)}$

$=\frac{\sqrt{7}+2}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( 2 \right)}^{2}}}$

$=\frac{\sqrt{7}+2}{7-4}$

$=\frac{\sqrt{7}+2}{3}$

$\Rightarrow \frac{1}{\sqrt{7}+2}=\frac{\sqrt{7}+2}{3}$

Rationalizing the denominator of  $\frac{1}{\sqrt{7}-2}$ produces $\frac{\sqrt{7}+2}{3}$.

Long Answer Questions (5 Marks)

It can be seen that performing further divisions will produce a reminder of $3$ periodically, after every six divisions.

Thus, $\frac{329}{400}=s0.8225$ - this is a terminating decimal.

4. If $\sqrt{5}=2.236$ and $\sqrt{3}=1.732$. Find the value of $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}$.

Ans: It is given that – 

$\sqrt{5}=2.236$

$\sqrt{3}=1.732$

Now, $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}$

$\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}=\left[ \frac{2}{\left( \sqrt{5}+\sqrt{3} \right)}\times \frac{\left( \sqrt{5}-\sqrt{3} \right)}{\left( \sqrt{5}-\sqrt{3} \right)} \right]+\left[ \frac{7}{\left( \sqrt{5}-\sqrt{3} \right)}\times \frac{\left( \sqrt{5}+\sqrt{3} \right)}{\left( \sqrt{5}+\sqrt{3} \right)} \right]$

$=\left[ \frac{2\left( \sqrt{5}-\sqrt{3} \right)}{\left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)} \right]+\left[ \frac{7\left( \sqrt{5}+\sqrt{3} \right)}{\left( \sqrt{5}-\sqrt{3} \right)\left( \sqrt{5}+\sqrt{3} \right)} \right]$

$=\left[ \frac{\left( 2\sqrt{5}-2\sqrt{3} \right)+\left( 7\sqrt{5}+7\sqrt{3} \right)}{\left( \sqrt{5}+\sqrt{3} \right)\left( \sqrt{5}-\sqrt{3} \right)} \right]$

$=\left[ \frac{2\sqrt{5}-2\sqrt{3}+7\sqrt{5}+7\sqrt{3}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}} \right]$

$=\left[ \frac{(2+7)\sqrt{5}+(7-2)\sqrt{3}}{5-3} \right]$

$=\left[ \frac{9\sqrt{5}+5\sqrt{3}}{2} \right]$

Since, $\sqrt{5}=2.236$ and $\sqrt{3}=1.732$

$=\left[ \frac{(9\times 2.236)+(5\times 1.732)}{2} \right]$

$=\left[ \frac{20.124+8.66}{2} \right]$

$=\left[ \frac{28.784}{2} \right]$

Thus, $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{7}{\sqrt{5}-\sqrt{3}}=14.392$

5. Find the value of $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}$, if $\sqrt{5}=2.236$ and $\sqrt{2}=1.414$.

$\sqrt{2}=1.414$

Now, $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}$

$\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}=\left[ \frac{3}{\left( \sqrt{5}+\sqrt{2} \right)}\times \frac{\left( \sqrt{5}-\sqrt{2} \right)}{\left( \sqrt{5}-\sqrt{2} \right)} \right]+\left[ \frac{7}{\left( \sqrt{5}-\sqrt{2} \right)}\times \frac{\left( \sqrt{5}+\sqrt{2} \right)}{\left( \sqrt{5}+\sqrt{2} \right)} \right]$

$=\left[ \frac{3\left( \sqrt{5}-\sqrt{2} \right)}{\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)} \right]+\left[ \frac{7\left( \sqrt{5}+\sqrt{2} \right)}{\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}+\sqrt{2} \right)} \right]$

$=\left[ \frac{\left( 3\sqrt{5}-3\sqrt{2} \right)+\left( 7\sqrt{5}+7\sqrt{2} \right)}{\left( \sqrt{5}+\sqrt{2} \right)\left( \sqrt{5}-\sqrt{2} \right)} \right]$

$=\left[ \frac{3\sqrt{5}-3\sqrt{2}+7\sqrt{5}+7\sqrt{2}}{{{\left( \sqrt{5} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}} \right]$

$=\left[ \frac{(3+7)\sqrt{5}+(7-3)\sqrt{2}}{5-2} \right]$

$=\left[ \frac{10\sqrt{5}+4\sqrt{2}}{3} \right]$

Since, $\sqrt{5}=2.236$ and $\sqrt{2}=1.414$

$=\left[ \frac{(10\times 2.236)+(4\times 1.414)}{3} \right]$

$=\left[ \frac{22.36+5.656}{3} \right]$

$=\left[ \frac{28.016}{3} \right]$

Thus, $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{7}{\sqrt{5}-\sqrt{2}}=\frac{28.016}{3}$

6. Simplify $\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}$

Ans: $\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}$

\[\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=\left[ \frac{2+\sqrt{5}}{2-\sqrt{5}}\times \frac{\left( 2+\sqrt{5} \right)}{\left( 2+\sqrt{5} \right)} \right]+\left[ \frac{2-\sqrt{5}}{2+\sqrt{5}}\times \frac{\left( 2-\sqrt{5} \right)}{\left( 2-\sqrt{5} \right)} \right]\]

\[=\left[ \frac{\left( 2+\sqrt{5} \right)\left( 2+\sqrt{5} \right)}{\left( 2-\sqrt{5} \right)\left( 2+\sqrt{5} \right)} \right]+\left[ \frac{\left( 2-\sqrt{5} \right)\left( 2-\sqrt{5} \right)}{\left( 2+\sqrt{5} \right)\left( 2-\sqrt{5} \right)} \right]\]

\[=\left[ \frac{{{\left( 2+\sqrt{5} \right)}^{2}}+{{\left( 2-\sqrt{5} \right)}^{2}}}{\left( 2-\sqrt{5} \right)\left( 2+\sqrt{5} \right)} \right]\]

Using the identities –

\[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

\[{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]

\[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]

\[=\left[ \frac{\left( {{\left( 2 \right)}^{2}}+{{\left( \sqrt{5} \right)}^{2}}+\left( 2\times 2\times \sqrt{5} \right) \right)+\left( {{\left( 2 \right)}^{2}}+{{\left( \sqrt{5} \right)}^{2}}-\left( 2\times 2\times \sqrt{5} \right) \right)}{{{\left( 2 \right)}^{2}}-{{\left( \sqrt{5} \right)}^{2}}} \right]\]

\[=\left[ \frac{\left( 4+5+\left( 4\sqrt{5} \right) \right)+\left( 4+5-\left( 4\sqrt{5} \right) \right)}{4-5} \right]\]

$=\left[ \frac{9+9}{-1} \right]$

$=\left[ \frac{18}{-1} \right]$

Thus, $\frac{2+\sqrt{5}}{2-\sqrt{5}}+\frac{2-\sqrt{5}}{2+\sqrt{5}}=\left( -18 \right)$

7. Find a and b, if $\frac{3-\sqrt{6}}{3+2\sqrt{6}}=a\sqrt{6}-b$

Ans: $\frac{3-\sqrt{6}}{3+2\sqrt{6}}=a\sqrt{6}-b$

Here, 

$LHS=\frac{3-\sqrt{6}}{3+2\sqrt{6}}$

$RHS=a\sqrt{6}-b$

Start by rationalizing the denominator in LHS

In order to rationalize the denominator, we multiply and divide $\frac{3-\sqrt{6}}{3+2\sqrt{6}}$ by $3+2\sqrt{6}$

$\frac{3-\sqrt{6}}{3+2\sqrt{6}}\times \frac{3-2\sqrt{6}}{3-2\sqrt{6}}=\frac{\left( 3-\sqrt{6} \right)\left( 3-2\sqrt{6} \right)}{\left( 3+2\sqrt{6} \right)\left( 3-2\sqrt{6} \right)}$

$=\frac{\left( 3\times 3 \right)-\left( 3\times 2\sqrt{6} \right)-\left( \sqrt{6}\times 3 \right)+\left( \sqrt{6}\times 2\sqrt{6} \right)}{{{\left( 3 \right)}^{2}}-{{\left( 2\sqrt{6} \right)}^{2}}}$

$=\frac{\left( 9 \right)-\left( 6\sqrt{6} \right)-\left( 3\sqrt{6} \right)+\left( 12 \right)}{9-24}$

$=\frac{\left( 21 \right)-\left( 9\sqrt{6} \right)}{-15}$

$=\frac{\left( 21 \right)}{-15}-\frac{\left( 9\sqrt{6} \right)}{-15}$

They are all divisible by $3$

$=-\frac{7}{5}+\frac{\left( 3\sqrt{6} \right)}{5}$

Thus, $LHS=\frac{3}{5}\sqrt{6}-\frac{7}{5}$

Comparing with RHS, we get – 

Thus, 

$a=\frac{3}{5}$

$b=\frac{7}{5}$

Important Questions for Class 9 Maths Chapter 1 - Free PDF Download

Class 9 is like beginning to your academics career, which is right before board 10th. Thus students need to be very serious regarding their studies during preparation. No matter what we learn in class 9, it is important to clarify your concepts better. Hence, clearing Number System concepts will help students further apply in electronics physics and higher maths. Thus it is better to build a good base in mathematics with these Important Questions Maths Class 9 Chapter 1.

Vedantu provides a free PDF to download for Class 9 Chapter 1 Important Questions such that students can prepare well according to the CBSE syllabus. CBSE is strict to its pattern and follows the same throughout the question paper set. Students need to understand these guidelines and find solutions with a proper explanation. This free PDF online will surely help students master their concepts and build a Number System base. This PDF covers all the important concepts in the form of question example to learn how to implement them during exams. Thus PDF proves to be magical for those who are weak in mathematics as they also get solutions to the concepts covered in the back exercise.

Number System Class 9 Important Questions

Before you begin practising Class 9 Chapter 1 Maths Important Questions, you need to know the different topics and subtopics to cover in the chapter. Chapter 1 of Mathematics Class 9 covers a total of 6 exercises with a small introduction of the number system, number lines, defining real numbers, natural numbers, whole numbers, rational, and irrational numbers. Also, students become familiar with the concepts of addition, subtraction, division, and multiplication of the real numbers. The last topic in the chapter will be covering the law of exponents in the real numbers.

Below are the mentioned section-wise topics and concepts that a student to prepare through important questions.

Exercise 1.1

Under exercise 1.1, students will become familiar with the basic understanding of rational and irrational numbers. There is also a revision on whole numbers, real number, integers, and natural numbers and definition. They will even know how to represent a number in the form of p/q, where q is not equal to 0.

Exercise 1.2

Further moving to in-depth study about rational and irrational numbers, there are questions on justification with true and false. Also, students become familiar with the concept of representing rational and irrational numbers on the number line. A new concept of constructing a square root spiral comes in exercise 1.2.

Exercise 1.3

Here the student will know how to represent fractions into decimal form and find if it is terminating or non-terminating. Thus the concepts of terminating and non-terminating fractions will further help to identify it is a rational or irrational number.

Exercise 1.4

A new concept of representing decimal expansion on the number line is introduced through exercise 1.4 mathematics class 9. Here students will learn about magnifying the number to the maximum requirement and representing it on the number line. These decimal places can be either terminating or non-terminating. Hence there are two different concepts in number line representation of decimal expansions.

Exercise 1.5

Now comes the basic calculations of different rational and irrational numbers. It includes addition, multiplication, subtraction, and division of the rational and irrational numbers. Questions will be based on such concepts, and thus you have to simplify the statement accordingly. Also, students become familiar with the concept of rationalising.

Exercise 1.6

Here students will learn to solve questions based with a number having power in fractional form. Also, it might cover the basic addition and subtraction of powers for in-depth conceptual and extra knowledge.

These are the six exercises which will be covered under Chapter 1 Maths Class 9 Important Questions. Hence students can prepare questions according to the concepts discussed above.

Chapter 1 Maths Class 9 Important Questions

According to the syllabus mentioned above exercise-wise, below are some important questions covered to let students prepare well for important questions for class 9 maths number system. These exercise-wise solutions will let students master each concept in detail. Below are some of the questions that are usually picked to set question paper as prepared by CBSE.

State if zero is a rational number. Justify your statement by representing it in p/q where q is not equal to 0 and p and q are both integers.

Find 5 different rational numbers between 5 and 6. Mention each step in detail.

Find out 5 different rational numbers between 12/11 and 10/11.

Justify your statement for the following terms stating true or false.

Rational numbers are real.

A number line having representation in the form of √m has m as a natural number.

A real number is always an irrational number.

Check if true or false. Square roots of all positive numbers will be irrational. Explain your answer statement with the help of an example.

Try to represent √5 on the number line.

Represent the following in the form of decimal expansion:

(i) 36/100  (ii) 1/11  (iii) 4⅛  (iv) 3/13  (v) 2/11  (vi) 329/400

Turn 2/7, 13/7, 4/7, 5/7, 6/7

into decimal expansions without actually doing any long division calculations. Here 1/7= 0.142857.

Express the following as fractional form p/q where q is not 0.

(i) \[0.\overline{6}\]

(ii) \[0.4 \overline{7}\]

(iii) \[0. \overline{001}\]

Represent 2.675 on the number line with number line magnification.

How will you visualise 6.2626…... on the number line up to 4 decimal places?

State if the numbers are rational or irrational.

(i) \[2 - \sqrt{5}\]  (ii)  \[(3 + \sqrt{23}) - \sqrt{23}\]  (iii) \[\frac{2\sqrt{7}}{7\sqrt{7}}\]  (iv) \[\frac{1}{\sqrt{2}}\]  (v) \[2 \pi\]

Simplify the following terms:

(i) \[(3 + \sqrt{3}) ( 2 + \sqrt{2})\] (ii) \[(3 + \sqrt{3}) (3 + \sqrt{3})\]

(iii) \[(\sqrt{5} + \sqrt{2})^{2}\] (iv)\[(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})\]

Rationalise the following terms with their denominators:

(i) \[\frac{1}{7}\] (ii) \[\frac{1}{\sqrt{7} - \sqrt{6}}\] (iii) \[\frac{1}{\sqrt{5} + \sqrt{2}}\] (iv) \[\frac{1}{\sqrt{7} - 2}\]

(i) \[64^{1}{2}\] (ii) \[32^{1}{5}\] (iii) \[125^{1}{3}\]

Class 9 Maths Chapter 1 Extra Questions

Find three rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$. 

Express 0.4323232 in the form of  $\frac{a}{b}$ where a and b are integers and b 0.

Simplify and find the value of $(729)^{1/6}$ .

Rationalise the denominator 1 9 + 5 + 6 .

Find 6 rational numbers between 4 and 6.

Simplify $\sqrt[3]{2}$+$\sqrt[4]{3}$ and $\sqrt{5}+\sqrt{2}$ .

Locate $\sqrt{5}$ on the number line.

Visualise the representation of 4.26 on the number line upto 3 decimal places.

Is 2 - 5   a rational number or irrational number?

Convert 0.45 into rational numbers.

Benefits of Important Questions for Class 9 Maths Number System

Number System Class 9 Important Questions PDF is prepared by Vedantu experts and is free of cost. Students can also schedule additional problems to prepare in-depth to clarify their concepts. These questions turn to be magical for those weak in mathematics or do not show interest in it. The questions are prepared precisely accordingly to CBSE guidelines for their question paper pattern such that students need not search them anywhere else.

Below are the mentioned reasons why students should refer to the Important Question of Maths Class 9 Chapter 1:

Students are free to access these important questions PDF. They need not pay for any study material on Vedantu website.

These Number System Class 9 Important Questions are helpful during exams and help students clarify their concepts of homework questions.

These crucial questions are available in PDF format, which can be easily downloaded through the website. Thus students need not unnecessarily waste their precious time in finding solutions to problems.

Students can print these PDF questions and solutions, which reduces the stress of preparing through soft copies.

All the questions and solutions to the questions are prepared according to CBSE guidelines. Thus it will help students to know question paper pattern.

There is a total of 100 marks for the Class 9 Mathematics paper where 20 marks are for internal assessment, and rest 80 is for the written exam. The marks weightage for Chapter 1 Maths Class 9 is 8 out of 80. Rest 72 marks are for rest of the syllabus of Class 9 Mathematics.

There are a total of 26 sums to be solved according to NCERT book for Chapter 1 Of Class 9 Maths in its six exercises. However, there are other Class 9 Maths Chapter 1 Important Questions prepared by the experts and solutions to each of them provided. Thus, students can build a good base in mathematics through this PDF of important questions available with Vedantu for free. According to exercise-wise, some questions describe different questions and situations that students generally encounter while sitting in the exam. Thus students need to struggle to find essential concepts in the chapter and solutions to each of them.

Important Related Links for CBSE Class 9 

FAQs on CBSE Class 9 Maths Important Questions for Chapter 1 - Number System

1. What number of questions are there in each exercise of Chapter 1 of Class 9 Maths?

Real numbers are used in the first chapter of Class 9 Mathematics. There are a total of six workouts. In Chapter 1, students will learn about several real number concepts. There are four questions in Exercise 1.1, four questions in Exercise 1.2, nine questions in Exercise 1.3, two questions in Exercise 1.4, five questions in Exercise 1.5, and three questions in Exercise 1.6. Students must rehearse all NCERT problems offered by Vedantu in order to understand Chapter 1 of Class 9 maths.

2. Is Chapter 1 of Class 9 Maths difficult to solve?

Students of Class 9 should understand the concepts of real numbers given in Chapter 1 of Class 9 Maths. It is not difficult to solve if students understand the basic concepts of real numbers. Students can take help from the NCERT solutions Class 9 Maths Chapter 1 given at Vedantu app and website. All concepts related to the real numbers are explained in a simple way for quick understanding of the students. 

3. Why are Class 9 Maths NCERT Solutions of Chapter 1 important?

Students should rehearse Class 9 Mathematics NCERT Answers for Chapter 1 to comprehend and practise real-world questions. The NCERT Answers are essential since students may face similar questions in their examinations. With the aid of these questions, students may simply tackle the actual numbers problems in the exam. Students may access complete NCERT Answers for Class 9 Mathematics Chapter 1 for free on the Vedantu platform.

4. What are five rational numbers between ⅖ and ⅗?

Students can find five or more rational numbers between the two given rational numbers easily. They can follow the given steps to find five rational numbers between ⅖ and ⅗. We have to multiply the numerator and denominator of the given rational numbers with the same number.

⅖ x 6/6 = 12/30

⅗ x 6/6= 18/30

Now, we can write five rational numbers between 12/30 and 18/30. Thus, the five rational numbers are 13/30, 14/30, 15/30, 16/30, 17/30.

5. What are the main topics covered in Chapter 1 of Class 9 Maths?

In Chapter 1 of Class 9 Mathematics, students will study about rational and irrational numbers. Students will learn about irrational numbers, how to represent real numbers on a number line, how to express real numbers in decimal form, different operations on real numbers, and real-world exponent rules. To comprehend the various principles of rational numbers in Class 9 Chapter 1, students should practise all NCERT problems offered in the textbook.

CBSE Class 9 Maths Important Questions

Cbse study materials.

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 9 Science Chapter 1 Matter in Our Surroundings

• Last modified on: 7 days ago
• Reading Time: 6 Minutes

Here we are providing case study questions for class 9 science chapter 12 sound. Students are suggested to go through each and every case study questions for better understanding of the chapter.

Case Study/Passage Based Questions:

Question 1:

Read the following passage and answer the questions given below.

Every matter is made up of tiny particles. These particles are so tiny that they can’t be seen with naked eyes.

The three characteristics shown by particles of matter are as follows:

(i) There are small voids between particles in a matter. This characteristic is the concept behind the solubility of a substance in other substances.

(ii) Particles of matter show continuous random movements, that is they possess kinetic energy. The spreading of ink in a beaker of glass, smell of agarbattis, etc. are few illustrations that show the movement of particles of a substance.

(iii) The particles of matter attract each other with a force called interparticle force of attraction. Read the given passage carefully and give the answer of the following questions:

Q 1. Spreading of fragrance of a burning incense stick in a room shows that:

a. particles of matter have spaces between them.

b. particles of matter attract each other.

c. particles of matter are constantly moving.

d. None of the above

Q 2. What happens when we add sugar to water?

a. Volume of water doubles.

b. Volume of water decreases

c. Volume of water remains the same.

Q 3. A stream of water cannot be cut by fingers. Which property of matter does this observation show?

a. Particles of matter attract each other.

b. Particles of matter have spaces between them.

c. Particles of matter are continuously moving.

Q 4. When we put some crystals of potassium permanganate in a beaker containing water, we observe that after some time, the whole water turns pink. This intermixing of particles of two different types of matter on their own is called:

a. Brownian motion

c. sublimation

d. diffusion

Q 5. Why is the rate of diffusion of liquids higher than that of solids?

a. In the liquid state, particles are tightly packed as compared to solids.

b. In the liquid state, particles move freely as compared to solids.

c. In solid state, particles have least force of attraction between the particles.

d. In solid state, particles cannot be compressed easily.

• (c) particles of matter are constantly moving.
• (c) Volume of water remains the same.
• (a) Particles of matter attract each other.
• (d) diffusion
• (b) In the liquid state, particles move freely as compared to solids

Related Posts

Download cbse books.

Exam Special Series:

• Sample Question Paper for CBSE Class 10 Science (for 2024)
• Sample Question Paper for CBSE Class 10 Maths (for 2024)
• CBSE Most Repeated Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Important Numericals Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
• Sample Question Papers for CBSE Class 12 Physics (for 2024)
• Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
• Sample Question Papers for CBSE Class 12 Maths (for 2024)
• Sample Question Papers for CBSE Class 12 Biology (for 2024)
• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• 10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online JEE Test Series for 499/- Only (Web + App) for 1 Year

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

Leave a Reply Cancel reply

Join our Online Test Series for CBSE, ICSE, JEE, NEET and Other Exams

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

CBSE Expert

CBSE Case Study Questions Class 9 Maths Chapter 11 Constructions PDF Download

CBSE Case Study Questions Class 9 Maths Chapter 11  are very important to solve for your exam. Class 9 Maths Chapter 11 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  case study-based   questions for Class 9 Maths Chapter 11  Constructions

CBSE Case Study Questions Class 9 Maths Chapter 11

Case Study 1: A group of students is learning about constructions in geometry. They encountered the following scenario:

Aman and Bhavya were given a task to construct a triangle with specific conditions. They made the following observations:

• Aman constructed a triangle with two sides measuring 5 cm and 6 cm, and the included angle measuring 60 degrees.
• Bhavya constructed a triangle with one side measuring 7 cm, and the adjacent angles measuring 45 degrees and 60 degrees.

Based on this information, the students were asked to analyze the constructions made by Aman and Bhavya. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The type of triangle constructed by Aman is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Answer: (c) Scalene

Q2. The type of triangle constructed by Bhavya is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Answer: (b) Isosceles

Q3. The measure of the third angle in the triangle constructed by Aman is: (a) 30 degrees (b) 45 degrees (c) 60 degrees (d) 90 degrees

Answer: (c) 60 degrees

Q4. The measure of the third angle in the triangle constructed by Bhavya is: (a) 45 degrees (b) 60 degrees (c) 75 degrees (d) 90 degrees

Answer: (a) 45 degrees

Q5. The construction made by Aman satisfies the following condition: (a) Side-side-side (SSS) (b) Angle-side-angle (ASA) (c) Side-angle-side (SAS) (d) None of the above

Answer: (b) Angle-side-angle (ASA)

Hope the information shed above regarding Case Study and Passage Based Questions for Case Study Questions Class 9 Maths Chapter 11 Constructions with Answers Pdf free download has been useful to an extent. If you have any other queries about Case Study Questions Class 9 Maths Chapter 11 Constructions and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible.

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Download India's best Exam Preparation App Now.

Key Features

• Revision Notes
• Important Questions
• Previous Years Questions
• Case-Based Questions
• Assertion and Reason Questions

No thanks, I’m not interested!