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NCERT Solutions for Class 9 Maths (PDF): Updated for 2022-2023

Ncert solutions for class 9 maths are available here for download in pdf. the chapter-wise solutions are best to prepare for the annual assessment in class 9 maths exam 2022-23..

NCERT Solutions for Class 9 Maths: Check the best explained NCERT Solutions for Class 9 Maths below. Download the chapter-wise solutions in PDF. These solutions have been updated for the latest edition of the NCERT Maths Book and are explained in the simplest manner. Therefore, following the Class 9th Maths NCERT Solutions by Jagran Josh will help students clear the concepts and score well in Class 9 Maths Exam 2022-23. 

Also Read| CBSE Class 9 Maths Revised Syllabus 2022-2023 

Links to Download Class 9 Maths NCERT Solutions are provided below:

Chapter 1: Number System

Chapter 2: Polynomials

Chapter 3: Coordinate Geometry

Chapter 4: Linear Equations in two Variables

Chapter 5: Introduction to Euclid's Geometry

Chapter 6: Lines and Angles

Chapter 7: Triangles

Chapter 8: Quadrilaterals

Chapter 9: Areas of Parallelograms And Triangles

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Heron's Formula

Chapter 13: Surface Areas And Volumes

Chapter 14: Statistics

Chapter 15: Probability

In Class 9, students are introduced to several new and complex concepts which are important to lay a strong foundation for higher classes. However, it also becomes difficult for students to grasp difficult topics and clear the concepts. But if you have good and reliable study resources then it becomes easier to understand the concepts and logic. NCERT Solutions by Jagran Josh form the best resource to remove all your doubts and clear the concepts. All these NCERT solutions have been explained by the subject matter experts to provide accurate and detailed solutions to class 9 students. These NCERT Solutions are prepared for the latest NCERT book for the new academic session 2021-22. 

• Each solution is framed in a way to involve the key concepts.
• Solutions have been rechecked/revised by subject experts.
• Chapter-wise solutions are available for free download.

Also, check Class 9 Mathematics NCERT Exemplar Problems

When working with Mathematics problems, it is quite essential to focus on understanding the process and logic that is involved. This helps you understand how you should approach similar problems in school tests or annual examinations. Following the NCERT solutions provided above in this article will help to cover the complete syllabus in an organised manner.

Follow the Class 9 Maths NCERT Book:

Setting up a strong foundation of concepts during Class 9 can be much effective in cracking the biggest test of a student’s life, i.e., the Class 10 th Board Exam. For this, following the NCERT textbooks is considered as the best practice as these books help to develop a strong foundation of all the concepts.

Importance of Class 9 Maths NCERT Textbooks and NCERT Solutions

CBSE affiliated schools follow NCERT textbooks as questions asked in CBSE Class 9th annual examination are based on the NCERT textbooks only. The contents of the NCERT textbooks i.e., concepts, definitions, theorems, etc., are presented in the precise form. You will find here to the point detail on all topics and concepts. Facts and examples given in NCERT textbooks are universally accepted and it is one of the main reasons why CBSE prescribes NCERT textbooks.

NCERT textbooks play an essential role to get a hold of every topic if studied thoroughly. With simple and easy language, they help to lay a better foundation of all tough and advanced level concepts. To achieve the maximum marks in exams, all students need to read the NCERT books thoroughly and solve the exercise questions given at the end of each chapter. This will surely help them to get the desired results in the exams.

Also, check:

NCERT Books for Class 9 - All Subjects

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Std 9 Maths Paper Solution 2023 PDF Download Gujarat Board

Gseb std 9 maths paper solution 2023 pdf download gujarat board.

The present (05-03-2022) GSEB (Gujarat Secondary Education Board) GSEB STD 9 Maths Paper Solution 2022 PDF Download Gujarat Board (First Language) and English (F.L) Subject Paper Solution 2022. STD 9 Maths Paper Solution 2022 Bunches of Candidate were showed up in this GSEB Std. tenth Examination.

Composed Examination Comprise Two Parts PART-An and PART-B. Section A Contains 16 Marks Multiple Choice Questions (MCQ). Applicant Need to Choice and Fill Correct Answers in OMR Sheet. In the wake of Completing SSC Maths Paper All are Looking For SSC Maths and English Medium Paper Solution 2022. We can gives answer key of all subject in this site stage. for true response key you should visit official sites

• Name of Examination : GSEB 9/STD 9/SSC/Guj Board Exam
• Official Name : GSEB i.e Gujarat Secondary Education Board
• Date of Exam : 05th March 2022
• Test Subject : Maths (F.L) 01 and English (F.L) (04)
• Official Answer Key/Solutions : Released (Very soon )
• Official Website : www.gseb.org

STD 9 Maths PAPER SOLUTION 2023

Gujarat Board additionally taken Following Subject Examination on fifth March 2022 and Candidate Eagerly hanging tight for Answer Sheet

A Leading Coaching Institutes Like Nib School additionally Published On GSEB SSC Answer Key 2022 on its site which is Marugujaratdaily.com Lots of Students don’t mindful with regards to that so we are likewise Credit to them for Great Work

GSEB STD 9 Maths Answer Key 2023 PDF Download

• Steps to Check GSEB SSC {9th} Maths Answer sheet 07-03-2022
• First Open the Official Website of GSEB i.e Gujarat Board www.gseb.org
• Explore and Click on the GSEB Answer Key 2022 Section
• Select the tenth Answer Key Section and Click on that
• Presently Open the “Std.9 SSC Part-An Answer Key” or Alternate Link to Check Answer Key
• Download PDF design Answer key and Verify Answers

STD 9 Maths Paper Solution 2023 PDF Download in Maths

Here you can download the GSEB STD 9 Maths Paper Solution 2022 PDF in Maths by click on the Download link given below.

STD 9 Maths Paper Solution 2023 PDF Download Link

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NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials

• NCERT Solutions
• Chapter 2 Polynomials

NCERT Solutions for Maths Chapter 2 Polynomials Class 9 - Free PDF Download

Vedantu specialists have put up NCERT Answers Class 9 Mathematics Chapter 2 Polynomials to satisfy the long-standing need of CBSE students studying for Board and other competitive Examinations. This answer has been rigorously reviewed in compliance with the CBSE's newly modified syllabus. CBSE Class 9 Mathematics Chapter 2 Polynomials solutions include a substantial number of solved questions that span the complete syllabus in the form of graded exercises and step-by-step explanations. Vedantu's goal is to clarify the chapter's key subject and to help students build problem-solving abilities.

Glance of NCERT Solutions for Class 9 Maths Chapter 2 Polynomials | Vedantu

Chapter 2 of Class 9 Maths deals with Polynomials, which are basically algebraic expressions built using variables (like x, y), constants (numbers like 2, 3), and exponents (whole numbers like $x^2, y^3$).

Learn about Degree of a Polynomial and types of Polynomials.

Polynomials are classified based on the highest exponent of the variable:

Linear Polynomial (degree 1) (e.g., 5x + 2)

Quadratic Polynomial (degree 2) (e.g., x^2 + 3x - 4)

Cubic Polynomial (degree 3) (e.g., 2x^3 - x^2 + 5x + 1)

Covered concepts duch as Degree of a Polynomial, Zero Polynomial, Operations on Polynomials and Monomial, Binomial, Trinomial.

This article contains chapter notes, formulas, exercise links and important questions for chapter 2 -  Polynomials. 

There are five exercises (33 fully solved questions) in Class 9th Maths Chapter 2 Polynomials.

Access Exercise Wise NCERT Solutions for Chapter 2 Maths Class 9

Exercises Under NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Exercise 2.1: This exercise covers the definition and basic concepts of polynomials. The questions in this exercise aim to familiarise students with terms like coefficients, exponents, degrees, and standard forms of polynomials. Students are also required to classify polynomials based on their degrees. They will have to solve problems related to the addition, subtraction, and multiplication of polynomials and also learn how to factorise polynomials.

Exercise 2.2: This exercise deals with the factors and zeros of polynomials. The questions in this exercise require students to find the factors and zeros of given polynomials. They will also learn how to use the factor theorem and remainder theorem to factorise polynomials and find their zeros.

Exercise 2.3: This exercise covers the division algorithm for polynomials. The questions in this exercise require students to divide a polynomial by another polynomial using the long division method. They will also learn how to use the remainder theorem to find the remainder when a polynomial is divided by another polynomial.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomial - PDF Download

Variables - the unknown value.

Have you ever wondered why children have different heights? Some children grow taller and some end up being shorter than average. To answer this question Scientists have come closer and researched the parameters in the form of variables that are the cause of height.

The word ‘variable’ is derived from the word ‘vary’ which means changing. Therefore, a variable can be any trait, condition, or factor that can change by only differing amounts or it is the unknown term whose value is not known. Example: A child’s height is dependent on the amount of protein and nutrients he or she consumes. Not only that, the height of kids is also dependent on their DNA which means if their parents are tall then there are more chances of them being tall whereas short parents usually have short kids. The height of the kid is also dependent on the rate of work or activities. It is believed that children with more activities like jumping, running, skipping, etc tend to grow faster. Thus, nutrients, DNA, and activities are the three variables that control the height in our body. These variables keep changing from body to body.

For example, while cooking dal we know that the quantity of water is thrice the number of lentils. That you can add 1 cup of lentils to three cups of water. This process can be expressed as,

Here, the quantity of lentils is variable. That means if the quantity of lentils changes then the quantity of water also changes.

In a World Full of Variables, You Will Always Find Constant.

There is one interesting thing about constants and that is this it never changes. A constant is actually a value that is a fixed number on its own. For example - In the equation 9 - x = 5, 9 and 5 are two constants whose values will not change whereas the value of x is not known. Thus, x is a variable.

Can Constant be a Coefficient To?

Since now we already know about variables, it is easier for us to understand the constant and coefficient. A coefficient is usually the number that is multiplied by the variable or letters. For example in ‘5x + y - 7’, 5 is a coefficient of x in the term 5x because it is a number that is multiplied by the unknown variable x. Also, in the term y, it can be considered as the coefficient of y because y can be written as 1xy.

The coefficient is the number that is always multiplied by the variables but constants are terms without variables. Therefore, coefficients cannot be called constants and vice versa. In the aforementioned example, -7 is constant.

Like terms are the terms having the same variables raised to the same power. In 5x + y - 7, no variable is common therefore no like terms. 

In 5a + 2b - 3a + 4 the terms like 5a and -3a are like terms whereas 4 is constant.

What is a Polynomial?

The word Polynomial is derived from the word poly ("many") and nominal ("term"). It is an expression consisting of many terms such that each term holds at least one variable. The variables can be raised to the power and further multiplied by a coefficient but the simplest polynomials hold one variable. The terms are separated by signs( + or - ). Also, the variables and numbers can be combined using addition, subtraction, multiplication or division but it can never be divided by a variable which means a term can never be like  2/x. A polynomial can also not have infinite terms. It always has a finite sum of terms with all variables having whole-number exponents and no variable as a denominator.

Polynomials are composed of the following:

Constants such as 3, −20, or ½, etc.

Variables such as g, h, x, y, etc.

Exponents such as 2 in y 2 or 5 in x5  etc

Examples of Polynomials: 5x 3 – 2x 2 + x – 13 and  x 2 y 3 + xy.

Degree of a Polynomial

It is simply the highest of the powers or exponents on the terms present in the algebraic expression.

Example: In 7x – 5, the first term is 7x, whereas the second term is -5. The power on the variable of the given first term is one and on the second term is zero. Since the highest exponent is one, the degree of the polynomial is also 1.

Types of Polynomials

Polynomials can be classified on the basis of

Number of Terms.

Degree of a polynomial.

Classification on the Basis of Terms

A polynomial either has one term, two terms, three terms, or more than three terms.

Monomials- ‘Mono’ stands for one and ‘mial’ stands for terms thus an algebraic expression with one term is called a monomial. 

Binomials- ‘Bi’ stands for two and ‘mial’ stands for terms therefore an algebraic expression with two, unlike terms is called binomials. 

Trinomials- ‘Tri’ stands for three and ‘mial’ stands for terms thus an algebraic expression with three unlike terms is called trinomials.

Classification on the Basis of Degrees

The Degree of Polynomial is considered as the highest value of the exponent in the expression because it is the largest exponent. We can also call it an order of the polynomial. While finding the degree of the polynomial, remember that the polynomial powers of the variables must be either in an ascending or descending order.

Linear Polynomial: If the expression holds degree 1 then we can call it a linear polynomial. 

Quadratic Polynomial: If the expression holds degree 2 then it can be called a quadratic polynomial.

Cubic Polynomial: If the expression holds degree 3 then it will be called a cubic polynomial.

Zeros of Polynomials

If the value of every coefficient of a variable is zero then it is called the zeros of a Polynomial. In order to find the relationship between the zeroes and coefficients of a given quadratic polynomial, we can find the zeros of the polynomial by the factorization method that is, by taking the sum and product of these zeros.

Operations on Polynomial

There are four main polynomial operations which are:

Addition of Polynomials

Subtraction of Polynomials

Multiplication of Polynomials

Division of Polynomials

NCERT Solutions Class 9 Maths Chapter 2 Polynomials All Exercises

NCERT Maths Class 9 Solutions Vedantu's polynomials provide a thorough grasp of this significant subject. Students can build a solid foundation in algebra by concentrating on important ideas such as polynomial expressions, degree of polynomials, and polynomial operations.It's important to pay attention to the step-by-step solutions provided in the NCERT Solutions, as they help clarify concepts and reinforce problem-solving techniques. Understanding polynomials is crucial as they form the basis for understanding more complex algebraic concepts. Approximately four to five questions from this chapter have usually been included in previous year's question papers. As a result, practicing a range of issues from NCERT Solutions and past test papers helps improve exam readiness and confidence.

Other Related Links for CBSE Class 9 Maths Chapter 2

Chapter-specific ncert solutions for class 9 maths.

Given below are the chapter-wise NCERT Solutions for Class 9 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 9 Maths Chapter 2 - Polynomials

1. What is Polynomial and How is It Classified on the Basis of the Number of Terms and Degrees ? 

A polynomial in a variable x is an algebraic expression  of the form

p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + . . . . .+ a₂x²+ a₁x + a₀ ,where a₀, a₁, a₂, . . . . ., aₙ are constants and aₙ ≠ 0.

In the above expression of  polynomial p(x),  a₀, a₁, a₂, . . . . ., aₙ are respectively the coefficients of variables  x⁰, x, x² , . . . . ., xⁿ, and the degree of polynomial p(x) is the non-negative (i.e. n 0) power n to which the variable x is raised in the expression.

Each of the expressions aₙxⁿ + aₙ₋₁xⁿ⁻¹ + . . . . .+ a₂x²+ a₁x + a₀  is called terms of the polynomial p(x).

Polynomial class 9 are classified on the basis of the number of terms and degrees they have as follows:

Classification of Polynomials Based on the Number of Terms:

Monomial: A polynomial having only one term is known as a monomial.

Example: 2x², – 3,  –3/2 (y) 

Binomial: A polynomial having two terms is known as a binomial.

Example: 2x² + 1, x – 3,  –3/2(y) + x

Trinomial: A polynomial having three terms is known as a trinomial.

Example: x² - 2x + 2

Classification of Polynomials Based on the Degrees of Its Variable:

Linear polynomial: A polynomial having degree one is known as a linear polynomial. 

Example: 2x + y, x – 3

Quadratic polynomial: A polynomial having degree two is known as a quadratic polynomial.

Example: 2x² + 1, x² - 2x + 2

Cubic polynomial: A polynomial of degree three is known as a cubic polynomial.

Example: x³ + 4x² + 7x - 3

2. Give the Difference between Remainder and Factor Theorem?

The remainder theorem states that if a polynomial p(x) is divided by (x - a), then the remainder is obtained by evaluating the expression p(a).

While, the Factor theorem states that (x - a) will be a factor of polynomial p(x) only if the remainder obtained by evaluating the expression p(a) equals zero i.e; p(a) = 0.

For Example: let p(x) = x² - 6x + 9, then find the remainder when it is divided by (x - 1).

So, according to the remainder theorem, remainder is obtained by evaluating the expression p(1)

p(1) = 1² - 6(1) + 9 

       = 1 - 6 + 9 = 4.

Therefore, the remainder is 4 when  p(x) = x² - 6x + 9 is divided by (x - 1).

Now, check whether (x - 3) is the factor of p(x) = x² - 6x + 9.

So, according to factor theorem  (x - 3) is the factor of p(x) = x² - 6x + 9 only if p(3) = 0.

p(3) = 3² - 6(3) + 9

        = 9 - 18 + 9 = 0.

Therefore,  (x - 3) is the factor of p(x) = x² - 6x + 9.

3. What is an Algebraic Identity and What are Its Uses?

An algebraic identity is an algebraic equation that is true for any values of the variables occurring in it.

Algebraic identities are used to factorize the algebraic expressions and to compute the products of some arithmetic expressions without multiplying them directly.

Most commonly used algebraic identities are, (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b², etc.

4. What are Polynomials in One Variable?

Polynomials are algebraic expressions in which variables and constants terms are connected by various arithmetic operators. For ‘Polynomials in one variable’ the terms of the polynomial have the same common variable, with numeric coefficients. In ‘Polynomials in one variable’, the variables are raised to powers and the degree of the equation can be determined with the highest power of the variable. Also, the degree of a polynomial is always a positive integer. Hence, if there is a variable term raised to a negative power in an algebraic expression, then it is not counted as a polynomial.

Examples of Polynomials: 2x³ - 7x² + 3x - 2, in this algebraic expression, there are four terms, and all the variable terms have positive integer exponents. So, this algebraic expression is a polynomial.

5. What do We Learn in the Class 9 Maths Chapter 2 Polynomials?

‘Polynomials’ is one of the most important chapters in the Maths syllabus of class 9. The basic highlights of this chapter are listed below.

Polynomial is an algebraic expression consisting of variables, preceded by coefficients, and connected by arithmetic operators. In this chapter, you will learn about polynomials in one variable, that is, the variable coefficient for all the terms in a polynomial, here, will be the same.

The variables along with their coefficients in the algebraic expression are called terms of the polynomial. Some variable terms in a polynomial are raised to exponents.

The degree of a polynomial refers to the greatest exponent of its variable term. If there is a constant term in the polynomial, then it has to be assumed that the exponential value for its variable coefficient is zero.

6. What are the Important Topics Discussed in Class 9 Maths Chapter 2 Polynomials?

Remainder theorem, degree of polynomials, factorization, zeros of a polynomial, algebraic identities, etc., are explained in this chapter. These concepts form the basis of higher mathematics so you must have a good knowledge of these concepts. For example, factorization is one of the most basic concepts of algebra, and you will find its application in other chapters as well.

You may expect sums for finding the degree of polynomials to carry fewer marks in the examination, and mostly these sums are in the compulsory part of the question paper. The sums of the NCERT maths book of class 9 are solved in a step-by-step manner on Vedantu and you may refer to it for a better understanding of all concepts of ‘Polynomials’.

7. How do I identify the degree of a polynomial?

The degree of polynomials in one variable is the highest power of the variable in the algebraic expression.

For example, X^2+5X+3. The degree of this equation is 2.

8. Are NCERT Solutions for Class 9 Maths Chapter 2 difficult to learn?

No, it is easy to score high in your exams if you regularly practice with Vedantu’s NCERT solutions for Class 9 Maths Chapter 2. The solutions are formulated by experienced subject matter experts who understand the CBSE curriculum. The solutions teach you how to present your answers in exams. Every minute detail is covered in a simple way for students to grasp the concepts easily. The solutions are updated according to the latest guidelines of the CBSE board.

9. What is a Polynomial?

Polynomials are expressions that contain one or more terms with coefficients that are not zero. Polynomial terms can be variables, constants, or both. A polynomial's exponents should always be whole numbers. The degree of a polynomial is the highest power in the polynomial. Polynomials are another way to represent real numbers. Further information may be found in Vedantu's NCERT answers for Class 9 Mathematics Chapter 2.

10. What is the Remainder Theorem?

When a polynomial p(x) is divided by the linear polynomial (x-a), then the remainder comes out to be p(a). The degree of the polynomial should be greater than one. For more details, refer to Vedantu’s NCERT solutions for Class 9 Maths Chapter 2. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu app as well for free of cost. 

11. What are the conditions for an expression to be Polynomial?

The following conditions are important to consider an expression as a polynomial:

All the coefficients should be real numbers.

The exponent should not be a negative number. 

For more details, refer to Vedantu’s NCERT solutions for Class 9 Maths Chapter 2. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu app as well for free of cost. 

NCERT Solutions for Class 9 Maths

Ncert solutions for class 9.

CBSE NCERT Solutions

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Class 9 Assignments Download Pdf

Please refer to Assignments for Class 9 for all subjects given below. We have given below free printable Assignments for Class 9 for easy and free download in PDF. The assignments have been developed based on the current year’s NCERT Books for Class 9. These Assignments for Grade 9 cover all critical topics which can come in your standard 9 class tests and exams. Free printable Assignments for CBSE Class 9 , school and class worksheets, and practice test papers have been developed by our experienced class 9 teachers. You can free download CBSE NCERT printable assignments for Class 9 with solutions and answers. All assignments and test sheets have been prepared by expert faculty as per the latest curriculum in Class 9. Students can simply click on the links below and download all Pdf assignments for class 9 for free. Latest Kendriya Vidyalaya Class 9 assignments and test papers are provided below.

Assignments for Class 9 Pdf Download

We have provided the largest collection of free CBSE NCERT KVS Assignments for Class 9 . You can download all free assignments in Pdf for standard 9th. Our expert faculty members have covered Class 9 questions and answers for all subjects as per the latest curriculum for the current academic year. All practice test papers and question banks for Class 9 subjects and CBSE Assignments for Class 9 will be really useful for students to properly prepare for the upcoming exams. Class 9th students are advised to free download in Pdf all printable question banks given below.

Subjectwise Assignments for Class 9 for Download in Pdf

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• By clicking on the links above you can access the largest collection of assignments for Grade 9, you will be able to find critical and scoring questions which can come in your exams and school tests.
• While you do the assignments, it is also important to go through all your CBSE textbooks for Class 9 .
• All assignments for standard 9 for all subjects have solutions and step by step answers. Students and parents will be able to understand then easily.
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• Class 9 practice question banks will help to further strengthen subject understanding which will help to score better rank in exam

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• Mathematics Olympiad

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Olympiad Questions & Sample Papers for Free

Olympiad exams are held with the motive of bringing more knowledge and confidence in the students. Olympiad questions are based on MCQ type questions. The total number of items is 35, that is to be done in 60 minutes for classes I-IV and classes V-XII; 50 questions are there. Olympiad's paper is divided into three to four parts depending upon the subject choice.

Olympiad is taken on two levels-

The initial level is the school level, while the second one is a national level competitive exam. It is a way through which a child's capability and this can also serve the school in a way that they can check on the fields that require more attention. It also enhances the reasoning and problem-solving skills of the students.

This exam is only meant for students of class I-XII. We provide sample papers and Olympiad questions to reduce the pressure on the students. Solving the sample papers thoroughly is one way of preparing for such competitive exams. It is essential to study the course thoroughly and then solve as many sample papers as we can. More practice will bring better results for the examination. Previous year question papers and Olympiad sample papers are also available to give you an idea of the paper pattern.

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https://www.practice-olympiad.com site provides free sample papers to help your child prepare better for the Olympiad Exams.

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Maths Olympiad is an examination curated with advanced level mathematics to work on a student's current potential. In such tests, the students are required to solve the given questions. Based on the problem solving, the student is rewarded with marks and accolades. To support in this particular situation, several coaching centres and websites are there that provide the necessary support.

We offer an ample amount of Math Olympiad Questions and Maths Olympiad Sample Papers for the student to be ready and practice their way to perfection. We also provide the international level of maths Olympiad questions. For providing extra assistance to prepare for the Olympiads, there is International Maths Olympiad Questions .

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Science Olympiads are designed with updated questions from every science field to test the science of a student at an advanced level. It can be an excellent way to examine knowledge in all sectors of science at the same time. The general format of an Olympiad exam is that the student must solve the given questions within an amount of specified time. Multiple efforts are made to introduce Olympiads for different age groups nationally or internationally. One of the official brands that are bringing together the Olympiad culture in our country, especially in science, is the NSO.

We have specially made available National Science Olympiad Question Papers, NSO Sample Papers , and NSO Question Papers to help the students prepare. This site will ensure that you are provided with all the advanced levels of various materials and assistance in solving them.

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Olympiads help students in various ways. Olympiads consist of multiple-choice questions, and students are required to select the answer they think is correct. Students are also required to carry their pen and pencils.

English Olympiads specifically will help you improve your grip over the language and even help you with your grammar and vocabulary.

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Hindi Olympiad is an exam that is commonly given by the students during their school time, and this is an important exam to test the Hindi abilities of the student. A student is allotted a set of questions, and they have to cover it in a specified period. You will find Hindi Olympiad Questions and sample papers at this site that will help the student to build a strong base in the Hindi language.

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GK Olympiad Questions organized by International General Knowledge Olympiad are the most prominent General Knowledge papers. It is held across the globe and in India, which analyses your child's brain's strengths and weaknesses and lets you recognize which position you have to work on for your child's betterment.

IGKO Sample Papers are provided every year for all primary and secondary classes, starting from 1st to 12th.

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Class 7 sample paper & practice questions for International Maths Olympiad (IMO) level 1 are given below. Syllabus for level 1 is also mentioned for these exams. You can refer these sample paper & quiz for preparing for the exam.

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What’s New for C++ Developers in Visual Studio 2022 17.10

May 22nd, 2024 7 3

We are happy to announce that Visual Studio 2022 version 17.10 is now generally available! This post summarizes the new features you can find in this release for C++. You can download Visual Studio 2022 from the Visual Studio downloads page  or upgrade your existing installation by following the  Update Visual Studio Learn page .

Standard Library and Compiler

We have incremented the minor version number of the MSVC toolset from 19.39 (VS 2022 v17.9) to 19.40 (VS 2022 v17.10). For more details, and some ways in which this will affect projects that assume that MSVC versions are all 19.3X for Visual Studio 2022 releases, see the  MSVC Toolset Minor Version Number 14.40 in VS 2022 v17.10 blog post .

We’ve made a host of changes to our standard library implementation, with the help of the community. As always, you can see the STL Changelog for full details. We implemented C++26’s P2510R3 Formatting Pointers , which brings the set of format specifiers for pointers when using std::format more in line with those that already exist for integers. We also implemented a few smaller features from C++26 and C++23, such as P2836R1 , which makes std::basic_const_iterator act in a more natural way with respect to implicit conversions. On the performance side, we improved the vectorized implementations of std::min_element , std::ranges::min and friends, made the copy/move assignment operators of std::expected trivial when expected, and more.

C++ Productivity and Game Development

You can now use Build Insights to view your template instantiation information. Template instantiation collection must be activated in Tools > Options > Build Insights.

Check out our blog post on Templates View or our recording from Pure Virtual C++ for more details:

You can now keep our Unreal Engine plugin needed for Unreal Engine Test Adapter running in the background, greatly reducing startup costs. This is an opt-in feature that can be activated via Tools > Options > Unreal Engine. Furthermore, we have added additional Unreal Engine Macros to be indented in accordance with the UE Code Style.

Cross-Platform

If you’re targeting Linux, be sure to check out our video on our most recent development features for Linux from Pure Virtual C++:

CMake Targets View

We have added support for pinning CMake targets in the CMake Targets View. There is a top-level folder now for Pinned Targets. You can pin any targets by right-clicking and selecting the Pin option in the context menu.

You can also unpin any target in the Pinned Targets folder by selecting Unpin.

Connection Manager

We introduced some UX updates and usability improvements to the Connection Manager. With these updates we provide a more seamless experience when connecting to remote systems and/or debugging failed connections. Check out our blog post for more details.

Core Editor

View and address pull request comments.

You can now view your GitHub and Azure DevOps pull request comments directly in your working file in Visual Studio. Enable the feature flag, “Pull Request Comments,” in Options > Environment > Preview Features and checkout the pull request branch to get started.

Generated Pull Request Descriptions

Similar to our generated  Git commit message feature , you can now get a first draft for your pull request description created by GitHub Copilot. You’ll need to verify you have an active GitHub Copilot subscription. Try it out by clicking the ‘Add AI Generated Pull Request Description’ sparkle pen icon within the Create a Pull Request window. Please share your feedback on this feature  here .

Image Hover Preview

If you hover over the path to an image, Visual Studio will now give you a small preview of the image itself, along with the size of the image in pixels and bytes. The size is capped to 500px wide and high.

New Text Formatting Options

You can now choose italic, bold, strikethrough, or underline styles for text formatting in the Options > Environment > Fonts and Colors settings:

Creating Conditional Breakpoints and Tracepoints from Expressions

You can now create a conditional breakpoint or tracepoint directly from an expression in your source code from the right-click menu. This works on property or field names and values from autos, locals, watch windows, or DataTips:

Attach to Process Dialog Revamp

The Attach to Process dialog has been improved for better functionality and user-friendliness. You can now easily switch between tree and list views, organize processes better with collapsible sections, and select code types with a simplified combobox. Moreover, the “Select/Track Window” feature is now easier to use, allowing two-way tracking: selecting a process highlights its window, and clicking on a window selects its process.

GitHub Copilot

We have unified the experience from the GitHub Copilot and Copilot Chat extensions and shipped them directly in Visual Studio. To install it, install the GitHub Copilot component in the Visual Studio Installer:

To use it, you’ll need an active GitHub Copilot subscription, and you can find the interface in the top-right corner of Visual Studio.

Send us your feedback

We are very much interested in your feedback to continue to improve this experience. The comments below are open. Feedback can also be shared through  Visual Studio Developer Community . You can also reach us on Twitter ( @VisualC ), or via email at  [email protected] .

Sy Brand C++ Developer Advocate, C++ Team

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Very much appreciate the investments you’ve made in the native code tool chain in this release. I’d really like to see more of the same, particularly in the fields of: Code Analysis, scaling the solution for very large projects, local ML based navigational help for large code bases.

With the new version of the MSVC toolset, please consider documenting the behavior of the default value, and add a “latest” value to the corresponding project configuration: https://developercommunity.visualstudio.com/t/Default-value-of-MSVC-Toolset-is-not-doc/10398953

Hi Sy, i see that have been added an image hover preview, could you bring attention from the extension team about this feature request? Its been on “roadmap” for incontable years, almost 100 upvotes and a lot of comments requesting it 🙏🙏 Image Watch

I see there has been some work done on the debugger, but it would be nice if you could let your colleagues on the debugger team know that it is impossible to debug x86 C++ code in Visual Studio 2022 on Intel Sapphire Rapids Xeon CPUs in Windows 11 since August 2023 when those CPUs were released. I submitted a bug report , but it’s “under consideration” seemingly forever. Apparently nobody except me uses workstation class CPUs with Intel AMX instruction set extensions.

too long to wait for std::runtime_format. please update!

Honestly, it’s quite worrying that your FE team was fired. Have you had any luck hiring them back as 50% of your compiler doesn’t exist (and the other 50% from the BE generates worse code compared to competitors from a decade ago)

The lack of C++23 “language” features will soon be a blocker for my industry in particular. We have a set of “reference” guidance and the decision on whether we’re allowed to use C++23 will be made in what looks like a faster timeline then you’ll be able to put the compiler back together again…

Exciting update! But I think the build tools version increase unfortunately broke CUDA with CMake. I’ve submitted https://developercommunity.visualstudio.com/t/VS-2022-1710-C-build-tools-144019/10668943 .

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• NCERT Solutions
• NCERT Class 9
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• Chapter 1: Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

Ncert solutions class 9 maths chapter 1 – cbse free pdf download.

Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 1 Number Systems

Download most important questions for class 9 maths chapter – 1 number systems.

In NCERT Solutions for Class 9 Maths Chapter 1 , students are introduced to several important topics that are considered to be very crucial for those who wish to pursue Mathematics as a subject in their higher classes. Based on these NCERT Solutions , students can practise and prepare for their upcoming CBSE exams, as well as equip themselves with the basics of Class 10. These Maths Solutions of NCERT Class 9 are helpful as they are prepared with respect to the latest update on the CBSE syllabus for 2023-24 and its guidelines.

• Chapter 1- Number Systems
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclids Geometry
• Chapter 6 Lines and Angles
• Chapter 7 Triangles
• Chapter 8 Quadrilaterals
• Chapter 9 Areas of Parallelograms and Triangles
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Heron’s Formula
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Inroduction to Probability

NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems

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

Exercise 1.1 page: 5.

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 in 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 in 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…}

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

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.

Exercise 1.2 Page: 8

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

(i) Every irrational number is a real number.

Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0.

i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….

Real numbers – The collection of both rational and irrational numbers are known as real numbers.

i.e., Real numbers = √2, √5, , 0.102…

Every irrational number is a real number, however, every real number is not an irrational number.

(ii) Every point on the number line is of the form √m where m is a natural number.

The statement is false since as per the rule, a negative number cannot be expressed as square roots.

E.g., √9 =3 is a natural number.

But √2 = 1.414 is not a natural number.

Similarly, we know that there are negative numbers on the number line, but when we take the root of a negative number it becomes a complex number and not a natural number.

E.g., √-7 = 7i, where i = √-1

The statement that every point on the number line is of the form √m, where m is a natural number is false.

(iii) Every real number is an irrational number.

The statement is false. Real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number.

Every irrational number is a real number, however, every real number is not irrational.

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

No, the square roots of all positive integers are not irrational.

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

3. Show how √5 can be represented on the number line.

Step 1: Let line AB be of 2 unit on a number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit.

Step 3: Join CA

Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem,

AB 2 +BC 2 = CA 2

2 2 +1 2 = CA 2 = 5

⇒ CA = √5 . Thus, CA is a line of length √5 unit.

Step 4: Taking CA as a radius and A as a center draw an arc touching

the number line. The point at which number line get intersected by

arc is at √5 distance from 0 because it is a radius of the circle

whose center was A.

Thus, √5 is represented on the number line as shown in the figure.

4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP 1 of unit length (see Fig. 1.9). Now draw a line segment P 2 P 3 perpendicular to OP 2 . Then draw a line segment P 3 P 4 perpendicular to OP 3 . Continuing in Fig. 1.9 :

Constructing this manner, you can get the line segment P n-1 Pn by square root spiral drawing a line segment of unit length perpendicular to OP n-1 . In this manner, you will have created the points P 2 , P 3 ,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …

Step 1: Mark a point O on the paper. Here, O will be the center of the square root spiral.

Step 2: From O, draw a straight line, OA, of 1cm horizontally.

Step 3: From A, draw a perpendicular line, AB, of 1 cm.

Step 4: Join OB. Here, OB will be of √2

Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C.

Step 6: Join OC. Here, OC will be of √3

Step 7: Repeat the steps to draw √4, √5, √6….

Exercise 1.3 Page: 14

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

= 0.36 (Terminating)

= 4.125 (Terminating)

(vi) 329/400

= 0.8225 (Terminating)

2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?

[Hint: Study the remainders while finding the value of 1/7 carefully.]

3. Express the following in the form p/q, where p and q are integers and q 0.

Assume that   x  = 0.666…

Then,10 x  = 6.666…

10 x  = 6 +  x

(ii) \(\begin{array}{l}0.4\overline{7}\end{array} \)

= (4/10)+(0.777/10)

Assume that  x  = 0.777…

Then, 10 x  = 7.777…

10 x  = 7 +  x

(4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 )

= (36/90)+(7/90) = 43/90

Assume that   x  = 0.001001…

Then, 1000 x  = 1.001001…

1000 x  = 1 +  x

4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Assume that x  = 0.9999…..Eq (a)

Multiplying both sides by 10,

10 x  = 9.9999…. Eq. (b)

Eq.(b) – Eq.(a), we get

10 x  = 9.9999

– x  = -0.9999…

_____________

The difference between 1 and 0.999999 is 0.000001 which is negligible.

Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.

Dividing 1 by 17:

There are 16 digits in the repeating block of the decimal expansion of 1/17.

6. Look at several examples of rational numbers in the form p/q (q ≠ 0), 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?

We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. For example:

1/2 = 0. 5, denominator q = 2 1

7/8 = 0. 875, denominator q =2 3

4/5 = 0. 8, denominator q = 5 1

We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.

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

We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:

• √3 = 1.732050807568
• √26 =5.099019513592
• √101 = 10.04987562112

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Three different irrational numbers are:

• 0.73073007300073000073…
• 0.75075007300075000075…
• 0.76076007600076000076…

9.  Classify the following numbers as rational or irrational according to their type:

√23 = 4.79583152331…

Since the number is non-terminating and non-recurring therefore, it is an irrational number.

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.

(iii) 0.3796

Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478

The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…

Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

Exercise 1.4 Page: 18

1. Visualise 3.765 on the number line, using successive magnification.

Exercise 1.5 Page: 24

1. Classify the following numbers as rational or irrational:

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23

(3 + √ 23) –√23 = 3+ √ 23–√23

Since the number 3/1 is in p/q form, ( 3 +√23)- √23 is rational.

(iii) 2√7/7√7

2√7/7√7 = ( 2/7)× (√7/√7)

We know that (√7/√7) = 1

Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

We know that, the value of = 3.1415

Hence, 2 = 2×3.1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

2. Simplify each of the following expressions:

(i) (3+√3)(2+√2)

(3+√3)(2+√2 )

Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(3-√3 )

(3+√3)(3-√3 ) = 3 2 -(√3) 2 = 9-3

(iii) (√5+√2) 2

(√5+√2) 2 = √5 2 +(2×√5×√2)+ √2 2

= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

(√5-√2)(√5+√2) = (√5 2 -√2 2 ) = 5-2 = 3

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π =c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

4. Represent (√9.3) on the number line.

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right angled triangle.

Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC

⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras theorem,

OD 2 =BD 2 +OB 2

⟹ (10.3/2) 2 = BD 2 +(8.3/2) 2

⟹ BD 2 = (10.3/2) 2 -(8.3/2) 2

⟹ (BD) 2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD 2  = 9.3

⟹ BD =  √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

5. Rationalize the denominators of the following:

Multiply and divide 1/√7 by √7

(1×√7)/(√7×√7) = √7/7

(ii) 1/(√7-√6)

Multiply and divide 1/(√7-√6) by (√7+√6)

= (√7+√6)/√7 2 -√6 2 [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]

= (√7+√6)/(7-6)

= (√7+√6)/1

(iii) 1/(√5+√2)

Multiply and divide 1/(√5+√2) by (√5-√2)

= (√5-√2)/(√5 2 -√2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]

= (√5-√2)/(5-2)

= (√5-√2)/3

(iv) 1/(√7-2)

Multiply and divide 1/(√7-2) by (√7+2)

1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√7 2 -2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]

= (√7+2)/(7-4)

Exercise 1.6 Page: 26

64 1/2 = (8×8) 1/2

= 8 1 [⸪2×1/2 = 2/2 =1]

32 1/5 = (2 5 ) 1/5

= 2 1 [⸪5×1/5 = 1]

(iii)125 1/3

(125) 1/3 = (5×5×5) 1/3

= 5 1 (3×1/3 = 3/3 = 1)

9 3/2 = (3×3) 3/2

= (3 2 ) 3/2

= 3 3 [⸪2×3/2 = 3]

(ii) 32 2/5

32 2/5 = (2×2×2×2×2) 2/5

= (2 5 ) 2⁄5

= 2 2 [⸪5×2/5= 2]

(iii)16 3/4

16 3/4 = (2×2×2×2) 3/4

= (2 4 ) 3⁄4

= 2 3 [⸪4×3/4 = 3]

(iv) 125 -1/3

125 -1/3 = (5×5×5) -1/3

= (5 3 ) -1⁄3

= 5 -1 [⸪3×-1/3 = -1]

3. Simplify :

(i) 2 2/3 ×2 1/5

2 2/3 ×2 1/5 = 2 (2/3)+(1/5) [⸪Since, a m ×a n =a m+n ____ Laws of exponents]

= 2 13/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/3 3 ) 7

(1/3 3 ) 7 = (3 -3 ) 7 [⸪Since,(a m ) n = a m x n ____ Laws of exponents]

(iii) 11 1/2 /11 1/4

11 1/2 /11 1/4 = 11 (1/2)-(1/4)

= 11 1/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 7 1/2 ×8 1/2

7 1/2 ×8 1/2 = (7×8) 1/2 [⸪Since, (a m ×b m = (a×b) m ____ Laws of exponents]

As the Number System is one of the important topics in Maths, it has a weightage of 8 marks in Class 9 Maths CBSE exams. On an average three questions are asked from this unit.

• One out of three questions in part A (1 marks).
• One out of three questions in part B (2 marks).
• One out of three questions in part C (3 marks).

This chapter talks about:

• Introduction of Number Systems
• Irrational Numbers
• Real Numbers and their Decimal Expansions
• Representing Real Numbers on the Number Line.
• Operations on Real Numbers
• Laws of Exponents for Real Numbers

List of Exercises in NCERT Solutions for Class 9 Maths Chapter 1:

Exercise 1.1 Solutions 4 Questions ( 2 long, 2 short)

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

Exercise 1.6 Solutions 3 Questions ( 3 long)

NCERT Solutions for Class 9 Maths Chapter 1- Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number System is the first chapter of Class 9 Maths. The Number System is discussed in detail in this chapter. The chapter discusses the Number Systems and their applications. The introduction of the chapter includes whole numbers, integers and rational numbers.

The chapter starts with the introduction of Number Systems in section 1.1, followed by two very important topics in sections 1.2 and 1.3

• Irrational Numbers – The numbers which can’t be written in the form of p/q.
• Real Numbers and their Decimal Expansions – Here, you study the decimal expansions of real numbers and see whether it can help in distinguishing between rational and irrational.

Next, it discusses the following topics.

• Representing Real Numbers on the Number Line – In this, the solutions for 2 problems in Exercise 1.4.
• Operations on Real Numbers – Here, you explore some of the operations like addition, subtraction, multiplication and division on irrational numbers.
• Laws of Exponents for Real Numbers – Use these laws of exponents to solve the questions.

Explore more about Number Systems and learn how to solve various kinds of problems only on  NCERT Solutions For Class 9 Maths . It is also one of the best academic resources to revise for your CBSE exams.

Key Advantages of NCERT Solutions for Class 9 Maths Chapter 1 – Number Systems

• These NCERT Solutions for Class 9 Maths help you solve and revise the whole CBSE syllabus of Class 9.
• After going through the step-wise solutions given by our subject expert teachers, you will be able to score more marks in the board exams.
• It follows NCERT guidelines.
• It contains all the important questions from the examination point of view.

The faculty have curated the solutions in a lucid manner to improve the problem-solving abilities of the students. For a more clear idea about Number Systems, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 9 Maths Number Systems

Disclaimer: 

Dropped Topics – 1.4 Representing real numbers on the number line.

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