case study class 9 maths chapter 1 with answers

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.

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FAQs on CBSE Case Study Questions for Class 9 Maths - Pdf

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

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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.

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• 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
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CBSE Class 9 Mathematics Case Study Questions

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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)

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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.

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16 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

I am Riddhi Shrivastava… These questions was very good.. That’s it.. ..

For challenging Mathematics Case Study Questions, seeking a writing elite service can significantly aid your research. These services provide expert guidance, ensuring your case study is well-researched, accurately analyzed, and professionally written. With their assistance, you can tackle complex mathematical problems with confidence, leading to high-quality academic work that meets rigorous standards.

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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.

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NCERT Solutions Class 9 Maths Chapter 1 - Number Systems

• NCERT Solutions
• Chapter 1 Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number System - Free PDF 2024-25

Chapter 1 number system class 9 delves into the principles covered under the topic of the number system. Vedantu offers an expert-curated NCERT answer for CBSE Class 9 Chapter 1. To ace your preparations, get the NCERT solution supplied by our professionals. The freely available number system class 9 PDF offers step-by-step solutions to the NCERT practice problems. The NCERT solutions PDF for chapter 1 maths class 9 contains the answers to all the Class 9 syllabus questions.

Glance of NCERT Solutions for Class 9 Maths Chapter 1 Number System | Vedantu

Class 9 chapter 1 maths dives into the fundamental building block of mathematics: Numbers! It introduces different types of number systems that we use for counting and calculations.

You will learn about various number systems like natural numbers (whole numbers starting from 1), whole numbers (including 0 with natural numbers), integers (whole numbers and their negatives), rational numbers (numbers expressible as a fraction p/q where q ≠ 0), and irrational numbers (numbers that cannot be expressed as a fraction, like the square root of 2).

The chapter explores properties like closure, commutativity, associativity, and distributivity for different operations (addition, subtraction, multiplication, division) on these number systems.

You will revisit how to represent these numbers on the number line, which is a visual aid to understand their relative positions and comparisons.

This article contains chapter notes, important questions, exemplar solutions and exercises links for Chapter 1 - Number System, which you can download as PDFs.

There are six exercises (27 fully solved questions) in class 10th maths chapter 3 Pair of Linear Equations in Two Variables.

Access Exercise Wise NCERT Solutions for Chapter 1 Maths Class 9

Exercises under NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

Exercise 1.1: This exercise covers basic concepts of the number system, such as natural numbers, whole numbers, integers, rational numbers, irrational numbers, etc. The questions in this exercise aim to familiarise students with these concepts and their properties.

Exercise 1.2: This exercise covers the representation of numbers in decimal form. The questions in this exercise require students to convert fractions into decimals, decimals into fractions, and perform basic operations such as addition, subtraction, multiplication, and division on decimals.

Exercise 1.3: This exercise deals with the representation of rational numbers on a number line. The questions in this exercise require students to mark the position of given rational numbers on a number line and identify the rational number represented by a given point on the number line.

Exercise 1.4: This exercise deals with the conversion of recurring decimals into fractions. The questions in this exercise require students to write recurring decimals as fractions and vice versa.

Exercise 1.5: This exercise covers the comparison of rational numbers. The questions in this exercise require students to compare given rational numbers using the concept of inequality, find rational numbers between two given rational numbers, and represent rational numbers on a number line.

NCERT Solutions Class 9 Maths Chapter 1 Number System - Free PDF Download

Exercise (1.1).

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

Ans: Remember that, according to the definition of rational number,

a rational number is a number that can be expressed in the form of  $\dfrac{p}{q}$, where $p$ and $q$ are integers and  $q\ne \text{0}$.

Now, notice that zero can be represented as $\dfrac{0}{1},\dfrac{0}{2},\dfrac{0}{3},\dfrac{0}{4},\dfrac{0}{5}.....$

Also, it can be expressed as $\dfrac{0}{-1},\dfrac{0}{-2},\dfrac{0}{-3},\dfrac{0}{-4}.....$

Therefore, it is concluded from here that $0$ can be expressed in the form of $\dfrac{p}{q}$, where $p$ and $q$ are integers.

Hence, zero must be a rational number.

2. Find any six rational numbers between $ {3}$ and $ {4}$. 

Ans: It is known that there are infinitely many rational numbers between any two numbers. Since we need to find $6$ rational numbers between $3$ and $4$, so multiply and divide the numbers by $7$ (or by any number greater than $6$)

Then it gives, 

$ 3=3\times \dfrac{7}{7}=\dfrac{21}{7} $ 

$  4=4\times \dfrac{7}{7}=\dfrac{28}{7} $

Hence, $6$ rational numbers found between $3$ and $4$ are $\dfrac{22}{7},\dfrac{23}{7},\dfrac{24}{7},\dfrac{25}{7},\dfrac{26}{7},\dfrac{27}{7}$.

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

Ans: It is known that there are infinitely many rational numbers between any two numbers.

Since here we need to find five rational numbers between $\dfrac{3}{5}$ and $\dfrac{4}{5}$,  so multiply and divide by $6$ (or by any number greater than $5$).

Then it gives,

$\dfrac{3}{5}=\dfrac{3}{5}\times \dfrac{6}{6}=\dfrac{18}{30}$,

$\dfrac{4}{5}=\dfrac{4}{5}\times \dfrac{6}{6}=\dfrac{24}{30}$.

Hence, $5$ rational numbers found between $\dfrac{3}{5}$ and $\dfrac{4}{5}$ are

$\dfrac{19}{30},\dfrac{20}{30},\dfrac{21}{30},\dfrac{22}{30},\dfrac{23}{30}$.

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

(i) Every natural number is a whole number. 

Ans: Write the whole numbers and natural numbers in a separate manner.

It is known that the whole number series is $0,1,2,3,4,5.....$. and the natural number series is $1,2,3,4,5...$.

Therefore, it is concluded that all the natural numbers lie in the whole number series as represented in the diagram given below.

Thus, it is concluded that every natural number is a whole number.

Hence, the given statement is true.

(ii) Every integer is a whole number.

Ans: Write the integers and whole numbers in a separate manner.

It is known that integers are those rational numbers that can be expressed in the form of $\dfrac{p}{q}$, where $q=1$.

Now, the series of integers is like $0,\,\pm 1,\,\pm 2,\,\pm 3,\,\pm 4,\,...$.

But the whole numbers are $0,1,2,3,4,...$.

Therefore, it is seen that all the whole numbers lie within the integer numbers, but the negative integers are not included in the whole number series.

Thus, it can be concluded from here that every integer is not a whole number.

Hence, the given statement is false.

(iii) Every rational number is a whole number.

Ans: Write the rational numbers and whole numbers in a separate manner.

It is known that rational numbers are the numbers that can be expressed in the form  $\dfrac{p}{q}$, where $q\ne 0$ and the whole numbers are represented as $0,\,1,\,2,\,3,\,4,\,5,...$

Now, notice that every whole number can be expressed in the form of $\dfrac{p}{q}$

as  \[\dfrac{0}{1},\text{ }\dfrac{1}{1},\text{ }\dfrac{2}{1},\text{ }\dfrac{3}{1},\text{ }\dfrac{4}{1},\text{ }\dfrac{5}{1}\],…

Thus, every whole number is a rational number, but all the rational numbers are not whole numbers. For example,

$\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},...$ are not whole numbers.

Therefore, it is concluded from here that every rational number is not a whole number.

Exercise (1.2)

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

(i) Every irrational number is a real number. 

Ans: Write the irrational numbers and the real numbers in a separate manner.

The irrational numbers are the numbers that cannot be represented in the form $\dfrac{p}{q},$ where $p$ and $q$ are integers and $q\ne 0.$

For example, $\sqrt{2},3\pi ,\text{ }.011011011...$ are all irrational numbers.

The real number is the collection of both rational numbers and irrational numbers.

For example, $0,\,\pm \dfrac{1}{2},\,\pm \sqrt{2}\,,\pm \pi ,...$ are all real numbers.

Thus, it is concluded that every irrational number is a real number.

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

Ans: Consider points on a number line to represent negative as well as positive numbers.

Observe that, positive numbers on the number line can be expressed as $\sqrt{1,}\sqrt{1.1,}\sqrt{1.2},\sqrt{1.3},\,...$, but any negative number on the number line cannot be expressed as $\sqrt{-1},\sqrt{-1.1},\sqrt{-1.2},\sqrt{-1.3},...$, because these are not real numbers.

Therefore, it is concluded from here that every number point on the number line is not of the form $\sqrt{m}$, where $m$ is a natural number.

(iii) Every real number is an irrational number. 

Real numbers are the collection of rational numbers (Ex: $\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{5},\dfrac{5}{7},$……) and the irrational numbers (Ex: $\sqrt{2},3\pi ,\text{ }.011011011...$).

Therefore, it can be concluded that every irrational number is a real number, but

every real number cannot be an irrational number.

2. Are the square roots of all positive integer numbers irrational? If not, provide an example of the square root of a number that is not an irrational number.

Ans: Square root of every positive integer does not give an integer.

For example: $\sqrt{2},\sqrt{3,}\sqrt{5},\sqrt{6},...$ are not integers, and hence these are irrational numbers. But $\sqrt{4}$ gives $\pm 2$ , these are integers and so, $\sqrt{4}$ is not an irrational number.

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

3. Represent $\sqrt{5}$ on the number line.

Ans: Follow the procedures to get $\sqrt{5}$ on the number line.

Firstly, Draw a line segment $AB$ of $2$ unit on the number line.

Secondly, draw a perpendicular line segment $BC$ at $B$ of $1$ units.

Thirdly, join the points $C$ and $A$, to form a line segment $AC$. 

Fourthly, apply the Pythagoras Theorem as 

$ A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} $

$  A{{C}^{2}}={{2}^{2}}+{{1}^{2}} $

$ A{{C}^{2}}=4+1=5 $

$ AC=\sqrt{5} $

Finally, draw the arc $ACD$, to find the number $\sqrt{5}$ on the number line as given in the diagram below.

Exercise (1.3)

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

(i) $\mathbf{\dfrac{ {36}}{ {100}}}$

Ans: Divide $36$ by $100$. 

$\,\,\,\,\,\,\,\,\,\, {0.36}$

$100 {\overline{)\;36\quad}}$

$\underline{\,\,\,\,\,\,\,\,\,-0\quad}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,360$

$\underline{\,\,\,\,\,\,\,\,\,\,-300\quad}$

$\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,600$

$\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,-600}$

$\underline{\,\,\,\,\,\,\,\,\,\,\,\,\quad 0 \,\,\,\,\,}$

So, $\dfrac{36}{100}=0.36$ and it is a terminating decimal number.

(ii) $\mathbf{\dfrac{ {1}}{ {11}}}$

Ans: Divide $1$ by $11$.

${\,\,\,\,\,\,\,\,0.0909..}$

$11 \, {\overline{)\;1\quad}}$

$\underline{\,\,\,\,\,\,\,-0\quad}$

$\,\,\,\,\,\,\,\,\,\,10$

$\underline{\,\;\;\,\,-0\quad}$

$\;\;\,\,\,\,100$

$\underline{\,\,\,\,\;-99}$

$\,\,\,\,\,\, \quad 10$

$\quad\underline{\;\;-0\quad}$

$\;\;\,\,\,\,\,\,\,\,100$

$\underline{\,\,\,\,\,\,\,\,\;-99}$

$\quad\,\,\,\,\,\,\,1\quad$

It is noticed that while dividing $1$ by $11$, in the quotient $09$ is repeated.

So, $\dfrac{1}{11}=0.0909.....$ or 

$\dfrac{1}{11}=0.\overline{09}$ 

and it is a non-terminating and recurring decimal number.

(iii)  $ \mathbf{{4}\dfrac{ {1}}{ {8}}}$

Ans: $4\dfrac{1}{8}=4+\dfrac{1}{8}=\dfrac{32+1}{8}=\dfrac{33}{8}$

Divide $33$ by $8$.

$\,\,\,\,\,{4.125}$

$8 {\overline{)\;33\quad}}$

$\underline{\,\,\,\,-32\quad}$

$\,\,\,\,\,\,\,\,\,\,\,\,10$

$\underline{\;\;\,\,\,\,-8\quad}$

$\;\;\,\,\,\,\,\,\,\,\,\,\,20$

$\underline{\,\,\,\,\,\,\,\,\,-16}$

$\;\quad\quad\,\,\,\,40$

$\quad\underline{\quad\,\,-40\quad}$

$\quad\underline{\quad\,\, \,\,\,\,0\quad}$

Notice that, after dividing $33$ by $8$, the remainder is found as $0$.

So, $4\dfrac{1}{8}=4.125$ and it is a terminating decimal number.

(iv)  $\mathbf{\dfrac{ {3}}{ {13}}}$

Ans: Divide $3$ by $13$.

$\quad \,\,{0.230769}$

$13 {\overline{)\;3\quad}}$

$\underline{\quad-0\quad}$

$\quad\quad 30$

$\underline{\;\,\quad-26\quad}$

$\;\quad\quad\,\,\,40$

$\underline{\quad\quad\,\,-39\quad}$

$\;\quad\quad\quad\;10$

$\quad\underline{\quad\quad -0\quad}$

$\quad{\quad\quad \quad 100}$

$\quad\quad\underline{\quad \,\, -91\quad}$

$\quad\quad \quad \,\,\,\quad90$

$\quad\quad\underline{\quad\,\,\,\,\,-78\quad}$

$\quad\quad\quad\quad \quad 120$

$\quad \quad\underline{\quad\quad\,\,-117\quad}$

$\quad\quad\underline{\quad \quad\quad\,\, 3\quad}$

It is observed that while dividing $3$ by $13$, the remainder is found as $3$ and that is repeated after each $6$ continuous divisions.

So, $\dfrac{3}{13}=0.230769.......$ or

$\dfrac{3}{13}=0.\overline{230769}$ 

(v)   $\mathbf{\dfrac{ {2}}{ {11}}}$

Ans: Divide $2$ by $11$.

$\quad \,\,{0.1818}$

$11 {\overline{)\;2\quad}}$

$\quad\quad20$

$\underline{\quad\;-11\quad}$

$\quad\quad \;\,90$

$\underline{\quad\,\,\,\, -88\;}$

$\;\quad\quad\;20$

$\quad\underline{\quad-11\quad}$

$\quad{\quad\quad  90}$

$\quad\underline{\,\,\quad -88}$

$\quad\quad\quad\,\,2\quad$

It can be noticed that while dividing $2$ by $11$, the remainder is obtained as $2$ and then $9$, and these two numbers are repeated infinitely as remainders.

So, $\dfrac{2}{11}=0.1818.....$ or 

$\dfrac{2}{11}=0.\overline{18}$ 

(vi) $\mathbf{\dfrac{ {329}}{ {400}}}$

Ans: Divide $329$ by $400$.

$\quad \quad{0.8225}$

$400 {\overline{)\;329\quad}}$

$\underline{\quad\,\,-0\quad}$

$\quad\quad3290$

$\underline{\quad\;-3200\quad}$

$\quad\quad\quad\;900$

$\underline{\quad\quad\quad-800\;}$

$\quad\quad\quad\quad\;1000$

$\quad\underline{\quad\quad\quad-800\quad}$

$\quad{\quad\quad\quad\quad\,\,2000}$

$\quad\underline{\quad\quad\quad\quad-2000\quad}$

$\quad\underline{\quad\quad\quad\quad\,\,\,\,\,\, 0 \quad}$

It can be seen that while dividing $329$ by $400$, the remainder is obtained as $0$.

So, $\dfrac{329}{400}=0.8225$ and is a terminating decimal number.

2. You know that $\dfrac{ {1}}{ {7}} {=0} {.142857}...$. Can you predict what the decimal expansions of $\dfrac{ {2}}{ {7}} {,}\dfrac{ {3}}{ {7}} {,}\dfrac{ {4}}{ {7}} {,}\dfrac{ {5}}{ {7}} {,}\dfrac{ {6}}{ {7}}$  are, without actually doing the long division? If so, how?

$\text{[}$Hint: Study the remainders while finding the value of $\dfrac{ {1}}{ {7}}$ carefully.$\text{]}$

Ans: Note that,  $\dfrac{2}{7},\dfrac{3}{7},\dfrac{4}{7},\dfrac{5}{7}$ and $\dfrac{6}{7}$ can be rewritten as $2\times \dfrac{1}{7},\text{ 3}\times \dfrac{1}{7},\text{ 4}\times \dfrac{1}{7},\text{ 5}\times \dfrac{1}{7},$ and $6\times \dfrac{1}{7}$

Substituting the value of $\dfrac{1}{7}=0.142857$ , gives 

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

$ 3\times \dfrac{1}{7} = 3\times .428571…= .428571...$

\[4\times \dfrac{1}{7}=4\times 0.142857...\]\[\text{=}\,\text{0}\text{.571428}...\]

$5\times \dfrac{1}{7}=5\times 0.71425...$  \[\text{=}\,\text{0}\text{.714285}...\]

$6\times \dfrac{1}{7}=6\times 0.142857...$\[\text{=}\,\text{0}\text{.857142}...\]

So, the values of $\dfrac{2}{7},\text{ }\dfrac{3}{7},\text{ }\dfrac{4}{7},\text{ }\dfrac{5}{7}$ and $\dfrac{6}{7}$ obtained without performing long division are

\[\dfrac{2}{7}=0.\overline{285714}\]

$\dfrac{3}{7}=0.\overline{428571}$

$\dfrac{4}{7}=0.\overline{571428}$

\[\dfrac{5}{7}=0.\overline{714285}\]

$\dfrac{6}{7}=0.\overline{857142}$

3. Express the following in the form \[\dfrac{ {p}}{ {q}}\], where $ {p}$ and $ {q}$ are integers and $ {q}\ne  {0}$.

(i) $\mathbf{ {0} {.}\overline{ {6}}}$

Ans: Let $x=0.\overline{6}$  

 $\Rightarrow x=0.6666$                                                   ….… (1)

 Multiplying both sides of the equation (1) by $10$, gives

$10x=0.6666\times 10$

$10x=6.6666$…..                 …… (2)

Subtracting the equation $\left( 1 \right)$ from $\left( 2 \right)$, gives

$ 10x=6.6666..... $

$ \underline{-x=0.6666.....} $

$  9x=6 $ 

$  9x=6 $

$  x=\dfrac{6}{9}=\dfrac{2}{3} $ 

So, the decimal number becomes

$0.\overline{6}=\dfrac{2}{3}$  and it is in the required  $\dfrac{p}{q}$ form.

(ii) $\mathbf{ {0} {.}\overline{ {47}}}$

Ans: Let  $x=0.\overline{47}$

$\text{   }\Rightarrow x=0.47777.....$                                             ……(a)

Multiplying both sides of the equation (a) by $10$, gives

$10x=4.7777.....$         ……(b)

Subtracting the equation $\left( a \right)$ from $\left( b \right)$, gives

$ 10x=4.7777..... $

$  \underline{-x=0.4777.....} $

$  9x=4.3 $

$x=\dfrac{4.3}{9}\times \dfrac{10}{10} $ 

$ \Rightarrow x=\dfrac{43}{90} $

So, the decimal number becomes 

$0.\overline{47}=\dfrac{43}{90}$  and it is in the required $\dfrac{p}{q}$ form.

(iii) $ \mathbf{{0} {.}\overline{ {001}}}$

Ans: Let $x=0.\overline{001} $           …… (1)

Since the number of recurring decimal number is $3$, so multiplying both sides of the equation (1) by $1000$, gives

$1000\times x=1000\times 0.001001.....$ …… (2)

Subtracting the equation (1) from (2) gives

$ 1000x=1.001001..... $

$  \underline{\text{    }-x=0.001001.....} $

$  999x=1 $

$\Rightarrow x=\dfrac{1}{999}$

Hence, the decimal number becomes 

$0.\overline{001}=\dfrac{1}{999}$ and it is in the $\dfrac{p}{q}$ form.

4. Express $ {0} {.99999}.....$ in the form of $\dfrac{ {p}}{ {q}}$ . Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense.

Let $x=0.99999.....$                                                             ....... (a)

Multiplying by $10$ both sides of the equation (a), gives

$10x=9.9999.....$                                                             …… (b)

Now, subtracting the equation (a) from (b), gives

$ 10x=9.99999..... $

$  \underline{\,-x=0.99999.....} $

$  9x=9 $ 

$\Rightarrow x=\dfrac{9}{9}$

$\Rightarrow x=1$.

$0.99999...=\dfrac{1}{1}$ which is in the $\dfrac{p}{q}$ form.

Yes, for a moment we are amazed by our answer, but when we observe that $0.9999.........$ is extending infinitely, then the answer makes sense.

Therefore, there is no difference between $1$ and $0.9999.........$ and hence these two numbers are equal.

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

Ans: Here the number of digits in the recurring block of $\dfrac{1}{17}$ is to be determined. So, let us calculate the long division to obtain the recurring block of $\dfrac{1}{17}$. Dividing $1$ by $17$ gives

$\quad\quad {0.0588235294117646}$

$17{\overline{)\quad1\quad\quad\quad\quad\quad\quad\quad\quad}}$

$\underline{\quad\,\,\,\,-0\quad}\qquad\qquad\qquad$

$\quad \quad \,\,\,10\qquad\qquad\quad\quad$

$\underline{\quad \quad -0\quad}\qquad\qquad\quad$

$\quad \quad \,\,\,\,\,\;100\qquad\qquad\qquad$

$\underline{\quad \quad \,\,-85\;}\qquad\qquad\quad$

$\quad\qquad\,\,\;150\qquad\qquad\quad$

$\quad\underline{\qquad-136\;}\qquad\qquad\quad$

$\quad{\quad\quad\quad 140}\qquad\qquad\;\;$

$\quad\underline{\qquad-136\quad}\qquad\quad$

${\quad \qquad \,\,\quad 40 \quad}\quad$

$\underline{\qquad \,\,\,\quad -34\;\;}\quad$

$\;\qquad \qquad\,\,60$

$\underline{\qquad \qquad-51}$

$\quad\quad \qquad \quad 90$

$\quad\;\;\underline{\quad \qquad-85}$

$\qquad\quad\;\quad\,\,\,\, 50$

$\quad\quad\;\;\underline{\,\,\quad\,\, -34}$

$\quad\quad\qquad \quad 160$

$\qquad\quad\;\underline{\quad-153}$

$\qquad\qquad\quad\;70$

$\qquad\quad\quad\;\;\underline{-68}$

$\quad\,\,\qquad\qquad 20$

$\qquad\qquad\quad\underline{-17}$

$\qquad\qquad\quad\quad\; 130$

$\qquad\qquad\quad\;\;\underline{-119}$

$\qquad\qquad\qquad\quad 110$

$\qquad\qquad\qquad\;\;\underline{-102}$

$\qquad\qquad\qquad\quad\quad\quad 80$

$\qquad\qquad\qquad\qquad\;\underline{-68}$

$\qquad\qquad\qquad\quad\quad\quad\; 120$

$\qquad\qquad\qquad\qquad\;\;\underline{-119}$

$\qquad\qquad\qquad\quad\quad\quad\; 1$

Thus, it is noticed that while dividing $1$ by $17$, we found $16$ number of digits in the

repeating block of decimal expansion that will continue to be $1$ after going through $16$ continuous divisions.

Hence, it is concluded that $\dfrac{1}{17}=0.0588235294117647.....$ or 

 $\dfrac{1}{17}=0.\overline{0588235294117647}$ and it is a recurring and non-terminating decimal number.

6. Look at several examples of rational numbers in the form $\dfrac{ {p}}{ {q}}\left(  {q}\ne  {0} \right)$, 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 such rational numbers $\dfrac{5}{2},\dfrac{5}{4},\dfrac{2}{5},\dfrac{2}{10},\dfrac{5}{16}$ of the form $\dfrac{p}{q}$ which have terminating decimal representations.

$ \dfrac{5}{2}=2.5 $

$ \dfrac{5}{4}=1.25 $ 

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

$ \dfrac{2}{10}=0.2 $

$ \dfrac{5}{16}=0.3125 $

In each of the above examples, it can be noticed that the denominators of the rational numbers have powers of $2,5$ or both.

So, $q$ must satisfy the form either ${{2}^{m}}$, or ${{5}^{n}}$, or  both ${{2}^{m}}\times {{5}^{n}}$ (where $m=0,1,2,3.....$ and $n=0,1,2,3.....$) in the form of $\dfrac{p}{q}$.

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

Ans: All the irrational numbers are non-terminating and non-recurring, because irrational numbers do not have any representations of the form of $\dfrac{p}{q}$ $\left( q\ne 0 \right)$, where $p$ and $q$are integers. For example: 

$\sqrt{2}=1.41421.....$,

$\sqrt{3}=1.73205...$

$\sqrt{7}=2.645751....$

are the numbers whose decimal representations are non-terminating and non-recurring.

8. Find any three irrational numbers between the rational numbers $\dfrac{ {5}}{ {7}}$ and $\dfrac{ {9}}{ {11}}$.

Ans: Converting  $\dfrac{5}{7}$and $\dfrac{9}{11}$ into the decimal form gives

$\dfrac{5}{7}=0.714285.....$ and 

$\dfrac{9}{11}=0.818181.....$

Therefore, $3$ irrational numbers that are contained between $0.714285......$ and $0.818181.....$

$ 0.73073007300073...... $ 

$  0.74074007400074...... $ 

$ 0.76076007600076...... $

Hence, three irrational numbers between the rational numbers $\dfrac{5}{7}$ and $\dfrac{9}{11}$ are

9. Classify the following numbers as rational or irrational:

(i) $\mathbf{\sqrt{ {23}}}$

Ans: The following diagram reminds us of the distinctions among the types of rational and irrational numbers.

After evaluating the square root gives

$\sqrt{23}=4.795831.....$ , which is an irrational number.

(ii) $\mathbf{\sqrt{ {225}}}$

Ans: After evaluating the square root gives

$\sqrt{225}=15$, which is a rational number.

That is, $\sqrt{225}$ is a rational number.

(iii) $ \mathbf{{0} {.3796}}$

Ans: The given number is $0.3796$. It is terminating decimal. 

So, $0.3796$ is a rational number.

(iv) $ \mathbf{{7} {.478478}}$

Ans: The given number is \[7.478478\ldots .\] 

It is a non-terminating and recurring decimal that can be written in the $\dfrac{p}{q}$ form.

Let      $x=7.478478\ldots .$                                   ……(a)

Multiplying the equation (a) both sides by $100$ gives

$\Rightarrow 1000x=7478.478478.....$                                               ……(b)

Subtracting the equation (a) from (b), gives

$ 1000x=7478.478478.... $

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

$ 999x=7471 $

$  \text{      }x=\dfrac{7471}{999} $

Therefore, $7.478478.....=\dfrac{7471}{999}$, which is in the form of $\dfrac{p}{q}$

So, $7.478478...$ is a rational number.

(v) $ \mathbf{{1} {.101001000100001}.....}$

Ans: The given number is \[1.101001000100001....\]

It can be clearly seen that the number \[1.101001000100001....\] is a non-terminating and non-recurring decimal and it is known that non-terminating non-recurring decimals cannot be written in the form of $\dfrac{p}{q}$.

Hence, the number \[1.101001000100001....\] is an irrational number.

Exercise (1.4)

1.  Classify the following numbers as rational or irrational:

(i) $ \mathbf{{2-}\sqrt{ {5}}}$

Ans: The given number is $2-\sqrt{5}$.

Here, $\sqrt{5}=2.236.....$ and it is a non-repeating and non-terminating irrational number.

Therefore, substituting the value of $\sqrt{5}$ gives

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

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

So, $2-\sqrt{5}$ is an irrational number.

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

Ans: The given number is $\left( 3+\sqrt{23} \right)-\left( \sqrt{23} \right)$.

The number can be written as

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

$  =3 $

$=\dfrac{3}{1}$, which is in the $\dfrac{p}{q}$ form and so, it is a rational number.

Hence, the number $\left( 3+\sqrt{23} \right)-\sqrt{23}$ is a rational number.

(iii) $\mathbf{\dfrac{ {2}\sqrt{ {7}}}{ {7}\sqrt{ {7}}}}$

Ans: The given number is $\dfrac{2\sqrt{7}}{7\sqrt{7}}$.

$\dfrac{2\sqrt{7}}{7\sqrt{7}}=\dfrac{2}{7}$, which is in the $\dfrac{p}{q}$  form and so, it is a rational number.

Hence, the number  $\dfrac{2\sqrt{7}}{7\sqrt{7}}$ is a rational number.

(iv) $\mathbf{\dfrac{ {1}}{\sqrt{ {2}}}}$

Ans: The given number is $\dfrac{1}{\sqrt{2}}$.

It is known that, $\sqrt{2}=1.414.....$ and it is a non-repeating and non-terminating irrational number.

Hence, the number $\dfrac{1}{\sqrt{2}}$ is an irrational number.

(v) $ \mathbf{{2\pi }}$

Ans: The given number is $2\pi $.

It is known that, $\pi =3.1415$ and it is an irrational number.

Now remember that, Rational $\times $ Irrational = Irrational.

Hence, $2\pi $ is also an irrational number.

2. Simplify each of the of the following expressions:

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

Ans: The given number is $\left( 3+\sqrt{3} \right)\left( 2+\sqrt{2} \right)$.

By calculating the multiplication, it can be written as

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

\[= 6 + 4 \sqrt{2} + 2 \sqrt{3}+ \sqrt{6}\]

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

Ans: The given number is $\left( 3+\sqrt{3} \right)\left( 3-\sqrt{3} \right)$.

By applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, the number can be written as

$\left( 3+\sqrt{3} \right)\left( 3-\sqrt{3} \right)={{3}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=9-3=6$.

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

Ans: The given number is ${{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}$.

Applying the formula ${{\left( a+b \right)}^{2}}={{a}^{2+}}2ab+{{b}^{2}}$, the number can be written as

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

 $=5+2\sqrt{10}+2$

 $=7+2\sqrt{10}$.

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

Ans: The given number is $\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}+\sqrt{2} \right)$.

Applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, the number can be expressed as

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

$ =3. $ 

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

Ans: It is known that, $\pi =\dfrac{22}{7}$, which is a rational number. But, note that this value of $\pi $ is an approximation.

On dividing $22$ by $7$, the quotient $3.14...$ is a non-recurring and non-terminating number. Therefore, it is an irrational number.

In order of increasing accuracy, approximate fractions are

$\dfrac{22}{7}$, $\dfrac{333}{106}$, $\dfrac{355}{113}$, $\dfrac{52163}{16604}$, $\dfrac{103993}{33102}$, and \[\dfrac{245850922}{78256779}\].

Each of the above quotients has the value $3.14...$, which is a non-recurring and non-terminating number.

Thus, $\pi $ is irrational.

So, either circumference $\left( c \right)$ or diameter $\left( d \right)$ or both should be irrational numbers.

Hence, it is concluded that there is no contradiction regarding the value of $\pi $ and it is made out that the value of $\pi $ is irrational.

4. Represent $\sqrt{ {9} {.3}}$ on the number line.

Ans: Follow the procedure given below to represent the number $\sqrt{9.3}$.

First, mark the distance $9.3$ units from a fixed-point $A$ on the number line to get a point $B$. Then $AB=9.3$ units.

Secondly, from the point $B$ mark a distance of $1$ unit and denote the ending point as $C$.

Thirdly, locate the midpoint of $AC$ and denote it as $O$.

Fourthly, draw a semi-circle to the centre $O$ with the radius $OC=5.15$ units. Then 

$ AC=AB+BC $ 

$  =9.3+1 $ 

$  =10.3 $

So, $OC=\dfrac{AC}{2}=\dfrac{10.3}{2}=5.15$.

Finally, draw a perpendicular line at $B$ and draw an arc to the centre $B$ and then let it meet at the semicircle $AC$ at $D$ as given in the diagram below.

5. Rationalize the denominators of the following:

(i) $\mathbf{\dfrac{ {1}}{\sqrt{ {7}}}}$

Ans: The given number is $\dfrac{1}{\sqrt{7}}$.

Multiplying and dividing by $\sqrt{7}$ to the number gives

$\dfrac{1}{\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}=\dfrac{\sqrt{7}}{7}$.

(ii) $\mathbf{\dfrac{ {1}}{\sqrt{ {7}} {-}\sqrt{ {6}}}}$

Ans: The given number is $\dfrac{1}{\sqrt{7}-\sqrt{6}}$.

Multiplying and dividing by $\sqrt{7}+\sqrt{6}$ to the number gives

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

Now, applying the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ to the denominator gives

$ \dfrac{1}{\sqrt{7}-\sqrt{6}}=\dfrac{\sqrt{7}+\sqrt{6}}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( \sqrt{6} \right)}^{2}}} $ 

$ =\dfrac{\sqrt{7}+\sqrt{6}}{7-6} $ 

$  =\dfrac{\sqrt{7}+\sqrt{6}}{1}. $

(iii) $\mathbf{\dfrac{ {1}}{\sqrt{ {5}} {+}\sqrt{ {2}}}}$

Ans: The given number is $\dfrac{1}{\sqrt{5}+\sqrt{2}}$.

Multiplying and dividing by $\sqrt{5}-\sqrt{2}$ to the number gives

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

Now, applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$  to the denominator gives

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

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

$ =\dfrac{\sqrt{5}-\sqrt{2}}{3}. $ 

(iv) $\mathbf{\dfrac{ {1}}{\sqrt{ {7}} {-2}}}$

Ans: The given number is $\dfrac{1}{\sqrt{7}-2}$.

Multiplying and dividing by $\sqrt{7}+2$ to the number gives

$\dfrac{1}{\sqrt{7}-2}=\dfrac{\sqrt{7}+2}{\left( \sqrt{7}-2 \right)\left( \sqrt{7}+2 \right)}\\$.

Now, applying the formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ to the denominator gives

$ \dfrac{1}{\sqrt{7}-2}=\dfrac{\sqrt{7}+2}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( 2 \right)}^{2}}} $

$ =\dfrac{\sqrt{7}+2}{7-4} $ 

$  =\dfrac{\sqrt{7}+2}{3}. $

Exercise (1.5)

1. Compute the value of each of the following expressions:

(i) $\mathbf{ {6}{{ {4}}^{\dfrac{ {1}}{ {2}}}}}$

Ans: The given number is \[{{64}^{\dfrac{1}{2}}}\].

By the laws of indices,

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$, where$a>0$.

$ {{64}^{\dfrac{1}{2}}}=\sqrt[2]{64} $

$  =\sqrt[2]{8\times \text{8}} $

$  =8. $

Hence, the value of ${{64}^{\dfrac{1}{2}}}$ is $8$.

(ii) $ \mathbf{{3}{{ {2}}^{\dfrac{ {1}}{ {5}}}}}$

Ans: The given number is ${{32}^{\dfrac{1}{5}}}$.

${{a}^{\dfrac{m}{n}}}=\sqrt[m]{{{a}^{m}}}$, where $a>0$

$ {{32}^{\dfrac{1}{5}}}=\sqrt[5]{32}$

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

$ =\sqrt[5]{{{2}^{5}}} $

Alternative Method:

By the law of indices ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, then it gives

$ {{32}^{\dfrac{1}{5}}}={{(2\times 2\times 2\times 2\times 2)}^{\dfrac{1}{5}}}$ 

$ ={{\left( {{2}^{5}} \right)}^{\dfrac{1}{5}}} $

$ ={{2}^{\dfrac{5}{5}}} $

Hence, the value of the expression ${{32}^{\dfrac{1}{5}}}$ is $2$.

(iii) $\mathbf{{12}{{ {5}}^{\dfrac{ {1}}{ {5}}}}}$

Ans: The given number is ${{125}^{\dfrac{1}{3}}}$.

By the laws of indices

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where$a>0$.

$ {{125}^{\dfrac{1}{3}}}=\sqrt[3]{125} $

$  =\sqrt[3]{5\times 5\times 5} $

$  =5. $

Hence, the value of the expression ${{125}^{\dfrac{1}{3}}}$ is $5$.

2. Compute the value of each of the following expressions:

(i) $\mathbf{{{ {9}}^{\dfrac{ {3}}{ {2}}}}}$

Ans: The given number is ${{9}^{\dfrac{3}{2}}}$.

 ${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where $a>0$.

$ {{9}^{\dfrac{3}{2}}}=\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 $

By the laws of indices, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, then it gives

$ {{9}^{\dfrac{3}{2}}}={{\left( 3\times 3 \right)}^{\dfrac{3}{2}}}$

$  ={{\left( {{3}^{2}} \right)}^{\dfrac{3}{2}}} $

$  ={{3}^{2\times \dfrac{3}{2}}} $

$ ={{3}^{3}} $

${{9}^{\dfrac{3}{2}}}=27.$

Hence, the value of the expression ${{9}^{\dfrac{3}{2}}}$ is $27$.

(ii) $\mathbf{{3}{{ {2}}^{\dfrac{ {2}}{ {5}}}}}$

Ans: We know that ${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where $a>0$.

We conclude that ${{32}^{\dfrac{2}{5}}}$ can also be written as

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

$  =2\times 2 $

$ =4 $ 

Therefore, the value of ${{32}^{\dfrac{2}{5}}}$ is $4$.

(iii) $\mathbf{{1}{{ {6}}^{\dfrac{ {3}}{ {4}}}}}$

Ans: The given number is ${{16}^{\dfrac{3}{4}}}$.

By the laws of indices, 

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$, where $a>0$.

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

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

$  =2\times 2\times 2 $

Hence, the value of the expression ${{16}^{\dfrac{3}{4}}}$ is $8$.

${{({{a}^{m}})}^{n}}={{a}^{mn}}$, where $a>0$.

$ {{16}^{\dfrac{3}{4}}}={{(4\times 4)}^{\dfrac{3}{4}}} $

$  ={{({{4}^{2}})}^{\dfrac{3}{4}}} $ 

$ ={{(4)}^{2\times \dfrac{3}{4}}} $

$ ={{({{2}^{2}})}^{2\times \dfrac{3}{4}}} $ 

$ ={{2}^{2\times 2\times \dfrac{3}{4}}} $

$ ={{2}^{3}} $

Hence, the value of the expression is ${{16}^{\dfrac{3}{4}}}=8$.

(iv) $\mathbf{{12}{{ {5}}^{ {-}\dfrac{ {1}}{ {3}}}}}$

Ans: The given number is ${{125}^{-\dfrac{1}{3}}}$.

By the laws of indices, it is known that 

${{a}^{-n}}=\dfrac{1}{{{a}^{^{n}}}}$, where $a>0$.

Therefore, 

$ {{125}^{-\dfrac{1}{3}}}=\dfrac{1}{{{125}^{\dfrac{1}{3}}}} $

$  ={{\left( \dfrac{1}{125} \right)}^{\dfrac{1}{3}}} $

$ =\sqrt[3]{\left( \dfrac{1}{125} \right)} $

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

$ =\dfrac{1}{5}. $

Hence, the value of the expression ${{125}^{-\dfrac{1}{3}}}$ is  $\dfrac{1}{5}$.

3. Simplify and evaluate each of the expressions:

(i)$\mathbf{{{ {2}}^{\dfrac{ {2}}{ {3}}}} {.}{{ {2}}^{\dfrac{ {1}}{ {5}}}}}$

Ans: The given expression is ${{2}^{\dfrac{2}{3}}}{{.2}^{\dfrac{1}{5}}}$.

By the laws of indices, it is known that

${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$, where $a>0$.

 ${{2}^{\dfrac{2}{3}}}{{.2}^{\dfrac{1}{5}}}={{(2)}^{\dfrac{2}{3}+\dfrac{1}{5}}}$

 $ ={{(2)}^{\dfrac{10+3}{15}}} $

 $ ={{2}^{\dfrac{13}{15}}}. $

Hence, the value of the expression ${{2}^{\dfrac{2}{3}}}{{.2}^{\dfrac{1}{5}}}$ is ${{2}^{\dfrac{13}{15}}}$.

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

Ans: The given expression is  ${{\left( {{\frac{ {1}}{ {{3}^3}}}} \right)}^{ {7}}}$.

It is known by the laws of indices that,

 ${{({{a}^{m}})}^{n}}={{a}^{mn}}$, where $a>0$.

 ${{\left( {{\frac{ {1}}{ {{3}^3}}}} \right)}^{ {7}}} =\left ( \dfrac{1}{3^{21}} \right )$

Hence, the value of the expression  ${{\left( {{\frac{ {1}}{ {{3}^3}}}} \right)}^{ {7}}}$ is  $\left ( \dfrac{1}{3^{21}} \right )$

(iii) $\dfrac{ {1}{{ {1}}^{\dfrac{ {1}}{ {2}}}}}{ {1}{{ {1}}^{\dfrac{ {1}}{ {4}}}}}$

Ans: The given number is $\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}$.

It is known by the Laws of Indices that

 $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, where $a>0$.

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

$ ={{11}^{\dfrac{2-1}{4}}} $ 

$  ={{11}^{\dfrac{1}{4}}}. $

Hence, the value of the expression $\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}$ is  ${{11}^{\dfrac{1}{4}}}$.

(iv) $\mathbf{{{ {7}}^{\dfrac{ {1}}{ {2}}}} {.}{{ {8}}^{\dfrac{ {1}}{ {2}}}}}$

Ans: The given expression is ${{7}^{\dfrac{1}{2}}}\cdot {{8}^{\dfrac{1}{2}}}$.

${{a}^{m}}\cdot {{b}^{m}}={{(a\cdot b)}^{m}}$, where $a>0$.

$ {{7}^{\dfrac{1}{2}}}\cdot {{8}^{\dfrac{1}{2}}}={{(7\times 8)}^{\dfrac{1}{2}}} $  $={{(56)}^{\dfrac{1}{2}}}. $

Hence, the value of the expression ${{7}^{\dfrac{1}{2}}}\cdot {{8}^{\dfrac{1}{2}}}$ is ${{(56)}^{\dfrac{1}{2}}}$.

Class 9 Maths Chapter 1 Solutions - Free PDF Download

The NCERT Solutions for Class 9 Maths Chapter 1, "Number Systems," serve as the first chapter of the Class 9 Maths curriculum. This chapter provides an in-depth discussion on Number Systems and their applications, starting with an introduction to whole numbers, integers, and rational numbers.

The chapter begins with an overview of Number Systems in section 1.1, followed by two crucial topics in sections 1.2 and 1.3:

Irrational Numbers : These are numbers that cannot be expressed in the form p/q.

Real Numbers and their Decimal Expansions : This section examines the decimal expansions of real numbers to differentiate between rational and irrational numbers.

Further, the chapter covers:

Representing Real Numbers on the Number Line: Solutions for two problems in Exercise 1.4 are provided.

Operations on Real Numbers: This section explores operations such as addition, subtraction, multiplication, and division involving irrational numbers.

Laws of Exponents for Real Numbers: These laws are used to solve various questions.

NCERT Solutions for Class 9 Maths Chapter 1 All Exercise

The class 9 maths chapter 1 PDF solutions by Vedantu provide a detailed and clear explanation of the concepts in the chapter. This chapter covers important topics like rational and irrational numbers, real numbers, and their decimal expansions. Understanding these foundational concepts is crucial for success in higher-level maths. When studying, focus on grasping the properties of different types of numbers and practising their operations. The solutions by Vedantu simplify these concepts with step-by-step explanations, making them easier to understand. In previous year question papers, typically 3 to 5 questions from this chapter are asked. These questions often test your understanding of number classification, representation of numbers on the number line, and converting between different forms of numbers. Therefore, practice is key to mastering this chapter and performing well in exams.

Other Related Links for CBSE Class 9 Maths Chapter 1

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 Class 9 Maths Chapter 1 - Number Systems

1. What all Comes Under the Purview of NCERT Maths Class 9 Chapter 1 Number Systems?

The subjects covered in NCERT mathematics class 9 chapter 1 Number Systems include a brief introduction to number systems using number lines, defining rational and irrational numbers using fractions, defining real numbers and declaring their decimal expansions. The chapter then returns to the number line to teach pupils how to express real numbers on it. In addition, the chapter teaches pupils how to add, subtract, multiply, and divide real numbers, or how to perform operations on real numbers. The rules of exponents for real numbers are a part of operations and are the final topic in class 9 mathematics chapter 1.

2. What are the Weightage Marks for Mathematics in Class 9?

The total mathematics paper in class 9 is 100 marks, like any other subject. Out of these 100 marks, 20 marks goes from internal assessments (pen and paper tests, multiple assessments, portfolios/project work and lab practicals for 5 marks each), and the remaining 80 marks are from the written test at the end of the school year. Out of these 80 marks, the chapter Number Systems comes for 8 marks, Algebra for 17 marks,  Coordinate Geometry for 4 marks, Geometry for 28 marks, Mensuration for 13 marks, and Statistics and Probability for 10 marks. All of these chapters’ respective marks total up to a cumulative 80 marks for the written paper.

3. How many sums are there in the NCERT Class 9 Chapter 1 Number System?

There are six exercises in the NCERT Class 9 Chapter 1 Number System. In the first exercise, Ex-1.1, there are 4 sums and in the second exercise, Ex-1.2, there are 3 sums. These first two exercises deal with the basic concepts of the number system, such as identifying the features of a rational number or an irrational number and locating them on the number line. In the third exercise, Ex-1.3, there are 9 sums, and most of them have sub-questions. The fourth exercise, Ex-1.4, comprises 2 sums, that deal with successive magnification for locating a decimal number on the number line. The fifth exercise, Ex-1.5, consists of 5 sums, on the concept of rationalization. The sixth exercise, Ex-1.6, consists of 3 sums, that have sub-questions. The sums in this exercise will require you to find the various roots of numbers.

4. Why should we download NCERT Solutions for Class 9 Maths Chapter 1?

Students should download NCERT Solutions for Class 9 Maths Chapter 1 from Vedantu (vedantu.com) to understand and learn the concepts of the Number System easily. These solutions are available free of cost on Vedantu (vedantu.com). Students must have a solid base of all concepts of Class 9 Maths if they want to score well in their exams. They can download the NCERT Solutions and other study materials such as important questions and revision notes for all subjects of Class 9. You can download these from Vedantu mobile app also.

5. Why are Class 9 Maths NCERT Solutions Chapter 1 important?

Some students find it difficult to study and score good marks in their Maths exam. They get nervous while preparing for it and goof up in their exams. However, if they utilise the best resources for studying, they can do well. This is why the Class 9 Maths NCERT Solutions Chapter 1 is important. The answers to all the questions from the back of each chapter are provided for the reference of students. 

6. Give an overview of concepts present in NCERT Solutions for Class 9 Maths Chapter 1?

The concepts in the NCERT Solutions for Class 9 Maths Chapter 1 include the introduction of number systems, rational and irrational numbers using fractions, defining real numbers, decimal expansions of real numbers, number line, representing real numbers on a number line, addition, subtraction, multiplication and division of real numbers and laws of exponents for real numbers. Chapter 1 of Class 9 Maths has a weightage of 8 marks in the final exam. 

7. Do I Need to Practice all Questions Provided in NCERT Solutions Class 9 Maths Number Systems?

Yes. Students should practice all the questions provided in the NCERT Solutions of the Number Systems chapter of Class 9 Maths, as they have been created with precision and accuracy, by expert faculty, for the students. Students can access them for free and also download them for offline use to reduce their screen time. The solutions are beneficial not only for exams but also for school homework.

8. Where can I get the NCERT Solutions for Class 9 Maths Chapter 1?

Students can download the NCERT Solutions for Class 9 Maths Chapter 1 from NCERT Solutions for Class 9 Maths Chapter 1. These are available free of cost on Vedantu (vedantu.com). These can be downloaded from the Vedantu app as well. The answers to all the questions from the 6 exercises of Chapter 1 Number Systems are provided in the NCERT Solutions. Students would also learn how to solve one question with different techniques if available. This will help them learn how to structure their answers in their Class 9 Maths exam. 

NCERT Solutions for Class 9 Maths

Ncert solutions for class 9.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 1 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 1 Number Systems NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.1

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.2

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.3

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.4

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5

NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.6

NCERT Solutions for Class 9 Maths Chapter 1 – Topic Discussion

Below we have listed the topics that have been discussed in this chapter. As Number System is one of the important topics in Maths, it has a weightage of 6 marks in class 9 Maths exams. 

• 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

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• Introduction to Euclid’s Geometry Class 9 Case Study Questions Maths Chapter 5

Last Updated on August 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 maths. In this article, you will find case study questions for CBSE Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry. It is a part of Case Study Questions for CBSE Class 9 Maths Series.

Table of Contents

Case Study Questions on Introduction to Euclid’s Geometry

A National Public School organised an education trip to a museum. Almost all the students of class IX went to the trip with their teacher of Mathematics. They saw many pictures of mathematicians and read about their contributions in the field of Mathematics. After visiting the museum, teacher asked the following questions from the students.

On the basis of the above information, solve the following questions:

Q 1. Pythagoras was a student of: a. Euclid b. Thales c. Archimedes d. Both a. and b.

Ans. (b) Pythagoras was a student of Thales. So, option (b) is correct.

Q 2. Name of the mathematician who is visible in the last picture, is: a. Euclid b. Pythagoras c. Thales d. None of these

Ans. (a) Euclid mathematician is visible in the last picture. So, option (a) is correct.

Q 3. Euclid stated that ‘A circle can be drawn with any centre and any radius’, is a/an: a. definition b. postulate c. axiom d. proof

Ans. (b) Euclid stated that ‘A circle can be drawn with any centre and any radius’ is postulate. So, option (b) is correct.

Q 4. In which country Thales belong to? a. Greece b. Egypt c. Babylonia d. Rome

Ans. (a) Thales belongs to Greece Country. So, option (a) is correct.

Q 5. Which of the following needs a proof? a. Definition b. Theorem c. Axiom d. Postulate

Ans. (b) Theorem needs a proof. So, option (b) is correct.

Understanding Euclid’s Geometry

Euclid’s Definitions

• A point is that which has no part.
• A line is breadthless length.
• The ends of a line are points.
• A straight line is a line which lies evenly with the points on itself.
• A surface is that which has length and breadth only.
• The edges of a surface are lines.
• A plane surface is a surface which lies evenly with the straight lines on itself.

A sentence which can be judged to be true or false, e.g., The sum of the angles of a quadrilateral is 360°, is a true statement and a line segment has one end point, is a false statement.

The basic facts taken for granted without proof. e.g., A line has infinitely many points.

A mathematical statement whose truth has been established (proved)

Euclid’s Axioms

• Things which are equal to the same thing are equal to one another.
• If equals are added to equals, the wholes are equal.
• If equals are subtracted from equals, the remainders are equal.
• Things which coincide with one another, are equal to one another.
• The whole is greater than the part.
• Things which are double of the same things, are equal to one another.
• Things which are halves of the same things, are equal to one another.

Postulate: The assumptions which are specific to geometry, e.g., Two points make a line.

Euclid’s Postulates

Postulate 1: A straight line may be drawn from any one point to any other point. Postulate 2: A terminated line can be produced indefinitely. Postulate 3: A circle can be drawn with any centre and any radius. Postulate 4: All right angles are equal to one another. Postulate 5: (Parallel Postulate): If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

• Quadrilaterals Class 9 Case Study Questions Maths Chapter 8
• Triangles Class 9 Case Study Questions Maths Chapter 7
• Lines and Angles Class 9 Case Study Questions Maths Chapter 6
• Linear Equations in Two Variables Class 9 Case Study Questions Maths Chapter 4
• Coordinate Geometry Class 9 Case Study Questions Maths Chapter 3

Polynomials Class 9 Case Study Questions Maths Chapter 2

Number systems class 9 case study questions maths chapter 1, topics from which case study questions may be asked.

• History-Geometry in India and Euclid’s geometry
• Euclid’s method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems.
• The five postulates of Euclid. Showing the relationship between axiom and theorem, for example: (Axiom) 1. Given two distinct points, there exists one and only one line through them. (Theorem) 2. (Prove) Two distinct lines cannot have more than one point in common.

An axiom generally is true for any field in science, while a postulate can be specific on a particular field.

Case study questions from the above given topic may be asked.

Download Customised White Label Study Materials in MS Word Format

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Frequently Asked Questions (FAQs) on Introduction to Euclid’s Geometry Case Study

Q1: what is euclid’s geometry.

A1: Euclid’s Geometry is a mathematical system attributed to the ancient Greek mathematician Euclid. It is based on a set of definitions, postulates (axioms), and propositions (theorems). This geometry primarily deals with the properties and relations of points, lines, surfaces, and solids in a two-dimensional and three-dimensional space.

Q2: Who was Euclid?

A2: Euclid was a Greek mathematician, often referred to as the “Father of Geometry.” He lived around 300 BCE in Alexandria, Egypt. His most famous work, Elements , is a collection of books that form the foundation of what is now known as Euclidean geometry.

Q3: What are Euclid’s Axioms?

A3: Euclid’s axioms are basic assumptions that are accepted as true without proof. These axioms form the foundation of Euclidean geometry. Some of the key axioms include: (i) A straight line segment can be drawn joining any two points. (ii) Any straight line segment can be extended indefinitely in a straight line. (iii) All right angles are equal to each other. (iv) A circle can be drawn with any center and any radius.

Q4: What is the difference between an axiom and a theorem in Euclid’s Geometry?

A4: An axiom is a statement accepted as true without proof, serving as a starting point for further reasoning and arguments. A theorem, on the other hand, is a statement that has been proven to be true based on axioms and previously established theorems.

Q5: How did Euclid contribute to the field of geometry?

A5: Euclid’s major contribution to geometry was the systematic organization of geometric knowledge into a logical and rigorous framework. His work, Elements , consists of 13 books that cover a wide range of topics, including plane geometry, number theory, and solid geometry. This work laid the foundation for the study of geometry for centuries.

Q6: Why is Euclid’s Geometry still important in modern mathematics?

A6: Euclid’s Geometry is important because it provides a clear and logical framework for understanding the properties of shapes and spaces. The principles laid out by Euclid are still used as the foundation for many areas of mathematics, including modern geometry, algebra, and calculus.

Q7: Are there any online resources or tools available for practicing Euclid’s Geometry case study questions?

A8: We provide case study questions for CBSE Class 9 Maths on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.

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NCERT Solutions for Class 9 Mathematics Chapter 1- Number Systems

Home » NCERT Solutions » NCERT Solutions for Class 9 Mathematics Chapter 1- Number Systems

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Mathematics is a subject which requires a lot of practice. . The more you practice the better you become. . Therefore, you must practice to perfection. There are plenty of  examples to practice with in Extramarks NCERT Solutions for Class 9 Mathematics Chapter 1.

The Chapter -Number Systems of Class 9 Mathematics covers all the fundamentals of Mathematics and will help students understand the core concepts covered in higher classes. . As Mathematics is totally based on  numbers, this  chapter tells about all the different types of numbers and various applications of numbers in Mathematics. If you are looking for a thorough knowledge of the concepts of the Chapter, Extramarks is the right platform to get the right amount of practice and to develop your mathematical abilities and be confident at an early age. 

You can avail of NCERT Solutions for Class 9 Mathematics Chapter 1 on the Extramarks website and turn your child into a smart learner. Number systems- Chapter 1 of Class 9 Mathematics comprises all the  fundamental concepts.. Based on the CBSE NCERT latest 2021-2022 syllabus, we have provided points to ponder as well as detailed solutions for the better understanding of the subject. It also encourages students to be curious and look for answers themselves.  Students are recommended  to use the NCERT solutions Class 9 Mathematics to realise their  true potential and to enjoy the entire process of learning and stay ahead of the competition.    

Visit the Extramarks website to keep yourself updated about the CBSE syllabus, NCERT Solutions and exam patterns. You may  also search for NCERT Solutions Class 10 to step up  your preparation and stay ahead of others.

Key Topics Covered In NCERT Solution for Class 9 Mathematics Chapter 1

Number system is entirely the study of numeracy, and hence the students must understand the concepts and enjoy the learning experience.   It will directly connect to  chapters like Quadratic equations, Sets etc in higher classes. As a result, students aiming for good grades should be able to identify different types of numbers, know their representation and identities and should know how to rationalise them efficiently.

In Extramarks NCERT Solutions for Class 9 Mathematics Chapter 1, students can expect all topics to be covered and explained in detail. The  chapter includes sections like real numbers and their decimal expansion, representing real numbers on the number line, operation on real numbers etc. For complete study material for NCERT Solutions Class 9, NCERT Solutions Class 10, NCERT Solutions Class 11, and NCERT Solutions Class 12, visit the Extramarks website and app which is trusted by students across India and  their  numbers have been growing by leaps and bounds because of the unshakable trust and faith these schools have in us. 

The key topics covered in NCERT Solutions of Class 9 Mathematics Chapter 1:

NCERT Solutions for Class 9 Mathematics Chapter 1 requires students to apply and correlate whatever they have learnt in their previous classes. . Students can also access NCERT Solutions for Mathematics Class 8 and Class 7  to review the  concepts studied last year or earlier.

1.1 Introduction

This Chapter on Number Systems begins with the basic introduction of numbers and their applications in our daily lives. Further, it categorises  numbers as Natural numbers, Whole numbers, Integers, Rational numbers and Irrational numbers. The various examples provided in the chapter help recognise different numbers, which can help easily recall the concepts in prior  Classes.

1.2 Irrational numbers

This section deals entirely with what makes a number irrational and how one can distinguish between rational and irrational numbers. Students have to keep in mind specific points while deciding it is an irrational number which they will find in our NCERT Solutions for Class 9 Mathematics Chapter 1. Students will also read about the set of numbers called real numbers.

At the end of this section, students will get a proper understanding of irrational as well as real numbers. Also, they will be available to locate certain square roots of numbers on the number line. 

1.3 Real numbers and their decimal expansion

In this section, first, you will learn about  decimal expansions of real numbers. Then you would evaluate whether you can distinguish between rational and irrational numbers based on the decimal expansion. You come across different cases and will illustrate them on the basis of examples.

1.4 Representing Real numbers on the number line

As learnt in the previous t section about the decimal expansion of real numbers, we will use it for application on the number line. The decimal expansion helps represent real numbers and get good practice with examples.  

After going through this section, you would be able to locate points of the number line with ease, learn to visualize points on the number line in a systematic way, learn to round off to the nearest decimal and know that a unique point represents every real number.

1.5 Operation on Real numbers

In the earlier Classes, we have learnt that rational numbers follow commutative, associative and distributive properties for mathematical operations, i.e. when you add, subtract, multiply or divide a rational number, you get a rational number. Likewise, this holds true for irrational numbers also. . 

The set of rational and irrational numbers is called real numbers. Hence, this applies to real numbers too. 

After completing this section, you will be able to carry out operations on non-terminating and non-recurring decimal expansions with the help of illustrative examples. Refer to our NCERT Solutions for Class 9 Mathematics Chapter 1 to get access to more solved questions based on Operations on Real Numbers. 

1.6 Law of exponents for real numbers

You are already acquainted with exponents and laws of exponents from your earlier Classes. In this section, we will specifically learn about the laws of exponents on real numbers. The application of laws of exponents remains the same in the case of real numbers. You have to learn to convert the square root or the cube root of the number into exponential form.

NCERT Solutions for Class 9 Mathematics Chapter 1 Exercise &  Solutions

Find NCERT Solutions for Class 9 Mathematics Chapter 1  on the Extramarks website. From a detailed analysis of the Chapter to short notes, you can find everything to level up your preparation and gear up your performance in the exams. You will get access to  all the questions on Number Systems once you access the NCERT Solutions for Class 9 Mathematics on our website.

Click on the below links to view exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 1:

•  Chapter 9: Exercise 1.1 Question and answers    
•  Chapter 9: Exercise 1.2 Question and answers
• Chapter 9: Exercise 1.3 Question and answers
• Chapter 9: Exercise 1.4 Question and answers
• Chapter 9: Exercise 1.5 Question and answers

Along with Class 9 Mathematics Solutions, students can explore NCERT Solutions on our Extramarks website for all primary and secondary  classes.

• NCERT Solutions Class 5 
• NCERT solutions Class 10
• NCERT solutions Class 11
• NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics 

NCERT Exemplar Class 9 Mathematics is an excellent resource  for students preparing for their 9th standard exams. The book consists of a variety of questions of different levels  of difficulty. It  encourages students to develop more interest in  Mathematics and get more significant insights into the  chapter to become proficient in facing challenging questions in the exams.

 NCERT Exemplar helps students to develop confidence during their preparation as they have questions of basic level  as well as advanced level. It has proved to be quite  beneficial for students, especially for  those preparing for various competitive exams. It covers  the entire chapters in detail  , which makes it fruitful for all curriculum students.

After referring to the NCERT Solutions and NCERT Exemplar, the students are confident   to solve all the complicated and tricky questions. As a result, students can easily switch  to more advanced and higher-level conceptual questions. By studying from the Exemplar, you can prepare well for entrance exams like Olympiad, NTSE and KVPY.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 1

In order to obtain a good score in exams, revision of previous concepts is a must. Hence, NCERT Solutions for Class 9 Mathematics Chapter 1 offers a complete solution for all problems. The key features of NCERT solutions are:  : 

• Mathematics experienced faculty and subject experts have designed Extramarks NCERT Solutions for Class 9 Mathematics Chapter 1. It is a thoroughly researched material made in sync with CBSE examination guidelines.
• Students have a very clear understanding of the concepts and overcome  all their doubts with the help of Extramarks NCERT solutions.
• After completing the NCERT Solutions for Class 9 Mathematics Chapter 1 students will be able to solve all the basic and advanced level problems with better  accuracy.The systemic and well-laid out balanced study plan boosts their performance naturally and effortlessly. 
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Yes, 0 is a rational number. It can be represented as (0/1), (0/2), (0/3) etc.

Find six rational numbers between 3 and 4.

There are infinite rational numbers between 3 and 4. 3 and 4 can be represented as 24/8 and 32/8 respectively. The rational numbers between 3 and 4 are 25/8, 26/8, 27/8, 28/8, 29/8, 30/8.

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

Q.4 State whether the following statements are true or false. Give reasons for your answers. (i) Every natural number is a whole number. (ii) Every integer is a whole number. (iii) Every rational number is a whole number.

(i) True; since the collection of whole numbers contains all natural numbers. (ii) False; since integers may be negative but whole numbers are positive. For example: – 5 is an integer it is not a whole number. (iii) False; as rational number may be a fraction but whole number may not be a fraction. For example: 4/5 is a rational number and it is not a whole number.

State whether the following statements are true or false. Justify your answers. (i) Every irrational number is a real number. (ii) Every point on the number line is of the form

where m is a natural number. (iii) Every real number is an irrational number.

(i) True; because real number is a collection of rational and irrational number. (ii) False; as negative numbers cannot be represented as the square root of any other number. (iii) False; as real numbers include both rational and irrational numbers i.e., irrational number is a part of real number. Therefore, every real number cannot be an irrational number.

Q.6 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: The square roots of 4 and 9 are 2 and 3 respectively.

how 5 can be represented on the number line .

Write the following in decimal form and say what kind of

decimal expansion each has: i   36 100    ii   1 11    iii  4 1 8    iv   3 13    v   2 11    vi   329 400

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?

Express the following in the form

p q , where p and q are integers and q¹0. i  0   ii  0.4 7 ¯    iii  0. 001 ¯

Express 0.99999, in the form

p q . Are you surprised by your answer? With your teacher and classmates discussway the answer makes sense.

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 .

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 ?

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

Three numbers whose decimal expansions are non- terminating non-recurring are as follows:

0.030030012003000050004123000…

0.01200012500003500050010008879000102003…

1.5200050040060080010030010040038001…

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

Classify the following numbers as rational or irrational:

i   23    ii   225   iii  0.3796  iv  7.478478 . ..    v  1.101001000100001 . ..

Visualise 3.765 on the number line, using successive magnification.

3.765 can be visualised as in the following steps.

4 . 26 ¯ on the number line , upto 4 decimal places .

Classify the following numbers as rational or irrational:

i  2 − 5    ii   3+ 23 − 23    iii   2 7 7 7 iv   1 2   v  2π

each of the following expressions : ( i ) ( 3 + 3 ) ( 2 + 2 ) ( ii ) ( 3 + 3 ) ( 3 − 3 ) ( iii ) ( 5 + 2 ) 2 ( iv ) ( 5 − 2 ) ( 5 + 2 )

, 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. Remember that when you measure a length with a scale or any other device, you only get an approximate rational value. So, you may not realise that either c or d is irrational.

9 . 3 on the number line .

Mark a line segment AB = 9.3 on number line. Further, take BC of 1 unit. Draw a semi-circle on AC as diameter. Draw a perpendicular to line AC passing through point B. Let it intersect the semi circle at D. Taking B as centre and BD as radius, draw an arc intersecting number line at E. BE =

Rationalise

the denominators of the following : ( i ) 1 7 ( i i ) 1 7 − 6 ( i i i ) 1 5 + 2 ( i v ) 1 7 − 2

:   i 64 1 2       ii 32 1 5       iii 125 1 3

i 9 3 2    ii 32 2 5    iii 16 3 4     iv 125 − 1 3

:      ( i )   2 2 3 . 2 1 5       ( ii )   ( 1 3 3 ) 7       ( iii )   11 1 2 11 1 4         ( iv ) 7 1 2 . 8 1 2

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Faqs (frequently asked questions), 1. where should i search for the ncert solutions for class 9 mathematics chapter 1 online.

There are plenty of  online platforms that provide  study materials for Class 9 Mathematics. However students should rely on only those study solutions that are prepared by subject   experts and strictly follow the latest CBSE curriculum. 

Students can refer to Extramarks, one of the leading  e-learning platforms which  has made it possible for students to access NCERT Solutions for Class 9 Mathematics Chapter 1 as they are prepared by Mathematics subject matter experts with decades of experience. Along with Class 9th Solutions, one can find NCERT Solutions right from Class 1 to Class 12 on our website.  Extramarks has built its credibility and is trusted by students as well as private and government schools across India.

2. How to prepare for NCERT Class 9 Mathematics Chapter 1?

Students should start studying Class 9 Mathematics from NCERT textbook first. They should be attentive in their class lectures. Along with the NCERT textbook, students should solve questions from NCERT Exemplars to build a strong foundation. 

We highly recommend students  also register on reliable online learning platforms such as Extramarks which strictly follows NCERT books and provides solved exercises and practice questions to step up their learning experience and eliminate “mathematics phobia” among students. The additional support of online learning and classes  will  allow  students to clear  their doubts and strengthen their base. To get good grades in exams students  must refer to multiple study resources, practise a lot of questions  and  stick to a study schedule and follow it rigorously to come out with flying colours. 

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• ICSE Class 9
• Class 9 Selina Solutions
• Class 9 Maths Concise Selina Solutions
• Chapter 1 Rational And Irrational Numbers

Concise Selina Solutions for Class 9 Maths Chapter 1- Rational And Irrational Numbers

Selina solutions for Class 9 Maths Chapter 1 Rational And Irrational Numbers are provided here. Class 9 is an important phase of a student’s life, the concepts which are taught in Class 9 are vital to be understood as these concepts are continued in Class 10. To score good marks in Class 9 mathematics examination, it is advised to solve questions provided in each exercise of all the chapters in the Selina book. These Selina solutions for Class 9 Maths help the students in understanding all the concepts in a better way. Download pdf of Class 9 Maths Chapter 1 Selina Solutions from the given links.

Download PDF of Selina Solutions for Class 9 Maths Chapter 1:- Download Here

Exercise 1(A) PAGE: 4

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

Yes, zero is a rational number.

It can be written in the form of 𝑝/q, where p and q are integers and q ≠ 0 ⇒ 0 = 0/1.

2. Are the following statements true or false? Give reasons for your answers.

(i) Every whole number is a natural number.

(ii) Every whole number is a rational number.

(iii) Every integer is a rational number.

(iv) Every rational 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.

i.e., Whole numbers = 0, 1, 2, 3…

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 …

∴ Every whole number is a rational number; however, every rational number is not 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…}

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

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

i.e., Whole numbers = 0, 1, 2, 3, …

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

∴ Every whole numbers are rational, however, every rational numbers are not whole numbers.

3. Arrange -5/9, 7/12, -2/3 and 11/18 in the ascending order of their magnitudes. Also, find the difference between the largest and the smallest of these rational numbers. Express this difference as a decimal fraction correct to one decimal place.

The given numbers are: -5/9, 7/12, -2/3 and 11/18

Now, the L.C.M of 9, 12 and 18 is 36

So, the given numbers are:

-5/9, 7/12, -2/3 and 11/18

= -5×4/9×4, 7×3/12×3, -2×12/3×12 and 11×2/18×2

= -20/36, 21/36, -24/36 and 22/36

Numbers in ascending order are:

-24/36, -20/36, 21/36 and 22/36

Hence, given numbers in ascending order are

-2/3, -5/9, 7/12 and 11/18

Now, to find the difference between the largest and smallest of the above number

Difference = 11/18 – (-2/3)

= 11/18 + 2/3

= 11/18 + (2×6)/(3×6)

= 11/18 + 12/18

= (11 + 12)/18

Now, to express this fraction as a decimal by correcting to one decimal place

Hence, 23/18 = 1.27777777… ≈ 1.3

4. Arrange 5/8, -3/16, -1/4 and 17/32 in the descending order of their magnitudes. Also, find the sum of the lowest and the largest of these rational numbers. Express the result obtained as a decimal fraction correct to two decimal places.

Given numbers are: 5/8, -3/16, -1/4 and 17/32

The L.C.M of 8, 16, 4 and 32 is 32

5/8, -3/16, -1/4 and 17/32

= 5×4/8×4, -3×2/16×2, -1×8/4×8 and 17×1/32×1

= 20/32, -6/32, -8/32 and 17/32

Numbers in descending order are:

20/32, 17/32, -6/32, -8/32

Hence, given numbers in descending order are

5/8, 17/32, -3/16 and -1/4

Now, to find the sum of the largest and the smallest of the above numbers

Sum = 5/8 + (-1/4)

= 5/8 – 1/4

= 5/8 – (1×2)/(4×2)

= 5/8 – 2/8

= (5 – 2)/8

Now, to express this fraction as a decimal by correcting to two decimal place

Hence, 3/8 = 0.375 ≈ 0.38

5. Without doing any actual division, find which of the following rational numbers have terminating decimal representation:

(ii) 23/125

(vii) 61/75

(viii) 123/250

(i) Given number is 7/16

16 = 2 x 2 x 2 x 2 = 2 4 = 2 4 x 5 0

So, 16 can be expressed as 2 m x 5 n

Hence, 7/16 is convertible into the terminating decimal

(ii) Given number is 23/125

125 = 5 x 5 x 5 = 5 3 = 2 0 x 5 3

So, 125 can be expressed as 2 m x 5 n

Hence, 23/125 is convertible into the terminating decimal

(iii) Given number is 9/14

14 = 2 x 7 = 2 1 x 7 1

So, 14 cannot be expressed as 2 m x 5 n

Hence, 9/14 is not convertible into the terminating decimal

(iv) Given number is 32/45

45 = 3 x 3 x 5 = 3 2 x 5 1

So, 45 cannot be expressed as 2 m x 5 n

Hence, 32/45 is not convertible into the terminating decimal

(v) Given number is 43/50

50 = 2 x 5 x 5 = 2 1 x 5 2

So, 50 can be expressed as 2 m x 5 n

Hence, 43/50 is convertible into the terminating decimal

(vi) Given number is 17/40

40 = 2 x 2 x 2 x 5 = 2 3 x 5 1

So, 40 can be expressed as 2 m x 5 n

Hence, 17/40 is convertible into the terminating decimal

(vii) Given number is 61/75

75 = 3 x 5 x 5 = 3 1 x 5 2

So, 75 cannot be expressed as 2 m x 5 n

Hence, 61/75 is not convertible into the terminating decimal

(viii) Given number is 123/250

250 = 2 x 5 x 5 x 5 = 2 1 x 5 3

So, 250 can be expressed as 2 m x 5 n

Hence, 123/250 is convertible into the terminating decimal

Exercise 1(B) PAGE: 13

1. State whether the following numbers are rational or not:

(i) (𝟐 + √𝟐) 𝟐

(ii) (𝟑 − √𝟑) 𝟐

(iii) (𝟓 + √𝟓)(𝟓 − √𝟓)

(iv) (√𝟑 − √𝟐) 𝟐

(v) (3/2√𝟐) 2

(vi) (√7/6√𝟐) 2

(i) (2 + √2) 2 = 2 2 + 2(2)(√2) + (√2) 2

= 4 + 4 √ 2 + 2

= 6 + 4 √ 2

Therefore, it is irrational

(ii) (3 – √ 3) 2 = (3) 2 – 2(3)( √ 3) + ( √ 3) 2

= 9 – 6 √ 3 + 3

= 12 – 6 √ 3

= 6(2 – √ 3)

Therefore, it is irrational.

(iii) (5 + √ 5)(5 – √ 5) = (5) 2 – ( √ 5) 2

Therefore, it is rational.

(iv) ( √ 3 – √ 2) 2 = ( √ 3) 2 – 2( √ 3)( √ 2) + ( √ 2) 2

= 3 – 2 √ 6 + 2

= 5 – 2 √ 6

(v) (3/2 √ 2) 2 = 3 2 /(2 √ 2) 2

= 9/(4 x 2)

(vi) ( √ 7/6 √ 2) 2 = ( √ 7) 2 /(6 √ 2) 2

= 7/(36 x 2)

2. Find the square of:

(ii) √3 + √2

(iii) √5 – 2

(iv) 3 + 2√5

(i) (3√2/5) 2 = (3√2) 2 /5 2

= (9 x 2)/25

On further implication, we get

(ii) (√3 + √2) 2 = (√3) 2 + (√2) 2 + 2(√3)(√2)

= 3 + 2 + 2√6

(iii) (√5 – 2) 2 = (√5) 2 + (2) 2 – 2(√5)(2)

= 5 + 4 – 4√5

= 9 – 4√5

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

= 9 + 12√5 + (4×5)

= 9 + 20 + 12√5

= 29 + 12√5

3. State, in each case, whether true or false:

(i) √𝟐 + √𝟑 = √𝟓

(ii) 2√4 + 2 = 6

(iii) 𝟑√𝟕 − 𝟐√𝟕 = √𝟕

(iv) 𝟐/7 is an irrational number.

(v) 𝟓/11 is a rational number.

(vi) All rational numbers are real numbers.

(vii) All real numbers are rational numbers.

(viii) Some real numbers are rational numbers.

(vii) False

(viii) True

4. Given universal set is {-6, -5¾, -√4, -3/5, -3/8, 0, 4/5, 1, 1⅔, √8, 3.01, π, 8.47}

From the given set, find:

(i) Set of Rational numbers

(ii) Set of irrational numbers

(iii) Set of integers

(iv) Set of non-negative integers

(i) First find the set of rational numbers

Rational numbers are numbers of the form p/q, where q ≠ 0

U = {-6, -5¾, -√4, -3/5, -3/8, 0, 4/5, 1, 1⅔, √8, 3.01, π, 8.47}

Here, -5¾, -3/5, -3/8, 4/5 and 1⅔ are of the from p/q

Therefore, they are rational numbers

The set of integers is a subset of rational numbers, -6, 0 and 1 are also rational numbers

Here, decimal numbers 3.01 and 8.47 are also rational numbers as they are terminating decimals

Also, -√4 = -2 as square root of 4 is 2

Thus, -2 belongs to the set of integers

From the above set, the set of rational numbers is Q,

Q = {-6, -5¾, -√4, -3/5, -3/8, 0, 4/5, 1, 1⅔, 3.01, 8.47}

(ii) First find the set of irrational numbers

Irrational numbers are numbers which are not rational

From the above subpart, we know that the set of rational numbers is Q,

Here the set of irrational numbers is the set of complement of the rational numbers over real numbers

The set of irrational numbers is U – Q = {√8, π}

(iii) First find the set of integers

Set of integers consists of zero, the natural numbers and their additive inverses

Set of integers is Z

Z = {…, -3, -2, -1, 0, 1, 2, 3, …}

Here, the set of integers is U ⋂ Z = {-6, -√4, 0, 1}

(iv) First find the set of non-negative integers

Set of non-negative integers consists of zero and the natural numbers

Set of non-negative integers is Z + and

Z + = {0, 1, 2, 3, …}

Set of integers is U ⋂ Z + = {0, 1}

5. Use method of contradiction to show that √𝟑 and √𝟓 are irrational.

Consider √3 and √5 as rational numbers

√3 = a/b and √5 = x/y (where a, b, x, y ∈ Z and b, y ≠ 0)

By squaring on both sides, we have

3 = a 2 /b 2 , 5 = x 2 /y 2

3b 2 = a 2 , 5y 2 = x 2 …. (a)

a 2 and x 2 are odd as 3b 2 and 5y 2 are odd.

a and x are odd …. (1)

Take a = 3c, x = 5z

By squaring on both sides

a 2 = 9c 2 , x 2 = 25z 2

Using equation (a)

3b 2 = 9c 2 , 5y 2 = 25z 2

By further simplification

b 2 = 3c 2 , y 2 = 5z 2

B 2 and y 2  are odd as 3c 2  and 5z 2  are odd.

b and y are odd …… (2)

Using equation (1) and (2) we know that a, b, x, y are odd integers.

a, b and x, y have common factors 3 and 5 which contradicts our assumption that a/b and x/y are rational

a, b and x, y do not have any common factors

a/b and x/y is not rational

√3 and √5 are irrational.

6. Prove that each of the following numbers is irrational:

(i) √3 + √2

(ii) 3 − √2

Consider √3 + √2 be a rational number.

√3 + √2 = x

(√3 + √2) 2 = x 2

(√3) 2 + (√2) 2 + 2(√3)(√2) = x 2

3 + 2 + 2√6 = x 2

5 + 2√6 = x 2

2√6 = x 2 – 5

√6 = (x 2 – 5)/2

x is a rational number.

x 2 is a rational number.

x 2  – 5 is a rational number.

(x 2 – 5)/2 is also a rational number.

Considering the equation, (x 2 – 5)/2 = √6

√6 is an irrational number

But, (x 2 – 5)/2 is a rational number

So, x 2 – 5 has to be an irrational number.

Then, x 2 should also be an irrational number.

Also, x must be an irrational number.

We assumed that x is a rational number

So, we arrive at a contradiction.

Hence, our assumption that √3 + √2 is a rational number is wrong.

Therefore, √3 + √2 is an irrational number.

Consider 3 − √2 as a rational number.

By squaring on both sides, we get

(3 – √2) 2 = x 2

(3) 2 + (√2) 2 – 2(3)(√2) = x 2

9 + 2 – 6√2 = x 2

11 – 6√2 = x 2

6√2 = 11 – x 2

√2 = (11 – x 2 )/6

x 2  is a rational number.

11 – x 2 is a rational number.

(11 – x 2 )/6 is also a rational number.

Considering the equation, √2 = (11 – x 2 )/6

√2 is an irrational number

But, (11 – x 2 )/2 is a rational number

So, 11 – x 2 has to be an irrational number.

Hence, our assumption that 3 − √2 is a rational number is wrong.

Therefore, 3 − √2 is an irrational number.

Consider √5 – 2 as a rational number.

(√5 – 2) 2 = x 2

(√5) 2 + (2) 2 – 2(√5)(2) = x 2

5 + 4 – 4√5 = x 2

9 – 4√5 = x 2

4√5 = 9 – x 2

√5 = (9 – x 2 )/4

9 – x 2 is a rational number.

(9 – x 2 )/4 is also a rational number.

Considering the equation, √5 = (9 – x 2 )/4

√5 is an irrational number

But, (9 – x 2 )/4 is a rational number

So, 9 – x 2 has to be an irrational number.

Hence, our assumption that √5 – 2 is a rational number is wrong.

Therefore, √5 – 2 is an irrational number.

7. Write a pair of irrational numbers whose sum is irrational.

√3 + 5 and √5 – 3 are irrational numbers whose sum is irrational.

Sum = (√3 + 5) + (√5 – 3)

= √3 + √5 + 2

Hence, the resultant is irrational.

8. Write a pair of irrational numbers whose sum is rational.

√3 + 5 and 4 – √3 are irrational numbers whose sum is rational.

Sum = (√3 + 5) + (4 – √3)

= √3 – √3 + 9

Hence, the resultant is rational.

9. Write a pair of irrational numbers whose difference is irrational.

√3 + 2 and √2 – 3 are irrational numbers whose sum is irrational.

Difference = (√3 + 2) – (√2 – 3)

= √3 – √2 + 2 + 3

= √3 – √2 + 5

10. Write a pair of irrational numbers whose difference is rational.

√5 – 3 and √5 + 3 are irrational numbers whose sum is irrational.

Difference = (√5 – 3) – (√5 + 3)

= √5 – √5 – 3 – 3

11. Write a pair of irrational numbers whose product is irrational.

Let us take two irrational numbers (5 + √2) and (√5 – 2)

Here the product = (5 + √2) × (√5 – 2)

By further calculation

= 5 √5 – 10 + √10 – 2√2 which is irrational.

12. Write a pair of irrational numbers whose product is rational.

Let us consider two irrational numbers (2√3 – 3 √2) and (2√3 + 3√2)

Here, the product = (2√3 – 3 √2) × (2√3 + 3√2)

By further calculation, we get

= (3√2) 2 – (2√3) 2

Therefore, the resultant is rational.

13. Write in ascending order:

(i) 3√5 = √(3 2 x 5) = √(9 x 5) = √45

4√3 = √(4 2 x 3) = √(16 x 3) √48

We know that, 45 < 48

So, √45 < √48

Therefore, 3√5 < 4√3

(ii) 2∛5 =∛(2 3 x 5) = ∛40

3∛2 = ∛(3 3 x 2) = ∛54

We know that, 40 < 54

So, ∛40 < ∛54

Therefore, 2∛5 < 3∛2

(iii) 6√5 = √(6 2 x 5) = √(36 x 5) = √180

7√3 = √(7 2 x 3) = √(49 x 3) = √147

8√2 = √(8 2 x 2) = √(128 x 2) = √128

We know that, 128 < 147 < 180

So, √128 < √147 < √180

Therefore, 8√2 < 7√3 < 6√5

14. Write in descending order:

(i) 2∜6 and 3∜2

(ii) 7√3 and 3√7

(i) It can be written as

2∜6 = ∜(2 4 x 6) = ∜96

3∜2 = ∜(3 4 x 2) = ∜162

Here, 162 > 96

So, ∜162 > ∜96

Therefore, 3∜2 > 2∜6

(ii) It can be written as

7√3 = √(7 2 x 3) = √(49 x 3) = √141

3√7 = √(3 2 x 7) = √(9 x 7) = √63

Here, 141 > 63

So, √141 > √63

Thus, 7√3 > 3√7

15. Compare:

(ii) √24 = (24) 1/2 and ∛35 = (35) 1/3

In order to make the powers ½ and 1/3 same,

We find L.C.M. of 2 and 3 i.e., 6

½ x 3/3 = 3/6 and 1/3 x 2/2 = 2/6

(24) 1/2 = (24) 3/6 = (24 3 ) 1/6 = (13824) 1/6

(35) 1/3 = (35) 2/6 = (35 2 ) 1/6 = (1225) 1/6

On comparing,

13824 > 1225

So, (13824) 1/6 > (1225) 1/6

√24 > ∛35

16. Insert two irrational numbers between 5 and 6.

Let’s write 5 and 6 as square root

Then, 5 = √25 and 6 = √36

Now, take the numbers

√25 < √26 < √27 < √28 < √29 < √30 < √31 < √32 < √33 < √34 < √35 < √36

Hence, any two irrational numbers between 5 and 6 is √29 and √30

17. Insert five irrational numbers between 2√5 and 3√3.

Here, 2√5 = √(2 2 x 5) = √(4 x 5) = √20 and

3√3 = √(3 2 x 3) = √(9 x 3) = √27

√20 < √21 < √22 < √23 < √24 < √25 < √26 < √27

Hence, any five irrational numbers between 2√5 and 3√3 are:

√21, √22, √23, √24 and √26

18. Write two rational numbers between √2 and√3.

Let us take any two rational numbers between 2 and 3 which are perfect squares

For example, let us consider 2.25 and 2.56

Now, we have

√2.25 = 1.5 and √2.56 = 1.6

√2 < √2.25 < √2.56 √3

√2 < 1.5 < 1.6 < √3

√2 < 15/10 < 16/10 < √3

√2 < 3/2 < 8/5 < √3

Hence, any two rational numbers between √2 and √3 are: 3/2 and 8/5

19. Write three rational numbers between √3 and √5.

Let us take any two rational numbers between 3 and 5 which are perfect squares

For example, let us consider 3.24, 3.61, 4, 4.41 and 4.84

√3.24 = 1.8, √3.61 = 1.9, √4 = 2, √4.41 = 2.1 and √4.84 = 2.2

√3 < √3.24 < √3.61 <√4 < √4.41 < √4.84 <√5

√3 < 1.8 < 1.9 < 2 < 2.1 < 2.2 < √5

√3 < 18/10 < 19/10 < 2 < 21/10 < 22/10 < √5

√3 < 9/5 < 19/10 < 2 < 21/10 < 11/5 < √5

Hence, any three rational numbers between √3 and √5 are: 9/5, 21/10 and 11/5

20. Simplify each of the following:

(i) It can be rewritten as 16 1/5 x 2 1/5

By further simplification, we have

= (2 4 ) 1/5 x 2 1/5

= 2 4/5 x 2 1/5

= 2 4/5 + 1/5

(ii) It can be rewritten as ∜3 5 /∜3

= (3) 1/4 x 5 /(3) 1/4

= 3 5/4 /3 1/4

= (3) 5/4 – ¼

(iii) (3 + √2) (4 + √7)

By further calculation,

= 3 × 4 + 3 × √7 + 4 × √2 + √2 × √7

= 12 + 3√7 + 4√2 + √14

(iv) (√3 – √2) 2

It can be written as

= (√3) 2 + (√2) 2 – 2 × √3 × √2

= 3 + 2 – 2 √6

= 5 – 2√6

Exercise 1(C) PAGE: 21

1. State, with reason, which of the following are surds and which are not:

(i) √180 = √(2 x 2 x 5 x 3 x 3) = 6√5

It is irrational

Therefore, √180 is a surd.

(ii) ∜27 = ∜(3 x 3 x 3)

Therefore, ∜27 is a surd

(iv) ∛64 = ∛(4 x 4 x 4) = 4

It is rational

Therefore, ∛64 is not a surd

(v) ∛25. ∛40 = ∛(25 x 40) = ∛(5 x 5 x 2 x 2 x 5 x 2) = 2 x 5 = 10

Therefore, ∛23. ∛40 is not a surd

(vi) ∛-125 = ∛(-5 x -5 x -5) = -5

Therefore, ∛-125 is not a surd

(vii) π is irrational.

Therefore, √π is not a surd.

(viii) 3 + √2 is irrational

2. Write the lowest rationalizing factor of:

(iii) √5 – 3

(iv) 7 – √7

(v) √18 – √50

(vi) √5 – √2

(vii) √13 + 3

(viii) 15 – 3√2

(ix) 3√2 + 2√3

5√2 × √2 = 5 × 2 = 10

It is rational.

Therefore, lowest rationalizing factor is √2.

√24 = √(2 x 2 x 2 x 3) = 2√6

Therefore, lowest rationalizing factor is √6.

(√5 – 3) (√5 + 3) = (√5) 2 – 3 2 = 5 – 9 = – 4

Therefore, lowest rationalizing factor is (√5 + 3).

(7 – √7) (7 + √7) = 49 – 7 = 42

Therefore, lowest rationalizing factor is (7 + √7).

√18 – √50 = √(2 x 3 x 3) – √(5 x 5 x 2)

= 3√2 – 5√2

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

Therefore, lowest rationalizing factor is √5 + √2.

(√13 + 3) (√13 – 3) = (√13) 2 – 3 2 = 13 – 9 = 4

Therefore, lowest rationalizing factor is √13 – 3.

15 – 3√2 = 3 (5 – √2)

= 3 (5 – √2) (5 + √2)

= 3 [5 2 – (√2) 2 ]

= 3 × [25 – 2]

Therefore, lowest rationalizing factor is (5 + √2).

3√2 + 2√3 = (3√2 + 2√3) (3√2 – 2√3)

= 9 × 2 – 4 × 3

Therefore, lowest rationalizing factor is 3√2 – 2√3.

3. Rationalize the denominators of:

(i) (3/√5) x (√5/√5) = 3√5/5

(ii) (2√3/√5) x (√5/√5) = 2√15/5

4. Find the values of ‘a’ and ‘b’ in each of the following:

5. Simplify:

(iv) x 2 + y 2 = xy

(iii) We know that

By substituting the values

= 161 – 72√5 + 161 + 72√5 + 1

7. If m = 1/(3 – 2√2) and n = 1/(3 + 2√2), find:

mn = (3 + √2)(3 – √2)

mn = 3 2 – (2√2) 2

8. If x = 2√3 + 2√2, find:

(ii) x + 1/x

(iii) (x + 1/x) 2

9. If x = 1 − √2, find the value of (x + 1/x) 3 .

It is given that

We should find the value of (x + 1/x) 3

So, x = 1 − √2, we get

Using the formula (a – b) (a + b) = a 2 – b 2

(x – 1/x) = (1 – √2) – (-(1 + √2))

= 1 – √2 + 1 + √2

By cubing on both sides, we get

(x – 1/x) 3 = 2 3

10. If x = 5 − 2√6, find: x 2 + 1/x 2

x = 5 − 2√6

We should find the value of (x 2 + 1/x 2 )

So, x = 5 − 2√6, we get

(x – 1/x) = (5 – 2√6) – (5 + 2√6)

= 5 – 2√6 – 5 – 2√6

= -4√6 … (2)

Consider (x – 1/x) 2

Using the equation (a – b) 2  = a 2 + b 2 – 2ab

(x – 1/x) 2 = x 2 + 1/x 2 – 2(x)(1/x)

(x – 1/x) 2 = x 2 + 1/x 2 – 2

(x – 1/x) 2 + 2 = x 2 + 1/x 2 … (3)

From equations (2) and (3), we get

x 2 + 1/x 2 = (-4√6) 2 + 2

11. Show that:

Using the formula a 2 – b 2 = (a + b) (a – b)

= 3 + √8 – √8 – √7 + √7 + √6 – √6 – √5 + √5 + 2

12. Rationalize the denominator of:

We know that,

Using the formula (a – b) 2 = a 2 + b 2 – 2ab

13. If √2 = 1.4 and √3 = 1.7, find the value of each of the following, correct to

one decimal place:

(i) 1/(√3 – √2)

(ii) 1/(3 + 2√2)

(iii) (2 – √3)/√3

= 1.7 + 1.4

= 3 – 2√2

= 3 – 2(1.4)

(3.4 – 3)/3 = 0.4/3

= 0.133333…

14. Evaluate:

(4 – √5)/(4 + √5) + (4 + √5)/(4 – √5)

Using the formula (a 2 – b 2 ) = (a + b) (a – b)

15. If (2 + √5)/(2 – √5) = x and (2 – √5)/(2 + √5) = y; find the value of x 2 – y 2 .

x 2 – y 2 = (-9 – 4√5) 2 – (-9 + 4√5) 2

Expanding using the formula, we get

= 81 + 72√5 + 80 – (81 – 72√5 + 80)

= 81 + 72√5 + 80 – 81 + 72√5 – 80

Exercise 1D PAGE: 22

1. Simplify:

2. Simplify:

Using the formula, a 2 – b 2 = (a + b) (a – b)

3. Evaluate, correct to one place of decimal. The expression 5/(√20 – √10), if √5 = 2.2 and √10 = 3.2.

= 5/(2√5 – √10)

= 5/[(2 x 2.2) – 3.2)]

= 5/(4.4 – 3.2)

[Note: In textual answer, the value of √20 has been directly taken, which is 4.5. Hence the answer 3.8!]

4. If x = √3 − √2. Find the value of:

(i) x + 1/x

(ii) x 2 + 1/x 2

(iii) x 3 + 1/x 3

(iv) x 3 + 1/x 3 – 3(x 2 + 1/x 2 ) + x + 1/x

(i) We have,

= (√3 – √2) + 1/(√3 – √2)

= 6√3 – 2√18 + 6√2 – 2√12

= 6√3 – 2√(9 x 2) + 6√2 – 2√(4 x 3)

= 6√3 – 2 x 3√2 + 6√2 – 2 x 2√3

= 6√3 – 6√2 + 6√2 – 4√3

= 6√3 – 4√3

= (√3 – √2) 2 + 1/(√3 – √2) 2

(iii) We have,

x 3 + 1/x 3

= (√3 – √2) 3 + 1/(√3 – √2) 3

We know that, (a – b) 3 = a 3 – b 3 – 3ab(a – b)

(√3 – √2) 3 = (√3) 3 – (√2) 3 – 3(√3)(√2)(√3 – √2)

= 3√3 – 2√2 – 3√6(√3 – √2)

= 3√3 – 2√2 – 3√18 + 3√12

= 3√3 – 2√2 – 3√(3 2 x 2) + 3√(2 2 x 3)

= 3√3 – 2√2 – 3 x 3√2 + 3 x 2√3

= 3√3 – 2√2 – 9√2 + 6√3

= 9√3 – 11√2

Now, (9√3 – 11√2) + 1/(9√3 – 11√2) = (9√3 – 11√2) + (9√3 + 11√2)

= 9√3 – 11√2 + 9√3 + 11√2

= 9√3 + 9√3

( iv) x 3 + 1/x 3 – 3(x 2 + 1/x 2 ) + x + 1/x

According to the results obtained in (i), (ii) and (iii), we get

x 3 + 1/x 3 – 3(x 2 + 1/x 2 ) + x + 1/x = 18√3 – 3(10) + 2√3

= 20√3 – 30

= 10(2√3 – 3)

5. Show that:

(i) Negative of an irrational number is irrational.

Let the irrational number be √2

Considering the negative of √2, we get -√2

We know that -√2 is an irrational number

Hence, negative of an irrational number is irrational

(ii) The product of a non-zero rational number and an irrational number is an irrational number.

Let the non-zero rational number be 3

Let the irrational number be √5

Then, according to the question

3 × √5 = 3√5 = 3 × 2.2 = 6.6, which is irrational

6. Draw a line segment of length √5 cm.

We know that, √5 = √(2 2 + 1 2 )

Which relates to: Hypotenuse = √[(side 1) 2 + (side 2) 2 ] … [Pythagoras theorem]

Hence, considering

Side 1 = 2 and Side 2 = 1,

We get a right-angled triangle such that:

∠𝐴 = 90°, AB = 2 cm and AC = 1 cm

7. Draw a line segment of length √𝟑 cm.

We know that, √3 = √(2 2 – 1 2 )

Which relates to: Hypotenuse= √[(side 1) 2 + (side 2) 2 ] … [Pythagoras theorem]

Hypotenuse 2 – Side 1 2 = Side 2 2

Hence, considering Hypotenuse = 2 cm and Side 1 = 1 cm,

We get a right-angled triangle OAB such that:

∠O = 90°, OB = 2 cm and AB = 1 cm

8. Draw a line segment of length √8 cm.

We know that, √8 = √(3 2 – 1 2 )

Which relates to: Hypotenuse = √[(side 1) 2 + (side 2) 2 ] … (Pythagoras theorem)

Hypotenuse 2 – (Side 1) 2 = (Side 2) 2

Hence, considering Hypotenuse = 3 cm and Side 1 =1 cm,

∠A = 90°, OB = 3 cm and AB=1 cm

9. Show that:

10. Show that:

(i) x 3 + 1/x 3 = 52, if x = 2 + √3

(ii) x 2 + 1/x 2 = 34, if x = 3 + 2√2

(i) We know that, (a + b) 3 = a 3 + b 3 + 3ab(a + b)

x 3 + 1/x 3 = (2 + √3) 3 + 1/(2 + √3) 3

Here, taking

(2 + √3) 3 = 2 3 + (√3) 3 + 3(2)(√3)( 2 + √3)

= 8 + 3√3 + 6√3(2 + √3)

= 8 + 3√3 + 12√3 + 6(√3) 2

= 8 + 3√3 + 12√3 + (6 x 3)

= 8 + 15√3 + 18

= 26 + 15√3

(ii) We know that, (a + b) 2 = a 2 + b 2 + 2ab

x 2 + 1/x 2 = (3 + 2√2) 2 + 1/(3 + 2√2) 2

= (9 + 8 + 2 x 3 x 2√2) + 1/(9 + 8 + 2 x 3 x 2√2)

= (17 + 12√2) + 1/(17 + 12√2)

– Hence, proved.

11. Show that x is rational if:

(i) x 2 = 6

(ii) x 2 = 0.009

(iii) x 2 = 27

𝑥 = √6 = 2.449 … which is irrational.

𝑥 = √0.009 = 0.0948 … which is irrational.

𝑥 = √27 = 5.1961 … which is irrational.

12. Show that x is rational if:

(i) x 2 = 16

(ii) x 2 = 0.0004

𝑥 = √16 = 4, which is rational.

𝑥 = √0.0004 = 0.02, which is rational.

13. Using the following figure, show that BD = √x.

Let’s assume AB = x, BC = 1 and AC = x + 1

Here, AC is diameter and O is the centre

OA = OC = OD = radius = (x + 1)/2

OB = OC – BC

= (x + 1)/2 – 1

= (x + 1 – 2)/2

= (x – 1)/2

Now, using Pythagoras theorem, we have

OD 2 = OB 2 + BD 2

Selina Solutions for Class 9 Maths Chapter 1- Rational And Irrational Numbers

Chapter 1, Rational And Irrational Numbers, contains 4 exercises and the Selina Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

1.1 Introduction

1.2 Rational Numbers

1.3 Properties of Rational Numbers

1.4 Decimal Representation of Rational Numbers

1.5 Irrational Numbers

1.6 Real Numbers

1.7 Surds (Radicals)

1.8 Rationalization

1.9 Simplifying an Expression by Rationalizing its Denominator

Numbers are the basics of mathematics. In lower classes the students would have learned the different types of numbers including natural numbers, whole numbers, integers, etc. Chapter 1 of Class 9 takes the students to the different sets of numbers, the Rational And Irrational Numbers. Read and learn the Chapter 1 of Selina textbook to learn more about Rational And Irrational Numbers along with the concepts covered in it. Solve the Selina Solutions for Class 9 effectively to score high in the examination.

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CBSE Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula PDF Download

CBSE Case Study Questions Class 9 Maths Chapter 12  are very important to solve for your exam. Class 9 Maths Chapter 12 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 12  Heron’s Formula

Case Study Questions Class 9 Maths Chapter 12

Case Study 1: A group of students is learning about Heron’s Formula for finding the area of a triangle. They encountered the following scenario:

Rohan and Kavya came across a triangular field in their village. They made the following observations:

• The lengths of the three sides of the triangular field are 8 meters, 12 meters, and 15 meters.
• The perimeter of the triangular field is 35 meters.

Based on this information, the students were asked to apply Heron’s Formula to find the area of the triangular field. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The semiperimeter of the triangular field is: (a) 8 meters (b) 12 meters (c) 15 meters (d) 17.5 meters

Answer: (d) 17.5 meters

Q2. Using Heron’s Formula, the area of the triangular field is: (a) 24 square meters (b) 30 square meters (c) 36 square meters (d) 40 square meters

Answer: (b) 30 square meters

Q3. The type of triangle formed by the sides of the field is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Answer: (c) Scalene

Q4. The length of the altitude corresponding to the side of 15 meters is: (a) 2 meters (b) 4 meters (c) 6 meters (d) 8 meters

Answer: (c) 6 meters

Q5. The lengths of the altitudes corresponding to the sides of 8 meters and 12 meters are: (a) 4 meters and 6 meters (b) 6 meters and 8 meters (c) 8 meters and 10 meters (d) 10 meters and 12 meters

Answer: (a) 4 meters and 6 meters

Case Study 2: A group of students is studying Heron’s Formula for finding the area of a triangle. They encountered the following scenario:

Neha and Mohan went on a field trip to a riverbank. They noticed a triangular piece of land that they wanted to measure and calculate its area. They made the following observations:

• Neha measured the lengths of the three sides of the triangular piece of land as 7 meters, 9 meters, and 11 meters.
• Mohan measured the lengths of the three sides of the same triangular piece of land as 10 meters, 12 meters, and 15 meters.

Based on this information, the students were asked to apply Heron’s Formula to find the area of the triangular piece of land. Let’s see if you can answer the questions correctly:

Q1. Using Neha’s measurements, the semiperimeter of the triangular piece of land is: (a) 13 meters (b) 16 meters (c) 19 meters (d) 23 meters

Answer: (c) 19 meters

Q2. Using Neha’s measurements, the area of the triangular piece of land is: (a) 24 square meters (b) 26 square meters (c) 28 square meters (d) 30 square meters

Answer: (a) 24 square meters

Q3. Using Mohan’s measurements, the semiperimeter of the triangular piece of land is: (a) 16 meters (b) 18 meters (c) 21 meters (d) 25 meters

Answer: (c) 21 meters

Q4. Using Mohan’s measurements, the area of the triangular piece of land is: (a) 40 square meters (b) 42 square meters (c) 45 square meters (d) 48 square meters

Answer: (b) 42 square meters

Q5. The measurements taken by Neha represent a triangle that is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Hope the information shed above regarding Case Study and Passage Based Questions for Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula 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 12 Heron’s Formula and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible.

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CBSE Class 9 Examination

The Central Board of Secondary Education (CBSE) conducts the Class 9 examination annually. Class 9 is a crucial stage in the student’s life where a strong foundation of concepts is laid down for the higher classes. CBSE is a well-renowned board that imparts quality education to Indian students. It aims at a well-rounded, strong, and vibrant school education system that fosters excellence in all aspects of a student's life.

Students should be aware of preparation tips and syllabi when preparing for CBSE Class 9. Keep reading for more information on the CBSE 9th exam details.

CBSE Class 9 Subjects

CBSE prepares a syllabus for each of the subjects to make a strong base for the student in order to grab the exact concept that is very much needed in the higher classes. There are mainly core subjects in the class 9 syllabus which is listed below

Other elective subjects that depend on the school as extra options include Information Technology, Physical Education, or a foreign language. Each subject’s curriculum is designed to enhance specific skills and make smooth transitions into class 10.

Latest CBSE Class 9 Syllabus

The curriculum of the CBSE Class 9 is structured in such a manner that will lay the basic fundamentals for Class 10. The CBSE Board follows the NCERT curriculum, which is mandatory for all affiliated schools.

The New Education Policy has also integrated coding and computational thinking for students from class 9 to build technical knowledge among students as part of 21st-century skills.

For efficiency, students should pick those subjects to take in Class 9 that they intend to take in Class 10. 

CBSE Class 9 Exam Pattern

The CBSE Class 9 exam pattern outlines the format in which the question papers are designed, including the number of questions, marking scheme, and exam duration. Understanding this pattern is crucial for systematic preparation.

Note - Internal Assessment is a Total Mark of 20 for each subject which is divided into different categories based on the school choices:

• Project Work : Assignments and projects related to different subjects.
• Class Participation : Activeness in classroom activities and discussions.
• Oral Tests : Oral tests to assess reading, speaking, and listening skills.
• Notebook Submission : Regularity and neatness in notebook submissions.

Recommended CBSE Class 9 Books

For effective CBSE Class 9 exam preparation, learning from NCERT textbooks provides a solid foundation of all basic concepts, so students should read the NCERT thoroughly line by line in order to achieve good marks in the examination.

Additional Reference Books for Practice: Expand your knowledge and sharpen your skills with these additional reference books, which Topper’s Students have widely used for comprehensive practice.

• Mathematics : RS Aggarwal, RD Sharma
• Science : Lakhmir Singh & Manjit Kaur
• Social Science : Arihant, Oswal

CBSE Class 9 Preparation and Strategy

Below are some tips that will ease the preparation of any student who is going to appear in the Class 9 exam, which in turn will give them more confidence to appear for the final exams. 

• Plan Smart: Draw out a balanced study timetable and focus on strong, as well as weak subjects.
• Distraction-Free Zone : Move your study to a no-noise area.
• Visual Aids : Maintain notes, flowcharts, and diagrams to help in simplification of concepts.
• Practice Papers: Solve past year’s papers to realize the pattern of questions that come in the exam and how to manage time.
• Mock Test: Take mock tests regularly for self-evaluation.
• Early Revision: Complete the syllabus a month prior to the examination so that you revise everything.
• Take breaks: Relax your mind with small breaks in between study sessions.
• 1.0 Exam in which Students can appear with class 9

After students have completed their Class 9 exam, they will be promoted to Class 10. However, there are other competitive exams too that students can appear for. These exams test a student’s intellectual, critical thinking, conceptual, and problem-solving skills. Below is the list of exams that students can take after their CBSE Class 9:

• International Science Olympiad (ISO)
• International Computer Olympiad (ICO)
• International Drawing Olympiad (IDO)
• Silverzone International Social Studies Olympiad (iSSO)
• International Maths Olympiad (IMO)
• English International Olympiad (EIO)
• National Essay Olympiad (NESO)
• General Knowledge International Olympiad (GKIO)

Table of Contents

• 0.1 CBSE Class 9 Subjects
• 0.2 Latest CBSE Class 9 Syllabus
• 0.3 CBSE Class 9 Exam Pattern
• 0.4 Recommended CBSE Class 9 Books
• 0.5 CBSE Class 9 Preparation and Strategy

Yes, NCERT books are the best for learning and practice because CBSE sets examination questions based on these books.

Ideally, it is better to study 4-5 hours daily to score good marks.

To score well in the examination, students should focus on clearing their basic concepts. Using the ALLEN CBSE Class 9 study material can help in building a strong foundation. Additionally, it's important to allocate regular time for revision to reinforce learning and ensure better retention of key concepts.

Class 9 Maths Case Study Questions of Chapter 2 Polynomials PDF Download

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Case study Questions in Class 9 Mathematics Chapter 2  are very important to solve for your exam. Class 9 Maths Chapter 2 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 2 Polynomials

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

These case study questions challenge students to apply their knowledge of polynomials in real-life scenarios, enhancing their problem-solving abilities. This article provides for the Class 9 Maths Case Study Questions of Chapter 2: Polynomials , enabling students to practice and excel in their examinations.

Polynomials Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 2 Polynomials

Case Study/Passage Based Questions

Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p(x) = 4x 2 + 12x + 5, which is the product of their individual shares.

Coefficient of x 2 in the given polynomial is (a) 2 (b) 3 (c) 4 (d) 12

Answer: (c) 4

Total amount invested by both, if x = 1000 is (a) 301506 (b)370561 (c) 4012005 (d)490621

Answer: (c) 4012005

The shares of Ankur and Ranjan invested individually are (a) (2x + 1),(2x + 5)(b) (2x + 3),(x + 1) (c) (x + 1),(x + 3) (d) None of these

Answer: (a) (2x + 1),(2x + 5)

Name the polynomial of amounts invested by each partner. (a) Cubic (b) Quadratic (c) Linear (d) None of these

Answer: (c) Linear

Find the value of x, if the total amount invested is equal to 0. (a) –1/2 (b) –5/2 (c) Both (a) and (b) (d) None of these

Answer: (c) Both (a) and (b)

One day, the principal of a particular school visited the classroom. The class teacher was teaching the concept of a polynomial to students. He was very much impressed by her way of teaching. To check, whether the students also understand the concept taught by her or not, he asked various questions to students. Some of them are given below. Answer them

Which one of the following is not a polynomial? (a) 4x 2 + 2x – 1 (b) y+3/y (c) x 3 – 1 (d) y 2 + 5y + 1

Answer: (b) y+3/y

The polynomial of the type ax 2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial

Answer: (a) Linear polynomial

The value of k, if (x – 1) is a factor of 4x 3 + 3x 2 – 4x + k, is (a) 1 (b) –2 (c) –3 (d) 3

Answer: (c) –3

If x + 2 is the factor of x 3 – 2ax 2 + 16, then value of a is (a) –7 (b) 1 (c) –1 (d) 7

Answer: (b) 1

The number of zeroes of the polynomial x 2 + 4x + 2 is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (b) 2

Case Study/Passage-Based Questions

Case Study 3. Amit and Rahul are friends who love collecting stamps. They decide to start a stamp collection club and contribute funds to purchase new stamps. They both invest a certain amount of money in the club. Let’s represent Amit’s investment by the polynomial A(x) = 3x^2 + 2x + 1 and Rahul’s investment by the polynomial R(x) = 2x^2 – 5x + 3. The sum of their investments is represented by the polynomial S(x), which is the sum of A(x) and R(x).

Q1. What is the coefficient of x^2 in Amit’s investment polynomial A(x)? (a) 3 (b) 2 (c) 1 (d) 0

Answer: (a) 3

Q2. What is the constant term in Rahul’s investment polynomial R(x)? (a) 2 (b) -5 (c) 3 (d) 0

Answer: (c) 6

Q3. What is the degree of the polynomial S(x), representing the sum of their investments? (a) 4 (b) 3 (c) 2 (d) 1

Answer: (c) 2

Q4. What is the coefficient of x in the polynomial S(x)? (a) 7 (b) -3 (c) 0 (d) 5

Answer: (b) -3

Q5. What is the sum of their investments, represented by the polynomial S(x)? (a) 5x^2 + 7x + 4 (b) 5x^2 – 3x + 4 (c) 5x^2 – 3x + 5 (d) 5x^2 + 7x + 5

Answer: (b) 5x^2 – 3x + 4

Case Study 4. A school is organizing a fundraising event to support a local charity. The students are divided into three groups: Group A, Group B, and Group C. Each group is responsible for collecting donations from different areas of the town.

Group A consists of 30 students and each student is expected to collect ‘x’ amount of money. The polynomial representing the total amount collected by Group A is given as A(x) = 2x^2 + 5x + 10.

Group B consists of 20 students and each student is expected to collect ‘y’ amount of money. The polynomial representing the total amount collected by Group B is given as B(y) = 3y^2 – 4y + 7.

Group C consists of 40 students and each student is expected to collect ‘z’ amount of money. The polynomial representing the total amount collected by Group C is given as C(z) = 4z^2 + 3z – 2.

Q1. What is the coefficient of x in the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 0

Answer: (b) 5

Q2. What is the degree of the polynomial B(y)? (a) 2 (b) 3 (c) 4 (d) 1

Answer: (b) 3

Q3. What is the constant term in the polynomial C(z)? (a) 4 (b) 3 (c) -2 (d) 0

Answer: (c) -2

Q4. What is the sum of the coefficients of the polynomial A(x)? (a) 2 (b) 5 (c) 10 (d) 17

Answer: (c) 10

Q5. What is the total number of students in all three groups combined? (a) 30 (b) 20 (c) 40 (d) 90

Answer: (c) 40

The Class 9 Maths Case Study Questions of Chapter 2: Polynomials serve as a valuable resource for students seeking to enhance their understanding of polynomial concepts and problem-solving skills. By practicing these case studies, students can strengthen their grasp of polynomials and their applications in real-life scenarios. Embrace the opportunity to engage with practical problems and excel in your mathematical journey.

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The number of zeroes of the polynomial x2 + 4x + 2 is The answer is too easy, i.e. 2

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Exercise 1.5 (Revised) - Chapter 1 - Number Systems - Ncert Solutions class 9 - Maths

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Chapter 1 - Number Systems | NCERT Solutions Class 9 Maths

Ex 1.5 Question 1.

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

(i) $64^{\frac{1}{2}}$ We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, 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 know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$.

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

Ex 1.5 Question 2.

Find:(i) $9^{\frac{3}{2}}$ (ii) $32^{\frac{2}{5}}$ (iii) $16^{\frac{3}{4}}$ (iv) $125^{\frac{-1}{3}}$

(i) $9^{\frac{3}{2}}$

We know that $a^{\frac{1}{n}}=\sqrt[n]{a}$, where $a>0$. $ \begin{aligned} & \sqrt[2]{(9)^3}=\sqrt[2]{3 \times 3 \times 3 \times 3 \times 3 \times 3} \\ & =3 \times 3 \times 3 \\ & =27 \end{aligned} $

Therefore, the value of $9^{\frac{3}{2}}$ will be 27 . (ii) $32^{\frac{2}{5}}$

We conclude that $32^{\frac{2}{5}}$ can also be written as $\sqrt[5]{(32)^2}=\sqrt[5]{(2 \times 2 \times 2 \times 2 \times 2)(2 \times 2 \times 2 \times 2 \times 2)}=2 \times 2$ $=4$

Therefore, the value of $32^{\frac{2}{5}}$ will be 4 . (iii) $16^{\frac{3}{4}}$

We conclude that $16^{\frac{3}{4}}$ can also be written as $\sqrt[4]{(16)^3}=\sqrt[4]{(2 \times 2 \times 2 \times 2)(2 \times 2 \times 2 \times 2)(2 \times 2 \times 2 \times 2)}$ $=2 \times 2 \times 2$ $ =8 $

Therefore, the value of $16^{\frac{3}{4}}$ will be 8 . (iv) $125^{\frac{-1}{3}}$

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

We conclude that $125^{\frac{-1}{3}}$ can also be written as $125^{\frac{1}{3}}$, or $\left(\frac{1}{125}\right)^{\frac{1}{3}}$.

We know that $\left(\frac{1}{125}\right)^{\frac{1}{3}}$ can also be written as $\sqrt[3]{\left(\frac{1}{125}\right)}=\sqrt[3]{\left(\frac{1}{5} \times \frac{1}{5} \times \frac{1}{5}\right)}$

$ =\frac{1}{5} \text {. } $

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

Ex 1.5 Question 3.

Simplify: (i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$ (ii) $\left(3^{\frac{1}{3}}\right)^7$ (iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$ (iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$

(i) $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}$

We know that $a^m \cdot a^n=a^{(m+n)}$.

We can conclude that $2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{2}{3}+\frac{1}{5}}$. $ 2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}=(2)^{\frac{10+3}{15}}=(2)^{\frac{13}{15}} $ (ii) $\left(3^{\frac{1}{3}}\right)^7$

We know that $a^m \times a^n=a^{m+n}$

We conclude that $\left(3^{\frac{1}{3}}\right)^7$ can also be written as $\left(3^{\frac{7}{3}}\right)$. (iii) $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

We know that $\frac{a^m}{a^n}=a^{m-n}$

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

Therefore, the value of $11^{\frac{1}{4}}$ will be $11^{\frac{1}{2}}$. (iv) $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}$

We know that $a^m \cdot b^m=(a \times b)^m$.

We can conclude that $7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(7 \times 8)^{\frac{1}{2}}$. $ 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}=(7 \times 8)^{\frac{1}{2}}=(56)^{\frac{1}{2}} $

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

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