case study questions number system

Class 8 Maths Chapter 1 Case Based Questions - Number System

Correct Answer is Option (a) Number system: A numeral system (or system of numeration) is a writing system for expressing numbers that is, a mathematical notation for representing numbers of a given set, using digits or other symbols in a consistent manner. The number the numeral represents is called its value.

Correct Answer is Option (a) a p + q a p  . a q  = a p + q

Correct Answer is Option (b) r – s is rational number.

Reason  : This statement is false. The difference between a rational number (r) and an irrational number (s) can be either rational or irrational. There is no general rule that says the difference of a rational and an irrational number must be rational. Counterexamples can easily be constructed to show that the difference could be either rational or irrational.

Correct Answer is Option (a)

Q7:  In a school 5 out of every 7 children participated in ‘Save wild life’ campaign organised by the school authorities. How many rational numbers are there between 5 and 7.

(a)  0 (b)  1 (c)  2 (d)  infinite

Correct Answer is Option (d) There are an infinite amount of irrational numbers between 5 and 7.

Q8: In a school 5 out of every 7 children participated in ‘Save wild life’ campaign organised by the school authorities. What fraction of the students participated in the campaign. (a)  2/7 (b) 5/7 (c)  4/7 (d)  7/7

Correct Answer is Option (b)

Number of students participated in campaign = 5

Number of total students = 7

Therefore, fraction of the students participated in the campaign. = 5/7

Q9: In a school 5 out of every 7 children participated in ‘Save wild life’ campaign organised by the school authorities. What kind of decimal expansion it has. (a) Terminating (b)  non terminating (c)  terminating repeating (d)  non terminating repeating

Correct Answer is Option (d) A non-terminating, non-repeating decimal is a decimal number that continues endlessly, with no group of digits repeating endlessly. Decimals of this type cannot be represented as fractions, and as a result are irrational numbers. Non-terminating, non-repeating decimals can be easily created by using a pattern.

Q10: In a school 5 out of every 7 children participated in ‘Save wild life’ campaign organised by the school authorities. Every rational number is a _______ number. (a) Prime (b)  Composite (c)  real (d)  even

Correct Answer is Option (c) Every rational number is a real number. It can be defined as any number that can be expressed in the p/q form where q ≠ 0. We can also say that any fraction falls under the class of rational numbers, where the denominator and numerator are integers and the denominator is not equal to zero.

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CBSE Class 9 Maths Case Study Questions PDF Download

Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 2023-24. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams.

Case study questions play a pivotal role in enhancing students’ problem-solving skills. By presenting real-life scenarios, these questions encourage students to think beyond textbook formulas and apply mathematical concepts to practical situations. This approach not only strengthens their understanding of mathematical concepts but also develops their analytical thinking abilities.

Table of Contents

CBSE Class 9th MATHS: Chapterwise Case Study Questions

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked. For Class 9 Maths Case Study Questions, there would be 5 case-based sub-part questions, wherein a student has to attempt 4 sub-part questions.

Chapterwise Case Study Questions of Class 9 Maths

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

Checkout: Class 9 Science Case Study Questions

And for mathematical calculations, tap Math Calculators which are freely proposed to make use of by calculator-online.net

The above  Class 9 Maths Case Study Question s will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Study Questions have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

Class 9 Maths Syllabus 2023-24

UNIT I: NUMBER SYSTEMS

1. REAL NUMBERS (18 Periods)

1. Review of representation of natural numbers, integers, and rational numbers on the number line. Rational numbers as recurring/ terminating decimals. Operations on real numbers.

2. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as √2, √3 and their representation on the number line. Explaining that every real number is represented by a unique point on the number line and conversely, viz. every point on the number line represents a unique real number.

3. Definition of nth root of a real number.

4. Rationalization (with precise meaning) of real numbers of the type

(and their combinations) where x and y are natural number and a and b are integers.

5. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)

UNIT II: ALGEBRA

1. POLYNOMIALS (26 Periods)

Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeros of a polynomial. Motivate and State the Remainder Theorem with examples. Statement and proof of the Factor Theorem. Factorization of ax2 + bx + c, a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem. Recall of algebraic expressions and identities. Verification of identities:

RELATED STORIES

and their use in factorization of polynomials.

2. LINEAR EQUATIONS IN TWO VARIABLES (16 Periods)

Recall of linear equations in one variable. Introduction to the equation in two variables. Focus on linear equations of the type ax + by + c=0.Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line.

UNIT III: COORDINATE GEOMETRY COORDINATE GEOMETRY (7 Periods)

The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations.

UNIT IV: GEOMETRY

1. INTRODUCTION TO EUCLID’S GEOMETRY (7 Periods)

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.

2. LINES AND ANGLES (15 Periods)

1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180O and the converse.

2. (Prove) If two lines intersect, vertically opposite angles are equal.

3. (Motivate) Lines which are parallel to a given line are parallel.

3. TRIANGLES (22 Periods)

1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).

2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).

3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).

4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence)

5. (Prove) The angles opposite to equal sides of a triangle are equal.

6. (Motivate) The sides opposite to equal angles of a triangle are equal.

4. QUADRILATERALS (13 Periods)

1. (Prove) The diagonal divides a parallelogram into two congruent triangles.

2. (Motivate) In a parallelogram opposite sides are equal, and conversely.

3. (Motivate) In a parallelogram opposite angles are equal, and conversely.

4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.

5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.

6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.

5. CIRCLES (17 Periods)

1. (Prove) Equal chords of a circle subtend equal angles at the center and (motivate) its converse.

2. (Motivate) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.

4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

5. (Motivate) Angles in the same segment of a circle are equal.

6. (Motivate) If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.

7. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.

UNIT V: MENSURATION 1.

1. AREAS (5 Periods)

Area of a triangle using Heron’s formula (without proof)

2. SURFACE AREAS AND VOLUMES (17 Periods)

Surface areas and volumes of spheres (including hemispheres) and right circular cones.

UNIT VI: STATISTICS & PROBABILITY

STATISTICS (15 Periods)

 Bar graphs, histograms (with varying base lengths), and frequency polygons.

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.

Benefits of Practicing CBSE Class 9 Maths Case Study Questions

Regular practice of CBSE Class 9 Maths case study questions offers several benefits to students. Some of the key advantages include:

• Deeper Understanding : Case study questions foster a deeper understanding of mathematical concepts by connecting them to real-world scenarios. This improves retention and comprehension.
• Practical Application : Students learn to apply mathematical concepts to practical situations, preparing them for real-life problem-solving beyond the classroom.
• Critical Thinking : Case study questions require students to think critically, analyze data, and devise appropriate solutions. This nurtures their critical thinking abilities, which are valuable in various academic and professional domains.
• Exam Readiness : By practicing case study questions, students become familiar with the question format and gain confidence in their problem-solving abilities. This enhances their readiness for CBSE Class 9 Maths exams.
• Holistic Development: Solving case study questions cultivates not only mathematical skills but also essential life skills like analytical thinking, decision-making, and effective communication.

Tips to Solve CBSE Class 9 Maths Case Study Questions Effectively

Solving case study questions can be challenging, but with the right approach, you can excel. Here are some tips to enhance your problem-solving skills:

• Read the case study thoroughly and understand the problem statement before attempting to solve it.
• Identify the relevant data and extract the necessary information for your solution.
• Break down complex problems into smaller, manageable parts to simplify the solution process.
• Apply the appropriate mathematical concepts and formulas, ensuring a solid understanding of their principles.
• Clearly communicate your solution approach, including the steps followed, calculations made, and reasoning behind your choices.
• Practice regularly to familiarize yourself with different types of case study questions and enhance your problem-solving speed.Class 9 Maths Case Study Questions

Remember, solving case study questions is not just about finding the correct answer but also about demonstrating a logical and systematic approach. Now, let’s explore some resources that can aid your preparation for CBSE Class 9 Maths case study questions.

Q1. Are case study questions included in the Class 9 Maths Case Study Questions syllabus?

Yes, case study questions are an integral part of the CBSE Class 9 Maths syllabus. They are designed to enhance problem-solving skills and encourage the application of mathematical concepts to real-life scenarios.

Q2. How can solving case study questions benefit students ?

Solving case study questions enhances students’ problem-solving skills, analytical thinking, and decision-making abilities. It also bridges the gap between theoretical knowledge and practical application, making mathematics more relevant and engaging.

Q3. How do case study questions help in exam preparation?

Case study questions help in exam preparation by familiarizing students with the question format, improving analytical thinking skills, and developing a systematic approach to problem-solving. Regular practice of case study questions enhances exam readiness and boosts confidence in solving such questions.

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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.

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

This method is not easy for me

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

MATHS PAAGAL HAI

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

Where is search ? bar

maths is love

Can I have more questions without downloading the app.

I love math

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CBSE Class 10 Maths Case Study Questions for Class 10 Maths Chapter 1 - Real Numbers (Published by CBSE)

Cbse class 10 maths cased study question bank for chapter 1 - real numbers is available here. this question bank is very useful to prepare for the class 10 maths exam 2021-2022..

The Central Board of Secondary Education has introduced the case study questions in class 10 exam pattern 2021-2022. The CBSE Class 10 questions papers of Board Exam 2022 will have questions based on case study. Therefore, students should get familiarised with these questions to do well in their board exam.

We have provided here case study questions for Class 10 Maths Chapter 1 - Real Numbers. These questions have been published by the CBSE board itself. Students must solve all these questions at the same time they finish with the chapter - Real numbers. 

Case Study Questions for Class 10 Maths Chapter 1 - Real Numbers

To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- section A and section B of grade X. There are 32 students in section A and 36 students in section B.

1. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

Answer: c) 288

2. If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is

Answer: b) 4

3. 36 can be expressed as a product of its primes as

a) 2 2 × 3 2

b) 2 1 × 3 3

c) 2 3 × 3 1

d) 2 0 × 3 0

Answer: a) 2 2 × 3 2

4. 7 × 11 × 13 × 15 + 15 is a

a) Prime number

b) Composite number

c) Neither prime nor composite

d) None of the above

Answer: b) Composite number

5. If p and q are positive integers such that p = ab 2 and q= a 2 b, where a , b are prime numbers, then the LCM (p, q) is

Answer: b) a 2 b 2

CASE STUDY 2:

A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.

1. In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum number participants that can accommodated in each room are

Answer: b) 12

2. What is the minimum number of rooms required during the event?

Answer: d) 21

3. The LCM of 60, 84 and 108 is

Answer: a) 3780

4. The product of HCF and LCM of 60,84 and 108 is

Answer: d) 45360

5. 108 can be expressed as a product of its primes as

a) 2 3 × 3 2

b) 2 3 × 3 3

c) 2 2 × 3 2

d) 2 2 × 3 3

Answer: d) 2 2 × 3 3

CASE STUDY 3:

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

Observe the following factor tree and answer the following:

1. What will be the value of x?

Answer: b) 13915

2. What will be the value of y?

Answer: c) 11

3. What will be the value of z?

Answer: b) 23

4. According to Fundamental Theorem of Arithmetic 13915 is a

a) Composite number

b) Prime number

d) Even number

Answer: a) Composite number

5. The prime factorisation of 13915 is

a) 5 × 11 3 × 13 2

b) 5 × 11 3 × 23 2

c) 5 × 11 2 × 23

d) 5 × 11 2 × 13 2

Answer: c) 5 × 11 2 × 23

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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• CBSE Maths Important Questions
• Class 9 Maths
• Chapter 1: Number System

Important Questions Class 9 Maths Chapter 1 Number System

Important questions for Class 9 Maths Chapter 1 Number system are given here. These practice questions will help the students to build a better understanding of the  Number system concept in Maths. These Class 9 Chapter 1 questions are important for CBSE Class 9 Maths exams. These important questions give an overview of the question types that are asked in the final exams and so students are able to face the exams more confidently.

Also Check:

• Important 2 Marks Questions for CBSE 9th Maths
• Important 3 Marks Questions for CBSE 9th Maths
• Important 4 Marks Questions for CBSE 9th Maths

Number System Important Questions For Class 9 (Chapter 1)

Below given important Number system questions for 9th class students will help them to get acquainted with a wide variation of questions and thus, develop problem-solving skills.

Q.1:  Find five rational numbers between 1 and 2.

We have to find five rational numbers between 1 and 2.

So, let us write the numbers with denominator 5 + 1 = 6 

Thus, 6/6 = 1, 12/6 = 2

From this, we can write the five rational numbers between 6/6 and 12/6 as:

7/6, 8/6, 9/6, 10/6, 11/6

Q.2:  Find five rational numbers between 3/5 and 4/5.

We have to find five rational numbers between 3/5 and 4/5.

So, let us write the given numbers by multiplying with 6/6, (here 6 = 5 + 1)

3/5 = (3/5) × (6/6) = 18/30

4/5 = (4/5) × (6/6) = 24/30

Thus, the required five rational numbers will be: 19/30, 20/30, 21/30, 22/30, 23/30

Q.3:  Locate √ 3 on the number line.

Construct BD of unit length perpendicular to OB (here, OA = AB = 1 unit) as shown in the figure.

By Pythagoras theorem,

OD = √(2 + 1) = √3

Taking O as the centre and OD as radius, draw an arc which intersects the number line at the point Q using a compass.

Therefore, Q corresponds to the value of √3 on the number line.

Q.4:  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, since the square root of a positive integer 16 is equal to 4. Here, 4 is a rational number.

Q.5:  Find the decimal expansions of 10/3, 7/8 and 1/7.

Therefore, 10/3 = 3.3333…

7/8 = 0.875

1/7 = 0.1428571…

Let x = 0.3333…. 

Multiply with 10,

10x = 3.3333…

Now, 3.3333… = 3 + x (as we assumed x = 0.3333…)

Thus, 10x = 3 + x

10x – x = 3

Therefore, 0.3333… = 1/3. Here, 1/3 is in the form of p/q and q ≠ 0.

Q.7:  What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

Thus, 1/17 = 0.0588235294117647….

Therefore, 1/17 has 16 digits in the repeating block of digits in the decimal expansion.

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

The given two rational numbers are 5/7 and 9/11.

5/7 = 0.714285714…..

9/11 = 0.81818181……

Hence, the three irrational numbers between 5/7 and 9/11 can be:

0.720720072000…

0.730730073000…

0.808008000…

Q.9:  Visualise 3.765 on the number line, using successive magnification.

Visualisation of 3.765 on the number line, using successive magnification is given below:

Q.10:  Add 2 √ 2+ 5 √ 3 and √ 2 – 3 √ 3.

(2√2 + 5√3) + (√2 – 3√3)

= 2√2 + 5√3 + √2 – 3√3

= (2 + 1)√2 + (5 – 3)√3

= 3√2 + 2√3

Q.11: Simplify: ( √3+√7) (√3-√7). 

(√3 + √7)(√3 – √7)

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

(√3 + √7)(√3 – √7) = (√3) 2  – (√7) 2

= 3 – 7

Q.12: Rationalise the denominator of 1/[7+3 √3].

1/(7 + 3√3)

By rationalizing the denominator,

= [1/(7 + 3√3)] [(7 – 3√3)/(7 – 3√3)]

= (7 – 3√3)/[(7) 2  – (3√3) 2 ]

= (7 – 3√3)/(49 – 27)

= (7 – 3√3)/22

Q.13:  Represent √( 9.3) on the number line.

Representation of √9.3 on the number line is given below:

Q.14: Simplify:

(i) 7 2/3 .7 1/5 (ii) 10 1/2 /10 1/4

(i) 7 2/3 .7 1/5

Bases are equal, so add the powers.

7 (2/3 + 1/5)

= 7 (10 + 3)/15

(ii) 10 1/2 /10 1/4

Bases are equal, so subtract the powers.

= 10 (1/2 – 1/4)

Q.15:   What is the product of a rational and an irrational number? a) Always an integer b) Always a rational number c) Always an irrational number d) Sometimes rational and sometimes irrational

Correct answer: Option (c)

Explanation:

The product of a rational and an irrational number is always an irrational number.

For example, 2 is a rational number and √3 is irrational. Thus, 2√3 is always an irrational number.

Q.16: What is the value of (256) 0.16  X (256) 0.09 ? a) 4 b) 16 c) 64 d) 256.25

Correct answer: Option (a)

(256) 0.16 x (256) 0.09 = (256) (0.16 + 0.09)

= (256) 0.25

= (256) (25/100)

= (256) (1/4)

= (4 4 ) (1/4)

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

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

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

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

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

Download CBSE Class 9 Maths Important Questions 2024-25 PDF

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

Important Topics Covered in Class 9 Maths Chapter 1

Introduction to number system

Irrational Number

Real Number and Their Decimal Expansion

Representation of Real Number on Number Line

Operations on Real Number

Laws of Exponents for Real Number

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

1 Marks Questions

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

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

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

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

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

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

0.04004000400004.... 

0.07007000700007....

0.13001300013000013.... 

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

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

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

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

0.73073007300073...

0.74074007400074.... 

0.76076007600076.... 

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

 $(i)\frac{3}{5}$             

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

$(iii)\frac{40}{27}   $    

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

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

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

(iv) Infinite

Ans: (iv) Infinite

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

(i) $\text{1}$

(ii) $\text{-1}$

(iii) $\text{2}$

(iv) none of these

Ans: (i) $1$

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

(i) $\text{9}$

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

(iii) $\text{3}$

Ans: (ii) $1/9$

8. Every natural number is

(i) not an integer

(ii) always a whole number

(iii) an irrational number

(iv) not a fraction

Ans: (ii) always a whole number

9. Select the correct statement from the following

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

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

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

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

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

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

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

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

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

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

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

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

(i) an irrational number

(ii) rational number

(iii) a natural number

(iv) a whole number

Ans: (i) an irrational number

12. A terminating decimal is

(i) a natural number

(ii) a rational number

(iii) a whole number

(iv) an integer.

Ans: (ii) a rational number

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

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

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

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

(iv) $\text{1}$

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

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

Ans: (ii) $10$

15. The sum of rational and an irrational number

(i) may be natural

(ii) may be irrational

(iii) is always irrational

(iv) is always rational

Ans: (iii) is always rational

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

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

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

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

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

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

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

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

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

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

(iv) None of these

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

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

(a) natural number

(b) irrational number

(c) rational number

(d) integer

Ans: (b) irrational number

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=0.\overline{45}$

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

(i) $\text{2}$

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

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

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

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

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

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

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

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

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

(i) natural number

(iii) integer

(iv) irrational number

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

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

2 Marks Questions

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

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

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

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

Therefore, zero is a rational number.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ans: According to Pythagoras theorem, we can conclude that

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

10x=9.99999.... 

x=0.99999.... 

_____________

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

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

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

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

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

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

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

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

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

(Image will be uploaded soon)

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

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

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

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

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

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

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

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

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

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

10. Represent $9.3$ on the number line.

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

Then $BD=\sqrt{9.3}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Rationalizing factor is $\sqrt{3}$

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

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

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

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

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

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

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

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

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

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

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

$\sqrt{7}=x+3$

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

$\therefore x+3$ is a rational number

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=11 $ 

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

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

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

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

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

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

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

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

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

$10x=24.178178178.... $

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

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

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

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

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

$9990x=24154 $

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$x=0.252525....$

Multiply both sides by 100

$100x=25.252525....$

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

Subtract (i) from (ii)

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

 $x=\frac{25}{99} $

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

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

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

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

 $=18+9\sqrt{6}+4\sqrt{6}+12 $

 $ =30+\left( 9+4 \right)\sqrt{6} $

 $=30+13\sqrt{6} $

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

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

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

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

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

3 Marks Questions

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

reasons for your answers.

i. Every natural number is a whole number.

Separately, consider whole numbers and natural numbers.

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

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

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

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

ii. Every integer is a whole number.

Ans: Separately, consider whole numbers and integers.

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

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

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

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

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

iii. Every rational number is a whole number.

Ans: Separately, consider whole numbers and rational numbers.

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

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

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

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

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

i. Every irrational number is a real number.

Ans: Separately, consider irrational numbers and real numbers.

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

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

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

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

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

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

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

iii. Every real number is an irrational number.

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

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

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

i. $0.\overline{6}$

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

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

Multiplying both sides by 10 we get

$10x=6.6666$             ......(b)

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

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

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

ii. $0.4\overline{7}$

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

Multiplying both sides by 10 we get 

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

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

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

iii. $0.\overline{001}$

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

Multiplying both sides by 1000 we get 

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

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

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

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

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

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

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

Therefore, we conclude that

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

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

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

$ \frac{5}{2}=2.5 $ 

$ \frac{5}{4}=1.25 $ 

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

$  \frac{5}{16}=0.3125 $ 

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

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

6. Classify the following numbers as rational or irrational:

i. $2-\sqrt{5}$

Ans: $2-\sqrt{5}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ans: $2\pi $

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

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

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

7. Simplify each of the following expression:

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

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

Applying distributive law,

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

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

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

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

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

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

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

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

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

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

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

$=7+2\sqrt{10}$

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

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

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

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

8. Find 

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

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

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

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

$=3\times 3\times 3 $

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

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

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

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

$=2\times 2 $ 

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

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

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

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

$ =2\times 2\times 2 $

 & =8 $

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

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

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

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

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

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

$=\frac{1}{5} $

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

9. Simplify

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ans: Let us assume that the given decimal as,

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

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

$ 10x=8.8888 $

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

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

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

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

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

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

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

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

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

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

$ =\frac{94+42\sqrt{5}}{4} $ 

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

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

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

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

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

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

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

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

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

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

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

Ans: We are asked to prove the expression,

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

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

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

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

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

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

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

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

Take $OA=AB=1$ unit.

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

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

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

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

$ OB=\sqrt{2}$ 

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

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

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

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

$OD=\sqrt{3} $

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

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

Ans: We are given the expression as,

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

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

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

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

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

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

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

$=6\times 6=36 $

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

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

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

Multiply by 10

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

Multiply by 1000

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

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

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

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

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

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

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

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

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

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

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

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

Representing the given numbers in decimal form we have,

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

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

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

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

20. Give an example of two irrational numbers whose

i. Sum is a rational number

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

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

ii. Product is a rational number

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

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

iii. Quotient is a rational number

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

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

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

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

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

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

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

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

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

$  =-5\times -0.318 $

$  =1.59 $ 

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

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

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

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

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

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

$=\frac{4-2\sqrt{3}}{16-12} $ 

$  =\frac{4-2\sqrt{3}}{4} $

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

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

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

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

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

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

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

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

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

Let us compare the terms in LHS and RHS.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=0.\overline{285714}$

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

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

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

$=0.\overline{428571}$

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

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

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

$=0.\overline{571428}$

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

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

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

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

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

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

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

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

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

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

In $\vartriangle OAB$, apply Pythagoras Theorem,

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

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

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

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

$\Rightarrow OB=\sqrt{2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Short Answer Questions (4 Marks)

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

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

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

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

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

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

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

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

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

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

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

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

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

Performing long division of $33$ by $8$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2. Classify the following as rational or irrational:

i. $\sqrt{23}$

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

ii. $\sqrt{225}$

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

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

iii. $0.3796$

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

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

Thus $0.3796$ is a rational number.

iv. $7.478478...$

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

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

Converting it into fraction 

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

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

Subtract equations $(2)-(1)$

$ 1000x=7478.478478... $

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

$ \text{  }999x=7471 $

Now, $999x=7471$

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

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

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

v. $1.101001000100001...$

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

3. Rationalize the denominator of the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Long Answer Questions (5 Marks)

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

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

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

Ans: It is given that – 

$\sqrt{5}=2.236$

$\sqrt{3}=1.732$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\sqrt{2}=1.414$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Using the identities –

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

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

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

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

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

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

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

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

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

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

Here, 

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

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

Start by rationalizing the denominator in LHS

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

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

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

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

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

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

They are all divisible by $3$

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

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

Comparing with RHS, we get – 

Thus, 

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

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

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

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

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

Number System Class 9 Important Questions

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

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

Exercise 1.1

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

Exercise 1.2

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

Exercise 1.3

Here the student will know how to represent fractions into decimal form and find if it is terminating or non-terminating. Thus the concepts of terminating and non-terminating fractions will further help to identify it is a rational or irrational number.

Exercise 1.4

A new concept of representing decimal expansion on the number line is introduced through exercise 1.4 mathematics class 9. Here students will learn about magnifying the number to the maximum requirement and representing it on the number line. These decimal places can be either terminating or non-terminating. Hence there are two different concepts in number line representation of decimal expansions.

Exercise 1.5

Now comes the basic calculations of different rational and irrational numbers. It includes addition, multiplication, subtraction, and division of the rational and irrational numbers. Questions will be based on such concepts, and thus you have to simplify the statement accordingly. Also, students become familiar with the concept of rationalising.

Exercise 1.6

Here students will learn to solve questions based with a number having power in fractional form. Also, it might cover the basic addition and subtraction of powers for in-depth conceptual and extra knowledge.

These are the six exercises which will be covered under Chapter 1 Maths Class 9 Important Questions. Hence students can prepare questions according to the concepts discussed above.

Chapter 1 Maths Class 9 Important Questions

According to the syllabus mentioned above exercise-wise, below are some important questions covered to let students prepare well for important questions for class 9 maths number system. These exercise-wise solutions will let students master each concept in detail. Below are some of the questions that are usually picked to set question paper as prepared by CBSE.

State if zero is a rational number. Justify your statement by representing it in p/q where q is not equal to 0 and p and q are both integers.

Find 5 different rational numbers between 5 and 6. Mention each step in detail.

Find out 5 different rational numbers between 12/11 and 10/11.

Justify your statement for the following terms stating true or false.

Rational numbers are real.

A number line having representation in the form of √m has m as a natural number.

A real number is always an irrational number.

Check if true or false. Square roots of all positive numbers will be irrational. Explain your answer statement with the help of an example.

Try to represent √5 on the number line.

Represent the following in the form of decimal expansion:

(i) 36/100  (ii) 1/11  (iii) 4⅛  (iv) 3/13  (v) 2/11  (vi) 329/400

Turn 2/7, 13/7, 4/7, 5/7, 6/7

into decimal expansions without actually doing any long division calculations. Here 1/7= 0.142857.

Express the following as fractional form p/q where q is not 0.

(i) \[0.\overline{6}\]

(ii) \[0.4 \overline{7}\]

(iii) \[0. \overline{001}\]

Represent 2.675 on the number line with number line magnification.

How will you visualise 6.2626…... on the number line up to 4 decimal places?

State if the numbers are rational or irrational.

(i) \[2 - \sqrt{5}\]  (ii)  \[(3 + \sqrt{23}) - \sqrt{23}\]  (iii) \[\frac{2\sqrt{7}}{7\sqrt{7}}\]  (iv) \[\frac{1}{\sqrt{2}}\]  (v) \[2 \pi\]

Simplify the following terms:

(i) \[(3 + \sqrt{3}) ( 2 + \sqrt{2})\] (ii) \[(3 + \sqrt{3}) (3 + \sqrt{3})\]

(iii) \[(\sqrt{5} + \sqrt{2})^{2}\] (iv)\[(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})\]

Rationalise the following terms with their denominators:

(i) \[\frac{1}{7}\] (ii) \[\frac{1}{\sqrt{7} - \sqrt{6}}\] (iii) \[\frac{1}{\sqrt{5} + \sqrt{2}}\] (iv) \[\frac{1}{\sqrt{7} - 2}\]

(i) \[64^{1}{2}\] (ii) \[32^{1}{5}\] (iii) \[125^{1}{3}\]

Class 9 Maths Chapter 1 Extra Questions

Find three rational numbers between $\frac{1}{3}$ and $\frac{1}{2}$. 

Express 0.4323232 in the form of  $\frac{a}{b}$ where a and b are integers and b 0.

Simplify and find the value of $(729)^{1/6}$ .

Rationalise the denominator 1 9 + 5 + 6 .

Find 6 rational numbers between 4 and 6.

Simplify $\sqrt[3]{2}$+$\sqrt[4]{3}$ and $\sqrt{5}+\sqrt{2}$ .

Locate $\sqrt{5}$ on the number line.

Visualise the representation of 4.26 on the number line upto 3 decimal places.

Is 2 - 5   a rational number or irrational number?

Convert 0.45 into rational numbers.

Benefits of Important Questions for Class 9 Maths Number System

Number System Class 9 Important Questions PDF is prepared by Vedantu experts and is free of cost. Students can also schedule additional problems to prepare in-depth to clarify their concepts. These questions turn to be magical for those weak in mathematics or do not show interest in it. The questions are prepared precisely accordingly to CBSE guidelines for their question paper pattern such that students need not search them anywhere else.

Below are the mentioned reasons why students should refer to the Important Question of Maths Class 9 Chapter 1:

Students are free to access these important questions PDF. They need not pay for any study material on Vedantu website.

These Number System Class 9 Important Questions are helpful during exams and help students clarify their concepts of homework questions.

These crucial questions are available in PDF format, which can be easily downloaded through the website. Thus students need not unnecessarily waste their precious time in finding solutions to problems.

Students can print these PDF questions and solutions, which reduces the stress of preparing through soft copies.

All the questions and solutions to the questions are prepared according to CBSE guidelines. Thus it will help students to know question paper pattern.

There is a total of 100 marks for the Class 9 Mathematics paper where 20 marks are for internal assessment, and rest 80 is for the written exam. The marks weightage for Chapter 1 Maths Class 9 is 8 out of 80. Rest 72 marks are for rest of the syllabus of Class 9 Mathematics.

There are a total of 26 sums to be solved according to NCERT book for Chapter 1 Of Class 9 Maths in its six exercises. However, there are other Class 9 Maths Chapter 1 Important Questions prepared by the experts and solutions to each of them provided. Thus, students can build a good base in mathematics through this PDF of important questions available with Vedantu for free. According to exercise-wise, some questions describe different questions and situations that students generally encounter while sitting in the exam. Thus students need to struggle to find essential concepts in the chapter and solutions to each of them.

Important Related Links for CBSE Class 9 

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

1. What number of questions are there in each exercise of Chapter 1 of Class 9 Maths?

Real numbers are used in the first chapter of Class 9 Mathematics. There are a total of six workouts. In Chapter 1, students will learn about several real number concepts. There are four questions in Exercise 1.1, four questions in Exercise 1.2, nine questions in Exercise 1.3, two questions in Exercise 1.4, five questions in Exercise 1.5, and three questions in Exercise 1.6. Students must rehearse all NCERT problems offered by Vedantu in order to understand Chapter 1 of Class 9 maths.

2. Is Chapter 1 of Class 9 Maths difficult to solve?

Students of Class 9 should understand the concepts of real numbers given in Chapter 1 of Class 9 Maths. It is not difficult to solve if students understand the basic concepts of real numbers. Students can take help from the NCERT solutions Class 9 Maths Chapter 1 given at Vedantu app and website. All concepts related to the real numbers are explained in a simple way for quick understanding of the students. 

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

Students should rehearse Class 9 Mathematics NCERT Answers for Chapter 1 to comprehend and practise real-world questions. The NCERT Answers are essential since students may face similar questions in their examinations. With the aid of these questions, students may simply tackle the actual numbers problems in the exam. Students may access complete NCERT Answers for Class 9 Mathematics Chapter 1 for free on the Vedantu platform.

4. What are five rational numbers between ⅖ and ⅗?

Students can find five or more rational numbers between the two given rational numbers easily. They can follow the given steps to find five rational numbers between ⅖ and ⅗. We have to multiply the numerator and denominator of the given rational numbers with the same number.

⅖ x 6/6 = 12/30

⅗ x 6/6= 18/30

Now, we can write five rational numbers between 12/30 and 18/30. Thus, the five rational numbers are 13/30, 14/30, 15/30, 16/30, 17/30.

5. What are the main topics covered in Chapter 1 of Class 9 Maths?

In Chapter 1 of Class 9 Mathematics, students will study about rational and irrational numbers. Students will learn about irrational numbers, how to represent real numbers on a number line, how to express real numbers in decimal form, different operations on real numbers, and real-world exponent rules. To comprehend the various principles of rational numbers in Class 9 Chapter 1, students should practise all NCERT problems offered in the textbook.

CBSE Class 9 Maths Important Questions

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Case Study Questions for Class 9 Maths Chapter 1 Real Numbers

• Last modified on: 1 year ago
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Case Study Questions:

Question 1:

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

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

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

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

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

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

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

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

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

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

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

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Class 9 Maths Case Study Questions of Chapter 1 Real Numbers

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

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

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

Real Numbers Case Study Questions With Answers

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

Case Study/Passage-Based Questions

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

Observe the following factor tree and answer the following:

1. What will be the value of x?

Answer: b) 13915

2. What will be the value of y?

Answer: c) 11

3. What will be the value of z?

Answer: b) 23

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

a) Composite number

b) Prime number

c) Neither prime nor composite

d) Even number

Answer: a) Composite number

5. The prime factorization of 13915 is

a) 5 × 11 3  × 13 2

b) 5 × 11 3  × 23 2

c) 5 × 11 2  × 23

d) 5 × 11 2  × 13 2

Answer: c) 5 × 112 × 23

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

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

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

Answer: (d) no value of n3

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

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

Answer: (c) for all n > 1

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

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

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

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

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

Answer:(a) always true

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

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

Answer: (d) 8

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

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CBSE MCQ for Class 9 Maths Chapter 1 Number System Free PDF

Guys, we are working very hard to provide you with TOPIC-WISE MCQs (as listed below). Till then, attached below is the Master PDF having all the topics. Hope you understand. Enjoy your preparation! All the Best! 

CBSE MCQ for Class 9 Maths Chapter 1 Number System PDF

The CBSE MCQ for Class 9 Maths Chapter 1 Number System are provided above, in detailed and free to download PDF format. The solutions are latest , comprehensive , confidence inspiring , with easy to understand explanation . To download NCERT Class 9 Solutions PDF for Free, just click ‘ Download pdf ’.

Other MCQ Questions for Maths Class 9th CBSE

• CBSE MCQ for Class 9 Maths Chapter 2 Polynomials
• CBSE MCQ for Class 9 Maths: Chapter 3 Coordinate Geometry
• CBSE MCQ for Class 9 Maths: Chapter 4 Linear Equations in Two Variables
• CBSE MCQ for Class 9 Maths Chapter 5 Euclid’s Geometry

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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