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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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## Class 10 Maths Case Study Questions of Chapter 1 Real Numbers

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Case study Questions in the Class 10 Mathematics Chapter 1 are very important to solve for your exam. Class 10 Maths Chapter 1 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 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 10 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 10 Maths Chapter 1 Real Numbers

Case Study/Passage-Based Questions

Case Study 1: 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. (i) For what value of n, 4 n ends in 0?

Answer: (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 n is a rational number?

Answer: (c) for all n > 1

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

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

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

Answer: (a) always true

(v) If n is any odd integer, then n2 – 1 is divisible by

Answer: (d) 8

Case Study 2: HCF and LCM are widely used in number system especially in real numbers in finding relationship between different numbers and their general forms. Also, product of two positive integers is equal to the product of their HCF and LCM Based on the above information answer the following questions.

(i) If two positive integers x and y are expressible in terms of primes as x =p 2 q 3 and y=p 3 q, then which of the following is true? (a) HCF = pq 2 x LCM (b) LCM = pq 2 x HCF (c) LCM = p 2 q x HCF (d) HCF = p 2 q x LCM

Answer: (b) LCM = pq2 x HCF

ii) A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left, then which of the following is correct if the number of marbles is p? (a) p is odd (b) p is even (c) p is not prime (d) both (b) and (c)

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

(iii) Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively. (a) 3 (b) 1 (c) 34 (d) 17

Answer: (d) 17

(iv) Find the least positive integer that on adding 1 is exactly divisible by 126 and 600. (a) 12600 (b) 12599 (C) 12601 (d) 12500

Answer: (b) 12599

(v) If A, B and C are three rational numbers such that 85C – 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by (a) 3 (b) 6 (c) 7 (d) 9

Answer: (a) 3

Case Study 3: Real numbers are an essential concept in mathematics that encompasses both rational and irrational numbers. Rational numbers are those that can be expressed as fractions, where the numerator and denominator are integers and the denominator is not zero. Examples of rational numbers include integers, decimals, and fractions. On the other hand, irrational numbers are those that cannot be expressed as fractions and have non-terminating and non-repeating decimal expansions. Examples of irrational numbers include √2, π (pi), and e. Real numbers are represented on the number line, which extends infinitely in both positive and negative directions. The set of real numbers is closed under addition, subtraction, multiplication, and division, making it a fundamental number system used in various mathematical operations and calculations.

Which numbers can be classified as rational numbers? a) Fractions b) Integers c) Decimals d) All of the above Answer: d) All of the above

What are rational numbers? a) Numbers that can be expressed as fractions b) Numbers that have non-terminating decimal expansions c) Numbers that extend infinitely in both positive and negative directions d) Numbers that cannot be expressed as fractions Answer: a) Numbers that can be expressed as fractions

What are examples of irrational numbers? a) √2, π (pi), e b) Integers, decimals, fractions c) Numbers with terminating decimal expansions d) Numbers that can be expressed as fractions Answer: a) √2, π (pi), e

How are real numbers represented? a) On the number line b) In complex mathematical formulas c) In algebraic equations d) In geometric figures Answer: a) On the number line

What operations are closed under the set of real numbers? a) Addition, subtraction, multiplication b) Subtraction, multiplication, division c) Addition, multiplication, division d) Addition, subtraction, multiplication, division Answer: d) Addition, subtraction, multiplication, division

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 1 Real Numbers with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 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 Case Study Questions for Class 10 Maths Real Numbers Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Real Numbers in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

## CBSE Case Study Questions for Class 10 Maths Real Numbers PDF

Mcq set 1 -, mcq set 2 -, checkout our case study questions for other chapters.

- Chapter 2: Polynomials Case Study Questions
- Chapter 3: Pair of Linear Equations in Two Variables Case Study Questions
- Chapter 4: Quadratic Equation Case Study Questions
- Chapter 5: Arithmetic Progressions Case Study Questions

## How should I study for my upcoming exams?

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Solve every question of NCERT by hand, without looking at the solution.

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

- Last modified on: 10 months ago
- Reading Time: 7 Minutes

Question 1:

HCF and LCM are widely used in number system especially in real numbers in finding relationship between different numbers and their general forms. Also, product of two positive integers is equal to the product of their HCF and LCM Based on the above information answer the following questions.

(i) If two positive integers x and y are expressible in terms of primes as x =p 2 q 3 and y=p 3 q, then which of the following is true? (a) HCF = pq 2 x LCM (b) LCM = pq 2 x HCF (c) LCM = p 2 q x HCF (d) HCF = p 2 q x LCM

(ii) A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left, then which of the following is correct if the number of marbles is p? (a) p is odd (b) p is even (c) p is not prime (d) both (b) and (c)

(iii) Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively. (a) 3 (b) 1 (c) 34 (d) 17

(iv) Find the least positive integer which on adding 1 is exactly divisible by 126 and 600. (a) 12600 (b) 12599 (C) 12601 (d) 12500

(v) If A, B and C are three rational numbers such that 85C – 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by (a) 3 (b) 6 (c) 7 (d) 9

Question 2:

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.

(i) 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? (a) 144 (b) 128 (c) 288 (d) 272

(ii) 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 (a) 2 (b) 4 (c) 6 (d) 8

(iii) 36 can be expressed as a product of its primes as (a) (b) (c) (d)

(iv) 7 is a (a) Prime number (b) Composite number (c) Neither prime nor composite (d) None of the above

(v) If p and q are positive integers such that p = a and q= b, where a , b are prime numbers, then the LCM (p, q) is (a) ab (b) a 2 b 2 (c) a 3 b 2 (d) a 3 b 3

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## CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024 will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

## Chapterwise Case Study Questions for Class 10 Mathematics

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.

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

- Class 10th Science Case Study Questions
- Assertion and Reason Questions of Class 10th Science
- Assertion and Reason Questions of Class 10th Social Science

## Class 10 Maths Syllabus 2024

Chapter-1 real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

## Chapter-2 Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

## Chapter-3 Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

## Chapter-4 Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

## Chapter-5 Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

## Chapter-6 Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

## Chapter-7 Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

## Chapter-8 Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

## Chapter-9 Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

## Chapter-10 Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

## Chapter-11 Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

## Chapter-12 Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

## Chapter-13 Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

## Chapter-14 Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

## Chapter-15 Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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- Real Numbers for Class 10

## Introduction to Real Numbers

In mathematics, rational numbers and irrational numbers are together obtained from the set of real numbers. The set of real numbers is represented by the letter R. Therefore, it indicates that every real number is either a rational number or an irrational number. In either case, it contains a non–terminating decimal depiction. In the instance of rational numbers, the decimal depiction is repeating (including repeating zeroes) and if the decimal depiction is non–repeating, it is an irrational number.

## Real Numbers

Real numbers in the number system are nothing but the combination of rational and irrational numbers. All the arithmetic operations are performed with these numbers and can be represented in the number line. Whereas imaginary numbers are the un-real numbers that cannot be represented in the number line and are commonly used to express a complex number.

Real numbers in Class 10 consist of some of the advanced concepts related to real numbers. Besides knowing what real numbers are, students can have a clear knowledge of the real numbers formulas and concepts like Euclid’s Division Lemma, Euclid’s Division Algorithm, and arithmetic fundamental theorem in class 10.

## Euclid’s Division Lemma

Euclid’s Division Lemma states that, if there are two positive integers a and b, then there is an occurrence of unique integers q and r, such that it satisfies the condition a = (b x q) + r, (such that 0 ≤ r < b).

Where a, b, q, r are the dividend, divisor, quotient, and remainder respectively.

## Fundamental Theorem of Arithmetic

According to the Fundamental Theorem of Arithmetic, every integer that is greater than 1 is either a prime number or is expressed in the form of primes. In other words, all natural numbers can be represented in the form of the product of its prime factors. Prime factors are the numbers that cannot be divisible by other numbers and are only divisible by 1 . For example, the number 56 can be written in the form of its prime factors as:

56 = 2³ × 7

For the number 56, the prime factors are 2 and 7.

## Irrational Numbers

The real numbers which cannot be expressed as simple fractions are called irrational numbers. It cannot be expressed in the terms of a ratio, such as p/q, such that p and q are integers, q≠0, and is a contradiction of rational numbers.

Irrational numbers are generally represented as R\Q, such that the backward slash symbol represents ‘set minus’. it can also be denoted as R – Q, which is the difference between real numbers and rational numbers.

The calculations of irrational numbers are quite complicated. For example, √7, √13, √53, etc., are irrational.

## Rational Numbers

The Rational numbers can be written in the form of p/q, where p and q are integers and q ≠ 0. If these numbers are solved further, it gives the result in decimals.

For example: 0.6, 7/3, -16.6, etc.

## Solved Examples

Find out the HCF of 867 and 255

Using the Euclid's division algorithm, we have

867 = 255 x 3 + 102

255 = 102 x 2 + 51

102= 51 x 2 + 0

Therefore, HCF of (867, 255) = 51

Example2:

Find out if 1009 is a prime or a composite number

Numbers are of two types - prime and composite. Prime numbers consist of only two factors namely 1 and the number itself while composite numbers consist of factors besides 1 and itself.

It can be observed that

7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) (taking 13 out as common)

= 13 x (77 + 1)

= 13 x 13 x 6

The provided expression consists of 6 and 13 as its factors. Thus, it is a composite number.

= (7 x 6 x 5 x 4 x 3 x 2 x 1) + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)

= 5 x ( 1008 +1)

1009 cannot be factorized any further. Thus, the given expression consists of 5 and 1009 as its factors. Therefore, it is a composite number.

## FAQs on Real Numbers for Class 10

1. Where Can I Find the Best Real Numbers Class 10 Solutions?

To prepare for board Class 10 Maths Chapters 1, you can refer to Vedantu. You will find complete Class 10 Maths Chapter 1 Solutions as well as NCERT Solutions Chapter 1 Real Numbers which are exclusively prepared by the expert faculty at Vedantu. These Class 10 Maths Ch 1 Solutions will help students in their board exam preparations. Vedantu provides step-by-step Solutions for Maths so as to aid the students in solving the problems easily.

Real Numbers Class 10 Solutions are designed in a way that will allow students to focus on preparing the solutions in a manner that is easy to understand. A detailed and step-wise explanation of each answer to the questions is provided in the exercises of these solutions.

2. What are the Benefits of Referring to Class 10th Maths Chapter 1?

Answers for the questions are provided for Real Numbers or the first chapter of Maths. Moreover, you will have access to detailed step-by-steps solutions provided in free PDF available at Vedantu. At Vedantu, students are introduced to ample important concepts which will be useful for those who wish to pursue Mathematics as a subject in their Class 11. Based on these solutions, students can prepare excellently for their upcoming Board Exams. These solutions are of great help and benefit to Class 10 Board students as the syllabus covered here follows NCERT guidelines.

All the positive integers are natural numbers, starting from 1 to infinity. Most Importantly, all the natural numbers are integers but all integers are not necessarily natural numbers. These are the sets of all positive counting numbers such as 1, 2, 3, 4, 5, 6, 7, 8, 9, ……..∞.

Real numbers are numbers that have both rational and irrational numbers. Rational numbers are integers (-2, 0, 1), fractions (1/2, 2.5), and irrational numbers (√3, 22/7 ), etc.

As per the Fundamental Theorem of Arithmetic, every integer greater than 1 is said to be either a prime number or is expressed in the form of primes. In other words, all natural numbers can be represented in the form of the product of its prime factors. Prime factors are numbers that cannot be divisible by other numbers and are only divisible by 1. For example, the number 24 can be written in the form of its prime factors as:

24 = 2³ × 3

Here, 2 and 3 are the prime factors of 24.

The real numbers which cannot be expressed as simple fractions are called irrational numbers. It cannot be expressed in terms of fractions or ratios p/q, where p and q are integers, q ≠ 0, and is a rational number contradiction.

Euclid’s Division Lemma states that, if two positive integers a and b are present, then there is a possible occurrence of unique integers q and r, such that it satisfies the condition

Where a = (b x q) + r, (such that 0 ≤ r < b).

Where a, b, q, r are the dividend, divisor (or HCF), quotient, and remainder respectively.

This method is repeated until the remainder becomes zero. The divisor is the H.C.F of the given set of numbers.

The steps to represent the real numbers on the number line is as follows

Step 1: Draw a horizontal line with arrows at the extreme ends and mark the center of the line as 0. The number 0 is referred to as the origin.

Step 2: Make a mark at equal intervals on both sides of the origin and label it with a definite scale.

Step 3: The positive numbers lie on the right side of the origin and the negative numbers lie on the left side of the origin.

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## CBSE Class 10 Maths Case Study

CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus.

These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.

## Class 10 Maths (Formula, Case Based, MCQ, Assertion Reason Question with Solutions)

In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.

Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too.

Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.

For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.

## CBSE Class 10th Case study Questions

Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.

Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.

## CBSE Class 10th Assertion Reasoning Questions

These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement?

Moreover, to solve the problem they need to look at the given options and then answer them.

## CBSE Class 10 Maths Case Based MCQ

CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.

Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.

## Class 10th Mathematics Multiple Choice Questions

Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.

Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.

The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.

To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.

In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.

## CBSE 10th 2024-25 : Science Official Competency Focused Practice Questions released by CBSE

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## Class 10 Maths Chapter 1 MCQ

Class 10 Maths Chapter 1 Real Numbers MCQ (Multiple Choice Objective Questions) with answers and complete explanation case study type questions for the first term examination 2024-25. The answers of 10th Maths Chapter 1 MCQ are given with explanation, so that students can understand easily. This page of Class 10 Maths MCQ contains the questions released by CBSE as well as extra questions for practice.

## Case Study – 1

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.

## 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?

Factors of 32 = 2 х 2 х 2 х 2 х 2 = 2⁵ Factors of 36 = 2 х 2 х 3 х 3 = 2² х 3² LCM of 32 and 36 = 2⁵ х 3² = 32 х 9 = 288 Hence, the correct option is (C).

- View Answer

## 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

Factors of 32 = 2 х 2 х 2 х 2 х 2 = 2⁵ Factors of 36 = 2 х 2 х 3 х 3 = 2² х 3² LCM of 32 and 36 = 2⁵ х 3² = 32 х 9 = 288 HCF (32, 36) = (32 х 36) / LCM = (32 х 36) / 288 = 4 Hence, the correct option is (B).

## 36 can be expressed as a product of its primes as

Factors of 36 = 2 х 2 х 3 х 3 = 2² х 3² Hence, the correct option is (A).

## 7 х 11 х 13 х 15 + 15 is a

7 х 11 х 13 х 15 + 15 = 15 х (7 х 11 х 13 + 1) = 15 х (Integer) It has more than two factor. So, it is a composite number. Hence, the correct option is (B).

## If p and q are positive integers such that p = ab² and q = a²b, where a, b are prime numbers, then the LCM (p, q) is

p = ab² q = a²b LCM = highest powers of common factors of ab² and a²b = a²b² Hence, the correct option is (B).

## 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.

## 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

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ HCF of 60, 84, and 108 = 2² х 3 = 12 Hence, the correct option is (B).

## What is the minimum number of rooms required during the event?

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ HCF of 60, 84, and 108 = 2² х 3 = 12 Number of room required for Hindi participants = 60/12 = 5 Number of room required for English participants = 84/12 = 7 Number of room required for Mathematics participants = 108/12 = 9 Total number of room required = 5 + 7 + 9 = 21 Hence, the correct option is (D).

## The LCM of 60, 84, and 108 is

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ LCM of 60, 84, and 108 = 2² х 3³ х 5 х 7 = 4 х 27 х 5 х 7 = 3780 Hence, the correct option is (A).

## The product of HCF and LCM of 60, 84, and 108 is

Factors of 60 = 2 х 2 х 3 х 5 = 2² х 3 х 5 Factors of 84 = 2 х 2 х 3 х 7 = 2² х 3 х 7 Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ HCF of 60, 84, and 108 = 2² х 3 = 12 LCM of 60, 84, and 108 = 2² х 3³ х 5 х 7 = 4 х 27 х 5 х 7 = 3780 Product of HCF and LCM of 60, 84, and 108 = 12 х 3780 = 45360 Hence, the correct option is (D).

## 108 can be expressed as a product of its primes as

Factors of 108 = 2 х 2 х 3 х 3 х 3 = 2² х 3³ Hence, the correct option is (D).

## Case Study – 3

Rohit Singh is a worker in a petrol pump. He along with the other co-workers, use to transfer petrol from tanker to storage. On Monday, there were two tankers containing 850 litres and 680 litres of petrol respectively.

## What is the maximum capacity of a container which can measure the petrol of either tanker in exact number of time?

The maximum capacity of the container is the HCF of 850 and 680. Factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 HCF of 850 and 680 = 2 х 5 х 17 = 170 Hence, the correct option is (C).

## If the product of two positive integers is equal to the product of their HCF and LCM is true then, the LCM (850, 680) is

Factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 HCF of 850 and 680 = 2 х 5 х 17 = 170 LCM (850, 680) = (850 х 680) / HCF = (850 х 680) / 170 = 3400 Hence, the correct option is (D).

## 680 can be expressed as a product of its primes as

Factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 Hence, the correct option is (C).

## 2 х 3 х 5 х 11 х 17 + 11 is a

2 х 3 х 5 х 11 х 17 + 11 = 11 х (2 х 3 х 5 х 17 + 1) = 11 х (Integer) It has more than two factor. So, it is a composite number. Hence, the correct option is (B).

## If p and q are positive integers such that p = a³b² and q = a²b³, where a, b are prime numbers, then the LCM (p, q) is

p = a³b² q = a²b³ LCM = highest powers of common factors of a³b² and a²b³ = a³b³ Hence, the correct option is (B).

## Case Study – 4

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:

## What will be the value of x?

X = 5 х 2783 = 13915 Hence, the correct option is (B).

## What will be the value of y?

Y = 2783/253 = 11 Hence, the correct option is (C).

## What will be the value of z?

Z = 253/11 = 23 Hence, the correct option is (B).

## According to Fundamental Theorem of Arithmetic 13915 is a

Because 13915 can be written into the product of primes. 13915 = 5 х 11 х 11 х 23 = 5 х 11² х 23 Hence, the correct option is (A).

## The prime factorisation of 13915 is

13915 = 5 х 11 х 11 х 23 = 5 х 11² х 23 Hence, the correct option is (C).

## Case Study – 5

We all know that morning walk is good for health. In a morning walk, three friends Anjali, Sofia, and Angelina step of together. There steps measure 80 cm, 85 cm, and 90 cm. respectively.

## What is the minimum distance each should walk so that they can cover the distance in complete steps?

The minimum distance covered by each in complete steps must be the LCM of 80 cm, 85 cm, and 90 cm. Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 LCM of 80, 85, and 90 = 2² х 3² х 5 х 17 = 12240 Now, 12240 cm = 122 m 40 cm Hence, the correct option is (B).

## What is the minimum number of steps taken by any of the three friends, when they meet again?

Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 LCM of 80, 85, and 90 = 2² х 3² х 5 х 17 = 12240 The step size of Angelina is maximum among these three. So, she will take minimum number of steps to cover the same distance. Number of steps = 12240/90 = 136 Hence, the correct option is (D).

## The HCF of 80, 85, and 90 is

Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 HCF of 80, 85, and 90 = 5 Hence, the correct option is (A).

## The product of HCF and LCM of 80, 85, and 90 is

Factors of 80 = 2 х 2 х 2 х 2 х 5 = 2⁴ х 5 Factors of 85 = 5 х 17 Factors of 90 = 2 х 3 х 3 х 5 = 2 х 3² х 5 HCF of 80, 85, and 90 = 5 LCM of 80, 85, and 90 = 2² х 3² х 5 х 17 = 12240 Therefore, the product of HCF and LCM of 80, 85, and 90 = 12240 х 5 = 61200 Hence, the correct option is (C).

## 90 can be expressed as a product of its primes as

Factors of 108 = 2 х 3 х 3 х 5 = 2 х 3² х 5 Hence, the correct option is (D).

Class 10 Maths Chapter 1 MCQ are given below. There are total of 5 questions with four choices. Only one option is correct and the explanation of correct answer is given below the questions. Every time the students will get a new set of five questions with different levels of questions. For any further discussion, please join the Discussion Forum.

## Euclid’s division lemma represents the pair of integers 25 & 7.

Because 7 goes into 25 thrice and leaves remainder 4, where 0 < 4 <7.

## For some integer p, every even integer is of the form

Because multiple 2 of any integer either odd or even will be always even.

## The least number that is divisible by all the numbers from 1 to 8 is

The LCM of numbers 1 to 8 is 840. We can write the factors of the numbers from 1 to 8 as follows: 1, 2, 3, 2 × 2, 5, 2 × 3, 7, 2 × 2 × 2. Therefore, LCM = 1 × 2 × 2 × 2 × 3 × 5 × 7 = 840

## When a number is divided by 7, its remainder is always

According to Euclid’s Division Lemma, If a = 7q + r, then r should be less than 7. In other words, remainder should always be less than divisor.

## If HCF (16, y) = 8 and LCM (16, y) = 48, then the value of y is

We know that: HCF × LCM = 16 × y So, 8 × 48 = 16 × y y = 8 × 48/16 = 24

## What are the important topics in Class 10 Maths Chapter 1 MCQ?

Euclid’ division lemma and the Fundamental Theorem of Arithmetic are the two main topics in 10th Maths chapter 1 Real Numbers. Now questions are designed on the basis of case study. So practice MCQ questions based on daily life events which will be more helpful in CBSE exams.

## In which of the four exercise of 10th Maths Chapter 1, are Case Study MCQ asked?

There are questions from each exercise of Chapter 1 of 10th Maths, but most of the MCQs can be formed from Exercise 1.4. Now CBSE introduces the questions based on CASE STUDY which may be asked from any section of class 10 Maths chapter 1.

## How many MCQ are required to be perfect in Chapter 1 of Class 10 Maths?

If your concepts are clear, the MCQs provide more confidence in that section. More practice means more to retain and better understanding with the concepts of topics.

## How many questions from Chapter 1 of Class 10 Maths asked in CBSE Board?

There is no classification of number of questions from the different chapters. There may be one, more than one or none from Chapter 1 Real Numbers of Class 10 Maths.

We are adding more questions frequently, so that students can have a good practice of Class 10 Maths Chapters. If you have suggestion or feedback about this page or website improvement, you are welcome. Important questions with solutions and answers will be added very soon for each chapter of class 10 Maths.

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## NCERT Solutions for Chapter 1 Real Numbers Class 10 Maths

(i) HCF of 135 and 225: Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get 225 = (135 × 1) + 90 Since, 90 ≠ 0, therefore, applying the Euclid’s lemma to 135 and 90, we have: 135 = (90 × 1) + 45 But 45 ≠ 0 ∴ Applying Euclid’s Lemma to 90 and 45, we get 90 = (45 × 2) + 0 Here, r = 0, so our process stops. Since, the divisor at the last step is 45, ∴ HCF of 225 and 135 is 45.

(ii) HCF of 196 and 38220: We start dividing the larger number 38220 by 196, we get 38220 = (196 × 195) + 0 ∵ r = 0 ∴ HCF of 38220 and 196 is 196.

(iii) HCF of 867 and 255: Here, 867 > 255 ∴ Applying Euclid’s Lemma to 867 and 255, we get

867 = (255 × 3) + 102 Since, 102 ≠ 0, therefore, applying the Euclid’s lemma to 255 and 102, we have: 255 = (102 × 2) + 51 But 51 ≠ 0 ∴ Applying Euclid’s Lemma to 102 and 51, we get 102 = (51 × 2) + 0 Here, r = 0, so our process stops. Since, the divisor at the last step is 51, ∴ HCF of 867 and 255 is 51.

Total number of members = 616 ∵ The total number of members are to march behind an army band of 32 members.

HCF of 616 and 32 is equal to the maximum number of columns such that the two groups can march in the same number of columns. ∴ Applying Euclid’s lemma to 616 and 32, we get 616 = (32 × 19) + 8 ∵ 8 ≠ 0 ∴ Again, applying Euclid’s lemma to 32 and 8, we get 32 = (8 × 4) + 0 ∵ r = 0 ∴ HCF of 616 and 32 is 8 Hence, the required number of maximum columns = 8.

Let us consider an arbitrary positive integer as ‘x’ such that it is of the form 3q, (3q + 1) or (3q + 2).

For x = 3q, we have x 2 = (3q) 2 ⇒ x 2 = 9q 2 = 3(3q 2 ) = 3m --- (1) [putting 3q 2 = m where m is an integer]

For x = 3q+1 x 2 = (3q + 1) 2 = 9q 2 + 6q + 1 = 3(3q 2 + 2q) + 1 = 3m + 1 --- (2) [putting 3q 2 + 2q = m, where m is an integer]

For x = 3q + 2 x 2 = (3q + 2) 2 = (9q 2 + 12q + 4) = (9q 2 + 12q + 3) + 1 = 3(3q 2 + 4q + 1) + 1 = 3m + 1 --- (3) [putting 3q 2 + 4q + 1 = m, where m is an integer]

From (1), (2) and (3) x 2 = 3m or 3m + 1 Thus, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Let us consider a positive integer 'x' such that it is in the form of 3q, (3q + 1) or (3q + 2) and 'b' = 3

Now x = bq + r where, q ≥ 0 and 0 ≤ r < 3

∴ x is of the form 3q, (3q + 1) or (3q + 2).

Note: For any positive integer, x and b = 3, x = 3q + r, where q is quotient and r is remainder such that 0 r If r = 0 then x = 3q If r = 1 then x = 3q + 1 If r = 2 then x = 3q + 2

For x = 3q x 3 = (3q) 3 = 27q 3 = 9 (3q 3 ) = 9m --- (1) [putting 3q 3 = m, where m is an integer]

For x = 3q + 1 x 3 = (3q + 1) 3 = 27q 3 + 27q 2 + 9q + 1 = 9 (3q 3 + 3q 2 + q) + 1 = 9m + 1 --- (2) [putting (3q 3 + 3q 2 + 1) = m, where m is an integer]

For x = 3q + 2 x 3 = (3q + 2) 3 = 27q 3 + 54q 2 + 36q + 8 = 9 (3q 3 + 6q 2 + 4q) + 8 = 9m + 8 ...(3) |putting (3q 3 + 6q 2 + 4q) = m, where m is an integer

From (1), (2), (3) we have: x 3 = 9m, (9m + 1) or (9m + 8) Thus, x 3 of any positive integer can be in the form 9m, (9m + 1) or (9m + 8)

(i) Using factor tree method, we have:

∴ 140 = 2 × 2 × 5 × 7 = 2 2 × 5 × 7

(ii) Using factor tree method, we have:

156 = 2× 2 × 3 × 13 = 2 2 × 3 × 13

(iii) Using factor tree method, we have:

3825 = 3 × 3 × 5 × 5 × 17

= 3 2 × 5 × 5 × 17

(iv) Using factor tree method, we have:

∴ 7429 = 17 × 19 × 23

(i) 26 and 91. Using factor-tree method, we have:

⇒ 26 = 2×13

⇒ 91 = 7×13

∴ LCM (26,91) = Product of highest power of all the factors = 2 × 7 × 13 = 182 ∴ HCF (26, 91) = Product of least powers of all common factors = 13 Verification:

Now, LCM × HCF = 182 × 13 = 2366 and 26 × 91 = 2366 i.e., LCM × HCF = Product of two numbers. (ii) 510 and 92. Using the factor tree method, we have:

⇒ 510 = 2×5×3×17

⇒ 92 = 2×2×23 ∴ LCM (510, 92) = 2 × 3 × 5 × 17 × 2 × 23 = 23460 ∴ HCF of 510 and 92 = 2

Verification:

LCM × HCF = 23460 × 2 = 46920 510 × 92 = 46920 i.e., LCM × HCF = Product of two numbers.

(iii) Using the factor tree method, we have:

⇒ 54 = 2×3×3×3×3 ⇒ LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024 ∴ HCF of 336 and 54 = 2 × 3 = 6 Verification:

LCM × HCF = 3024 × 6 = 18144 Also 336 × 54 = 18144 Thus, LCM × HCF = Product of the numbers

⇒ 12 = 2×2×3

⇒ 21 = 3×7 ∴ HCF of 12, 15 and 21 is 3 LCM of 12, 15 and 21 is 2 × 2 × 3 × 5 × 7 or 420

(ii) We have:

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29 ∴ HCF (17, 23, 29) is 1 LCM (17, 23, 29) = 17 × 23 × 29 = 11339

(iii) We have:

∴ HCF (8, 9, 25) is 1 LCM (8, 9, 25) is 2 × 2 × 2 × 3 × 3 × 5 × 5 or 1800

Given that HCF (306, 657) = 9, find LCM (306, 657).

Here, HCF of 306 and 657 = 9 We have: LCM × HCF = Product of the numbers ∴ LCM × 9 = 306 × 657 ⇒ LCM = 34 × 657/9 = 22338 Thus, LCM of 306 and 657 is 22338.

⇒ 18 = 2 × 3 × 3

and 12 = 2 × 2 × 3 ∴ LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36 Thus, they will meet again at the starting point after 36 minutes.

Let √5 be a rational number. ∴ We have to find two integers a and b (where, b ≠ 0 and a and b are coprime)

such that a/b = √5 ⇒ a = √5.b ---(1) Squaring both sides, we have a 2 = 5b 2 ∴ 5 divides a 2 ⇒ 5 divides a ---(2) [∵ a prime number ‘p’ divides a 2 then ‘p’ divides ‘a’, where ‘a’ is positive integer.] ∴ a = 5c, where c is an integer. ∴ Putting a = 5c in (1), we have 5c = √5. b or (5c) 2 = 5b 2 ⇒ 25c 2 = 5b 2 ⇒ 5c 2 = b 2 ⇒ 5 divides b 2 ⇒ 5 divides b ---(3) From (2) and (3), a and b have at least 5 as a common factor. i.e., a and b are not coprime. ∴ Our supposition that √5 is rational is wrong. Hence, √5 is irrational

Let 3+2√5 is rational. ∴ We can find two coprime integers ‘a’ and ‘b’

such that [3+2√5] = a/b, where b ≠ 0

⇒ (1) is a rational ⇒ √5 is a rational But this contradicts the fact that √5 is irrational. ∴ Our supposition is wrong. 3+2√5 is an irrational.

Since, the division of two integers is rational. ∴ 2a/b is a rational. From (1), √2 is a rational number which contradicts the fact that √2 is irrational. ∴ Our assumption is wrong. Thus, 1√2 is irrational.

(ii) Let us suppose that 7√5 is rational. Let there be two coprime integers ‘a’ and ‘b’. Such that 7√5 = a/b , where b ≠ 0 Now, = 7√5 = a/b

This contradicts the fact that √5 is irrational. ∴ We conclude that 7√5 is irrational.

(iii) Let us suppose that 6+√2 is rational. ∴ We can find two coprime integers ‘a’ and ‘b’ (b ≠ 0),

such that, 6+√2 = a/b

From (1), √2 is a rational number, which contradicts the fact that √2 is an irrational number. ∴ Our supposition is wrong. ⇒ 6+ √2 is an irrational number.

(i) 13/3125 ∵ 3125 = 5 × 5 × 5 × 5 × 5 = 5 5 = 2 0 × 5 5 , [ ∵ 2 0 = 1] which is of the form (2 n · 5 m ) ∴ 13/3125 is a terminating decimal. i.e., 13/3125 will have a terminating decimal expansion.

(ii) 17/8 ∵ 8 = 2 × 2 × 2 = 2 3 = 1 × 2 3 = 5 0 × 2 3 , which is of the form 5 m · 2 n ∴ 17/8 will have a terminating decimal expansion.

(iii) 64/455 ∵ 455 = 5 × 7 × 13, which is not of the form 2 n · 5 m ∴ 64/455 will have a non-terminating repeating decimal expansion.

(iv) 15/1600 ∵ 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2 6 × 5 2 , which is of the form 2 n · 5 m ∴ 15/1600 will have a terminating decimal expansion.

(v) 29/343 ∵ 343 = 7 × 7 × 7 = 73, which is not of the form 2 n ·5 m . ∴ 29/343 will have a non-terminating repeating decimal expansion.

(vi) 23/2 3 5 2 Let p/q = 23/2 3 5 2 i.e., q = 2 3 · 5 2 , which is of the form 2n · 5m. ∴ 23/2 3 5 2 will have a terminating decimal expansion.

(vii) 129/2 2 5 7 7 5 Let p/q = 129/2 2 5 7 7 5 i.e., q = 2 2 .5 7 .7 5 , which is not of the form 2 n .5 m . ∴ 129/2 2 5 7 7 5 will have a non-terminating repeating decimal expansion.

(viii) 6/15

∴ 6/15 will have a terminating decimal expansion.

(ix) 35/50 ∵ 50 = 2 × 5 × 5 = 21 × 52, which is of the form 2 n ·5 m . ∴ 35/50 will have a terminating decimal expansion.

(x) 77/210 ∵ 210 = 2 × 3 × 5 × 7 = 2 1 · 3 1 · 5 1 · 7 1 , which is not of the form of 2 n · 5 m . ∴ 77/ 210 will have a non-terminating repeating decimal expansion.

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. (i) 13/ 3125 (ii) 17/18 (iii) 64/455 (iv) 15/1600

(v) 29/343 (vi) 23/2 3 5 3 (vii) 129/2 2 5 7 7 5 (viii) 16/15 (ix) 35/50 (x) 77/ 210

(i) 13/3125 We have,

(iii) 64/455 represents non-terminating repeating decimal expansion.

(iv) 15/1600

(v) 29/343, represents a non-terminating repeating decimal expansion.

(ix) 35/50 35/50 = 0.7

(x) 77/ 210 , represents a non-terminating repeating decimal expansion.

The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form , what can you say about the prime factors of q? (i) 43.123456789 (ii) 0.120120012000120000...

(i) 43.123456789 ∵ The given decimal expansion terminates. ∴ It is a rational of the form p/q ⇒ p/q = 43.123456789

Here, p = 43123456789 and q = 2 9 × 5 9 ∴ Prime factors of q are 2 9 and 5 9 .

(ii) 0.120120012000120000 ..... ∵ The given decimal expansion is neither terminating nor non-terminating repeating, ∴ It is not a rational number.

## NCERT Solutions for Chapter 4 The Age of Industrialisation Class 10 History

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## Class 10th Maths - Real Number Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Real Number, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

## QB365 - Question Bank Software

Real number case study questions with answer key.

10th Standard CBSE

Final Semester - June 2015

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?

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

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

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

(v) If n is any odd integer, then n2 - 1 is divisible by

Real numbers are extremely useful in everyday life. That is probably one of the main reasons we all learn how to count and add and subtract from a very young age. Real numbers help us to count and to measure out quantities of different items in various fields like retail, buying, catering, publishing etc. Every normal person uses real numbers in his daily life. After knowing the importance of real numbers, try and improve your knowledge about them by answering the following questions on real life based situations. (i) Three people go for a morning walk together from the same place. Their steps measure 80 cm, 85 cm, and 90 cm respectively. What is the minimum distance travelled when they meet at first time after starting the walk assuming that their walking speed is same?

(ii) In a school Independence Day parade, a group of 594 students need to march behind a band of 189 members. The two groups have to march in the same number of columns. What is the maximum number of columns in which they can march?

(iii) Two tankers contain 768litres and 420 litres of fuel respectively. Find the maximum capacity of the container which can measure the fuel of either tanker exactly.

(iv) The dimensions of a room are 8 m 25 cm, 6 m 75 crn and 4 m 50 cm. Find the length of the largest measuring rod which can measure the dimensions of room exactly.

(v) Pens are sold in pack of 8 and notepads are sold in pack of 12. Find the least number of pack of each type that one should buy so that there are equal number of pens and notepads

In a classroom activity on real numbers, the students have to pick a number card from a pile and frame question on it if it is not a rational number for the rest of the class. The number cards picked up by first 5 students and their questions on the numbers for the rest of the class are as shown below. Answer them. (i) Suraj picked up \(\sqrt{8}\) and his question was - Which of the following is true about \(\sqrt{8}\) ?

(ii) Shreya picked up 'BONUS' and her question was - Which of the following is not irrational?

(iii) Ananya picked up \(\sqrt{5}\) -. \(\sqrt{10}\) and her question was - \(\sqrt{5}\) -. \(\sqrt{10}\) _________is number.

(iv) Suman picked up \(\frac{1}{\sqrt{5}}\) and her question was - \(\frac{1}{\sqrt{5}}\) is __________ number.

(v) Preethi picked up \(\sqrt{6}\) and her question was - Which of the following is not irrational?

Decimal form of rational numbers can be classified into two types. (i) Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form \(\frac{p}{\sqrt{q}}\) where p and q are co-prime and the prime faetorisation of q is of the form 2 n ·5 m , where n, mare non-negative integers and vice-versa. (ii) Let x = \(\frac{p}{\sqrt{q}}\) be a rational number, such that the prime faetorisation of q is not of the form 2 n 5 m , where n and m are non-negative integers. Then x has a non-terminating repeating decimal expansion. (i) Which of the following rational numbers have a terminating decimal expansion?

(ii) 23/(2 3 x 5 2 ) =

(iii) 441/(2 2 x 5 7 x 7 2 ) is a_________decimal.

(iv) For which of the following value(s) of p, 251/(2 3 x p 2 ) is a non-terminating recurring decimal?

(v) 241/(2 5 x 5 3 ) is a _________decimal.

HCF and LCM are widely used in number system especially in real numbers in finding relationship between different numbers and their general forms. Also, product of two positive integers is equal to the product of their HCF and LCM. Based on the above information answer the following questions. (i) If two positive integers x and yare expressible in terms of primes as x = p2q3 and y = p3 q, then which of the following is true?

(ii) A boy with collection of marbles realizes that if he makes a group of 5 or 6 marbles, there are always two marbles left, then which of the following is correct if the number of marbles is p?

(iii) Find the largest possible positive integer that will divide 398, 436 and 542 leaving remainder 7, 11, 15 respectively.

(iv) Find the least positive integer which on adding 1 is exactly divisible by 126 and 600.

(v) If A, Band C are three rational numbers such that 85C - 340A :::109, 425A + 85B = 146, then the sum of A, B and C is divisible by

(ii) Find the LCM of 60, 84 and 108.

(iii) Find the HCF of 60, 84 and 108.

(iv) Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

(v) Based on the above (iv) conditions, find the minimum number of rooms required for all the participants and officials.

(b) Find the total number of stacks formed.

(c) How many stacks of Mathematics books will be formed?

(d) If the thickness of each English book is 3 cm, then the height of each stack of English books is

(e) If each Hindi book weighs 1.5 kg, then find the weight of books in a stack of Hindi books.

## *****************************************

Real number case study questions with answer key answer keys.

(i) (d) : For a number to end in zero it must be divisible by 5, but 4 n = 22 n is never divisible by 5. So, 4 n never ends in zero for any value of n. (ii) (c) : We know that product of two rational numbers is also a rational number. So, a 2 = a x a = rational number a 3 = a 2 x a = rational number a 4 = a 3 x a = rational number ................................................ ............................................... a n = a n-1 x a = rational number. (iii) (d): Let x = 2m + 1 and y = 2k + 1 Then x 2 + y 2 = (2m + 1) 2 + (2k + 1) 2 = 4m 2 + 4m + 1 + 4k 2 + 4k + 1 = 4(m 2 + k 2 + m + k) + 2 So, it is even but not divisible by 4. (iv) (a): Let three consecutive positive integers be n, n + 1 and n + 2. We know that when a number is divided by 3, the remainder obtained is either 0 or 1 or 2. So, n = 3p or 3p + lor 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3. If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3. So, we can say that one of the numbers among n, n + 1 and n + 2 Wi always divisible by 3. (v) (d): Any odd number is of the form of (2k +1), where k is any integer. So, n 2 - 1 = (2k + 1)2 -1 = 4k 2 + 4k For k = 1, 4k 2 + 4k = 8, which is divisible by 8. Similarly, for k = 2, 4k 2 + 4k = 24, which is divisible by 8. And for k = 3, 4k 2 + 4k = 48, which is also divisible by 8. So, 4k 2 + 4k is divisible by 8 for all integers k, i.e., n 2 - 1 is divisible by 8 for all odd values of n.

(i) (b): Here 80 = 2 4 x 5, 85 = 17 x 5 and 90 = 2 x 3 2 x 5 L.C.M of 80, 85 and 90 = 2 4 x 3 x 3 x 5 x 17 = 12240 Hence, the minimum distance each should walk when they at first time is 12240 cm. (ii) (c): Here 594 = 2 x 3 3 x 11 and 189 = 3 3 x 7 HCF of 594 and 189 = 3 3 = 27 Hence, the maximum number of columns in which they can march is 27. (iii) (c) : Here 768 = 2 8 x 3 and 420 = 2 2 x 3 x 5 x 7 HCF of 768 and 420 = 2 2 x 3 = 12 So, the container which can measure fuel of either tanker exactly must be of 12litres. (iv) (b): Here, Length = 825 ern, Breadth = 675 cm and Height = 450 cm Also, 825 = 5 x 5 x 3 x 11 , 675 = 5 x 5 x 3 x 3 x 3 and 450 = 2 x 3 x 3 x 5 x 5 HCF = 5 x 5 x 3 = 75 Therefore, the length of the longest rod which can measure the three dimensions of the room exactly is 75cm. (v) (a): LCM of 8 and 12 is 24. \(\therefore \) The least number of pack of pens = 24/8 = 3 \(\therefore \) The least number of pack of note pads = 24/12 = 2

(i) (b): Here \(\sqrt{8}\) = 2 \(\sqrt{2}\) = product of rational and irrational numbers = irrational number (ii) (c): Here, \(\sqrt{9}\) = 3 So, 2 + 2 \(\sqrt{9}\) = 2 + 6 = 8 , which is not irrational. (iii) (b): Here. \(\sqrt{15}\) and \(\sqrt{10}\) are both irrational and difference of two irrational numbers is also irrational. (iv) (c): As \(\sqrt{5}\) is irrational, so its reciprocal is also irrational. (v) (d): We know that \(\sqrt{6}\) is irrational. So, 15 + 3. \(\sqrt{6}\) is irrational. Similarly, \(\sqrt{24}\) - 9 = 2. \(\sqrt{6}\) - 9 is irrational. And 5 \(\sqrt{150}\) = 5 x 5. \(\sqrt{6}\) = 25 \(\sqrt{6}\) is irrational.

(i) (c): Here, the simplest form of given options are 125/441 = 5 3 /(3 2 x 7 2 ), 77/210 = 11/(2 x 3 x 5), 15/1600 = 3/(2 6 x 5) Out of all the given options, the denominator of option (c) alone has only 2 and 5 as factors. So, it is a terminating decimal. (ii) (b): 23/(2 3 x 5 2 ) = 23/200 = 0.115 (iii) (a): 441/(2 2 x 5 7 x 7 2 ) = 9/(2 2 x 5 7 ), which is a terminating decimal. (iv) (d): The fraction form of a non-terminating recurring decimal will have at least one prime number other than 2 and 5 as its factors in denominator. So, p can take either of 3, 7 or 15. (v) (a): Here denominator has only two prime factors i.e., 2 and 5 and hence it is a terminating decimal.

(i) (b): LCM of x and y = p 3 q 3 and HCF of x and y = p 2 q Also, LCM = pq 2 x HCF. (ii) (d): Number of marbles = 5m + 2 or 6n + 2. Thus, number of marbles, p = (multiple of 5 x 6) + 2 = 30k + 2 = 2(15k + 1) = which is an even number but not prime (iii) (d): Here, required numbers = HCF (398 - 7, 436 - 11,542 -15) = HCF (391,425,527) = 17 (iv) (b): LCMof126and600 = 2 x 3 x 21 x 100= 12600 The least positive integer which on adding 1 is exactly divisible by 126 and 600 = 12600 - 1 = 12599 (v) (a): Here 8SC - 340A = 109 and 425A + 85B = 146 On adding them, we get 85A + 85B + 85C = 255 ~ A + B + C = 3, which is divisible by 3.

(i) (d): Total number of participants = 60 + 84 + 108 = 252 (ii) (d): 60 = 22 x 3 x 5 84 = 22 x 3 x 7 108 = 22 x 33 LCM(60, 84, 108) = 22 x 33 x 5 x 7 = 3780 (iii) (a): 60 = 22 x 3 x 5 84 = 22 x 3 x 7 108 = 22 x 33 HCF(60, 84, 108) = 22 x 3 = 12 (iv) (c): Minimum number of rooms required for all the participants = 252/12 = 21 (v) (d): Minimum number of rooms required for all = 21 + 1 = 22

(a) (ii) 96 = 2 5 x 3 240 = 2 4 x3 x5 (b) (iii) Total number of books = 96 +240+336=672 Number of books in each stack = 48 \(\therefore\) Number of stacks formed -= \(\frac{672}{48}=14\) (c) (i) Number of mathmatics books = 336 Number of stacks of mathematics books formed = \(\frac{336}{48}\) = 7 (d) (iv) Number of books in each stack of english books = 48 Thickness of each english book = 3 cm \(\therefore\) Height of each stack of english books = (48X3) cm = 144cm (e) (iii) Number of books in a stack of hindi books = 48 Weight of each hindi book = 1.5kg \(\therefore\) The weight of books in a stack of hindi books = (48X1.5)kg = 72kg

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## Chapter 1 Class 10 Real Numbers

Click on any of the links below to start learning from Teachoo ...

Updated for NCERT 2023-2024 Book.

Answers to all exercise questions and examples are solved for Chapter 1 Class 10 Real numbers. Solutions of all these NCERT Questions are explained in a step-by-step easy to understand manner

In this chapter, we will study

- What is a Real Number
- What is Euclid's Division Lemma , and
- How to find HCF (Highest Common Factor) using Euclid's Division Algorithm
- Then, we study Fundamental Theorem of Arithmetic, which is basically Prime Factorisation
- And find HCF and LCM using Prime Factorisation
- We also use the formula of HCF and LCM of two numbers a and b HCF × LCM = a × b
- Then, we see what is an Irrational Number
- and Prove numbers irrational (Like Prove √ 2, √ 3 irrational)
- We revise our concepts about Decimal Expansion (Terminating, Non-Terminating Repeating, Non Terminating Non Repeating)
- And find out Decimal Expansion of numbers without performing long division

Click on an NCERT Exercise below to get started.

Or you can also check the Concepts from the concept Wise. The chapter is divided into concepts, and first each concept is explained. And then, the questions of the concept is solved, from easy to difficult. This is the Teachoo way of learning.

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- Real Numbers For Class 10

## Real Numbers Class 10 Notes: Chapter 1

Cbse real numbers class 10 notes:- download pdf here, class 10 maths chapter 1 real number notes.

CBSE Class 10 Maths Chapter 1 Real Numbers Notes are provided here in detail. As we all know, any number, excluding complex numbers, is a real number. Positive and negative integers, irrational numbers, and fractions are all examples of real numbers. To put it another way, any number found in the real world is a real number. Numbers can be found all around us. Natural numbers are being used to count objects, integers are used to measure temperature, rational numbers are used to represent fractions, irrational numbers are used to calculate the square root of a number, etc. These various types of numbers form a collection of real numbers. Here, we are going to learn what a real number is, Euclid’s division algorithm, the fundamental theorem of arithmetic, methods of finding LCM and HCF and the complete explanation of rational and irrational numbers with examples.

## Real Numbers

Positive integers, negative integers, irrational numbers, and fractions are all examples of real numbers. In other words, we can say that any number is a real number, except for complex numbers. Examples of real numbers include -1, ½, 1.75, √2, and so on. In general,

- Real numbers constitute the union of all rational and irrational numbers.
- Any real number can be plotted on the number line.

To know more about real numbers, visit here .

Students can refer to the short notes and MCQ questions along with a separate solution pdf of this chapter for quick revision from the links below.

- Real Numbers Short Notes
- Real Numbers MCQ Practice Questions
- Real Numbers MCQ Practice Solutions

## Euclid’s Division Lemma

- Euclid’s Division Lemma states that given two integers a and b , there exists a unique pair of integers q and r such that a = b × q + r a n d 0 ≤ r < b .
- This lemma is essentially equivalent to : dividend = divisor × quotient + remainder
- In other words, for a given pair of dividend and divisor, the quotient and remainder obtained are going to be unique.

## For more information on Euclid’s Division Lemma, watch the below video

To know more about Euclid’s Division Lemma, visit here .

## Euclid’s Division Algorithm

- Euclid’s Division Algorithm is a method used to find the H.C.F of two numbers, say a and b where a> b.
- We apply Euclid’s Division Lemma to find two integers q and r such that a = b × q + r a n d 0 ≤ r < b .
- If r = 0, the H.C.F is b; else, we apply Euclid’s division Lemma to b (the divisor) and r (the remainder) to get another pair of quotient and remainder.
- The above method is repeated until a remainder of zero is obtained. The divisor in that step is the H.C.F. of the given set of numbers.

## For more information on Euclid’s Division Algorithm, watch the below video

## The Fundamental Theorem of Arithmetic

## Prime Factorisation

- Prime Factorisation is the method of expressing a natural number as a product of prime numbers.
- Example: 36 = 2 × 2 × 3 × 3 is the prime factorisation of 36.

## Fundamental Theorem of Arithmetic

- The Fundamental Theorem of Arithmetic states that the prime factorisation for a given number is unique if the arrangement of the prime factors is ignored.
- Example: 36 = 2 × 2 × 3 × 3 OR, 36 = 2 × 3 × 2 × 3
- Therefore, 36 is represented as a product of prime factors (Two 2s and two 3s) ignoring the arrangement of the factors.

To know more about the Fundamental Theorem of Arithmetic, visit here .

## Method of Finding LCM

As we know, the smallest of the common multiples of two or more numbers is called the lowest common multiple (LCM). Example: To find the Least Common Multiple ( L.C.M ) of 36 and 56,

- 36 = 2 × 2 × 3 × 3 56 = 2 × 2 × 2 × 7
- The common prime factors are 2 × 2
- The uncommon prime factors are 3 × 3 for 36 and 2 × 7 for 56.
- LCM of 36 and 56 = 2 × 2 × 3 × 3 × 2 × 7 which is 504

To know more about LCM, visit here .

## Method of Finding HCF

We know that the greatest number that divides each of the given numbers without leaving any remainder is the highest common factor (HCF) of two or more given numbers. H.C.F can be found using two methods – Prime factorisation and Euclid’s division algorithm.

- Given two numbers, we express both of them as products of their respective prime factors. Then, we select the prime factors that are common to both the numbers
- Example – To find the H.C.F of 20 and 24 20 = 2 × 2 × 5 and 24 = 2 × 2 × 2 × 3
- The factor common to 20 and 24 is 2 × 2 , which is 4, which in turn is the H.C.F of 20 and 24.
- It is the repeated use of Euclid’s division lemma to find the H.C.F of two numbers.

- The required HCF is 6 .

To know more about HCF, visit here .

## For more information on HCF and LCM, watch the below video

To know more about the Properties of HCF and LCM, visit here .

## Product of Two Numbers = HCF X LCM of the Two Numbers

- For any two positive integers a and b, a × b = H . C . F × L . C . M .
- Example – For 36 and 56, the H.C.F is 4 and the L.C.M is 504 36 × 56 = 2016 4 × 504 = 2016 Thus, 36 × 56 = 4 × 504
- Let us consider another example: For 5 and 6, the H.C.F is 1 and the L.C.M is 30 5 × 6 = 30 1 × 30 =30 Thus, 5 × 6 = 1 × 30
- The above relationship, however, doesn’t hold true for 3 or more numbers

To know the Relationship between LCM and HCF, visit here .

## Applications of HCF & LCM in Real-World Problems

L.C.M can be used to find the points of common occurrence. For example,ringing of bells that ring with different frequencies, the time at which two persons running at different speeds meet, and so on.

## For more information on Applications Of LCM, watch the below video

## Revisiting Irrational Numbers

Irrational numbers.

Any number that cannot be expressed in the form of p/q (where p and q are integers and q ≠ 0 .) is an irrational number. Examples √2 , π , e and so on.

To know more about Irrational Numbers, visit here .

## Number theory: Interesting results

- If a number p (a prime number) divides a 2 , then p divides a. Example: 3 divides 6 2 i.e 36, which implies that 3 divides 6.
- The sum or difference of a rational and an irrational number is irrational
- The product and quotient of a non-zero rational and irrational number are irrational.
- √p is irrational when ‘p’ is a prime. For example, 7 is a prime number, and √7 is irrational. The above statement can be proved by the method of “Proof by contradiction”.

To know more about Number theory, visit here .

## Proof by Contradiction

In the method of contradiction, to check whether a statement is TRUE (i) We assume that the given statement is TRUE. (ii) We arrive at some result which contradicts our assumption, thereby proving the contrary. Eg: Prove that √7 is irrational. Assumption: √7 is rational. Since it is rational √7 can be expressed as √7 = a/b , where a and b are co-prime Integers, b ≠ 0. On squaring, a 2 /b 2 = 7 ⇒ a 2 = 7 b 2 . Hence, 7 divides a. Then, there exists a number c such that a=7c. Then, a 2 = 49 c 2 . Hence, 7 b 2 = 49 c 2 or b 2 = 7 c 2 . Hence 7 divides b. Since 7 is a common factor for both a and b, it contradicts our assumption that a and b are coprime integers. Hence, our initial assumption that √7 is rational is wrong. Therefore, √7 is irrational.

## Revisiting Rational Numbers and Their Decimal Expansions

Rational numbers.

Rational numbers are numbers that can be written in the form p/q, where p and q are integers and q ≠ 0 . Examples -1/2 , 4/5, 1 , 0 , − 3 and so on.

To know more about Rational Numbers, visit here .

## Terminating and Non-Terminating Decimals

Terminating decimals are decimals that end at a certain point. Example: 0.2, 2.56 and so on. Non-terminating decimals are decimals where the digits after the decimal point don’t terminate. Example: 0.333333….., 0.13135235343… Non-terminating decimals can be : a) Recurring – a part of the decimal repeats indefinitely (0 . 142857 142857 …. ) b) Non-recurring – no part of the decimal repeats indefinitely. Example: π = 3.1415926535…

To know more about terminating and non-terminating decimals, click here .

## Check if a given rational number is terminating or not

If a/b is a rational number, then its decimal expansion would terminate if both of the following conditions are satisfied : a) The H.C.F of a and b is 1. b) b can be expressed as a prime factorisation of 2 and 5 i.e b = 2 m × 5 n where either m or n, or both can = 0. If the prime factorisation of b contains any number other than 2 or 5, then the decimal expansion of that number will be recurring

1/40 = 0.025 is a terminating decimal, as the H.C.F of 1 and 40 is 1, and the denominator (40) can be expressed as 2 3 × 5 1 .

3/7 = 0.428571 is a recurring decimal as the H.C.F of 3 and 7 is 1 and the denominator (7) is equal to 7 1

## Real Numbers for Class 10 Solved Examples

Find the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively.

First, subtract the remainder from the number.

(i.e) 70-5 = 65

125-8 = 117.

Thus, we need to find the largest number that divides 65 and 117 and leaves the remainder 0.

To find the largest number, take the HCF of 65 and 117.

Finding HCF of 65 and 117.

65 = 5×13

117 = 3×3×13.

Hence, HCF (65, 117) = 13.

Therefore, the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively is 13.

Find the LCM of 306 and 657, given that HCF (306, 657) = 9.

Given that, HCF (306, 657) = 9.

We know that HCF × LCM = Product of Numbers

Hence, 9×LCM = 306×657

9×LCM = 201042

LCM = 201042/9

LCM = 22338.

Therefore, LCM of 306 and 657 is 22338.

Prove that 1/√2 is an irrational number.

To prove 1/√2 is an irrational number.

Now, let us take the opposite assumption.

(i.e) Take 1/√2 is a rational number.

We know that rational numbers are the numbers that can be written in the form of p/q, where q is not equal to 0. (p and q are two co-prime numbers)

Hence, 1/√2 = p/q.

Now, simplify the above equation by multiplying √2 on both sides.

1 = (p√2)/q

q = p√2

Hence, we get q/p = √2.

Here, p and q are integers, and hence q/p is a rational number.

But, √2 is an irrational number.

Hence, our assumption is wrong.

Therefore, 1/√2 is an irrational number.

Hence, proved.

## Frequently Asked Questions on Class 10 Real Numbers

What is euclid’s division algorithm, what does real numbers for class 10 explains, is 10i a real number, what does the fundamental theory of arithmetic explain.

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## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed the fundamental theorem of arithmetic and Euclid’s division lemma in details. CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

## CBSE Class 10 Maths Chapter Real Numbers Solutions

Below we have given the answers to all the questions present in Real Numbers in our NCERT Solutions for Class 10 Maths chapter 1. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

## NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1

## NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2

## NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.3

## NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.4

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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.

REAL NUMBERS- CASE STUDY CASE STUDY 1. 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 ...

In this post, you will get CASE Study Questions of Chapter 1 (Real Numbers) of Class 10th. These Case study Questions are based on the Latest Syllabus for 2020- 21 of the CBSE Board. Chapter 1 (Real numbers)

CBSE 10th Standard Maths Subject Real Number Case Study Questions With Solution 2021 Answer Keys. Case Study Questions. (i) (d) : For a number to end in zero it must be divisible by 5, but 4 n = 22 n is never divisible by 5. So, 4 n never ends in zero for any value of n. (ii) (c) : We know that product of two rational numbers is also a rational ...

Show Answer. (v) If A, B and C are three rational numbers such that 85C - 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by. (a) 3. (b) 6. (c) 7. (d) 9. Show Answer. Case Study 3: Real numbers are an essential concept in mathematics that encompasses both rational and irrational numbers.

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Real Numbers in order to fully complete your preparation.They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!. I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams.

Here, we have provided case-based/passage-based questions for Class 10 Maths Chapter 1 Real Numbers. Case Study/Passage-Based Questions. Question 1: 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 ...

Real Numbers Case Study Questions (CSQ's) Practice Tests. Timed Tests. Select the number of questions for the test: Select the number of questions for the test: TopperLearning provides a complete collection of case studies for CBSE Class 10 Maths Real Numbers chapter. Improve your understanding of biological concepts and develop problem ...

Students looking for Case Study on Real Numbers Class 10 Maths can use this page to download the PDF file. The case study questions on Real Numbers are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Real Numbers case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 ...

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers: Download PDF for Free and study offline. Clear doubts on Real Numbers of Class 10 Maths and excel in your exam. Register at BYJU'S for NCERT Solutions! ... Case (i): When r = 0, then, x 2 = (3q) 3 = 27q 3 = 9(3q 3)= 9m; where m = 3q 3.

Case Study Questions for Class 10 Maths Chapter 1 Real Numbers. Question 1: HCF and LCM are widely used in number system especially in real numbers in finding relationship between different numbers and their general forms. Also, product of two positive integers is equal to the product of their HCF and LCM. Based on the above information answer ...

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam of 2022-23. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards. ... Chapter-1 Real Numbers. Starting with an introduction to ...

Solution: Numbers are of two types - prime and composite. Prime numbers consist of only two factors namely 1 and the number itself while composite numbers consist of factors besides 1 and itself. It can be observed that. 7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) (taking 13 out as common) = 13 x (77 + 1) = 13 x 78. = 13 x 13 x 6.

Q.7: Give an example to show that the product of a rational number and an irrational number may be a rational number. Q.8: Prove that √3 - √2 and √3 + √5 are irrational. Q.9: Express 7/64, 12/125 and 451/13 in decimal form. Q.10: Find two irrational numbers lying between √2 and √3.

Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022. CBSE Class 10 Maths Chapter Wise Case Study. Maths Chapter 1 Real Number Case Study. Maths Chapter 2 Polynomial Case Study. Maths Chapter 3 Pair of Linear Equations in Two Variables Case Study. Maths Chapter 4 Quadratic Equations Case Study.

Class 10 Maths Chapter 1 Real Numbers MCQ (Multiple Choice Objective Questions) with answers and complete explanation case study type questions for the first term examination 2024-25. The answers of 10th Maths Chapter 1 MCQ are given with explanation, so that students can understand easily.

∵ The total number of members are to march behind an army band of 32 members. HCF of 616 and 32 is equal to the maximum number of columns such that the two groups can march in the same number of columns. ∴ Applying Euclid's lemma to 616 and 32, we get 616 = (32 × 19) + 8 ∵ 8 ≠ 0 ∴ Again, applying Euclid's lemma to 32 and 8, we get

Class 10th Maths - Real Number Case Study Questions and Answers 2022 - 2023. QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Real Number, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

Updated for NCERT 2023-2024 Book. Answers to all exercise questions and examples are solved for Chapter 1 Class 10 Real numbers. Solutions of all these NCERT Questions are explained in a step-by-step easy to understand manner. In this chapter, we will study. Click on an NCERT Exercise below to get started.

CBSE Class 10 Maths Chapter 1 Real Numbers Notes are provided here in detail. As we all know, any number, excluding complex numbers, is a real number. Positive and negative integers, irrational numbers, and fractions are all examples of real numbers. To put it another way, any number found in the real world is a real number.

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed the fundamental theorem of arithmetic and Euclid's division lemma in details.

Answer. 38. Euclid's division lemma states that for two positive integers a and b, there exist unique integer q and r such that a = bq + r, where r must satisfy. (a) a < r < b. (b) 0 < r ≤ b. (c) 1 < r < b. (d) 0 ≤ r < b. Answer. We hope the given MCQ Questions for Class 10 Maths Real Numbers with Answers will help you.

A Client Case Study One of our clients, a premier window and door installer with a new franchise in Raleigh, North Carolina, faced a challenging competitive environment dominated by established ...

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