case study of real numbers class 10

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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

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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 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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• NCERT Solutions
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• Chapter 1: Real Numbers

NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers are provided here. These solutions are prepared by our expert faculty to help students in their board exam preparations. They solve and provide the NCERT Solutions for Maths to aid the students in solving the problems easily. They also focus on preparing the solutions in such a way that it is easy to understand for the students. A detailed and step-wise explanation is given for each question given in the exercises of NCERT books.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 1 Real Numbers

Download most important questions for class 10 maths chapter – 1 real numbers.

Answers to the questions present in Real Numbers are given in the first chapter of Maths Solutions of NCERT Class 10. Here, students are introduced to several important concepts that will be useful for those who wish to pursue mathematics as a subject in their Class 11. Based on these solutions of Class 10 NCERT , students can prepare for their upcoming board exam. These solutions are helpful as they are in accordance with the CBSE Syllabus for 2023-24.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Exercise 1.1 Chapter 1 Real Numbers
• Exercise 1.2 Chapter 1 Real Numbers
• Exercise 1.3 Chapter 1 Real Numbers
• Exercise 1.4 Chapter 1 Real Numbers

Download PDF of NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers

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Access Answers to NCERT Class 10 Maths Chapter 1 – Real Numbers

Exercise 1.1 page: 7.

1. Use Euclid’s division algorithm to find the HCF of:

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

As you can see from the question, 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,

135 = 90 × 1 + 45

Again, 45 ≠ 0, repeating the above step for 45, we get,

90 = 45 × 2 + 0

The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

Hence, the HCF of 225 and 135 is 45.

In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,

38220 = 196 × 195 + 0

We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.

Hence, the HCF of 196 and 38220 is 196.

As we know, 867 is greater than 255. Let us apply now Euclid’s division algorithm on 867, to get,

867 = 255 × 3 + 102

Remainder 102 ≠ 0, therefore taking 255 as divisor and applying the division lemma method, we get,

255 = 102 × 2 + 51

Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,

102 = 51 × 2 + 0

The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,255) = HCF(255,102) = HCF(102,51) = 51.

Hence, the HCF of 867 and 255 is 51.

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Number of army contingent members = 616

Number of army band members = 32

If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF(616, 32), gives the maximum number of columns in which they can march.

By using Euclid’s algorithm to find their HCF, we get,

Since, 616>32, therefore,

616 = 32 × 19 + 8

Since, 8 ≠ 0, therefore, taking 32 as new divisor, we have,

32 = 8 × 4 + 0

Now we have got remainder as 0, therefore, HCF (616, 32) = 8.

Hence, the maximum number of columns in which they can march is 8.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, x = 3q, 3q+1 and 3q+2

Now as per the question given, by squaring both the sides, we get,

x 2 = (3q) 2 = 9q 2 = 3 × 3q 2

Let 3q 2 = m

Therefore, x 2 = 3m ……………………..(1)

x 2 = (3q + 1) 2 = (3q) 2 +1 2 +2×3q×1 = 9q 2 + 1 +6q = 3(3q 2 +2q) +1

Substitute, 3q 2 +2q = m, to get,

x 2 = 3m + 1 ……………………………. (2)

x 2 = (3q + 2) 2 = (3q) 2 +2 2 +2×3q×2 = 9q 2 + 4 + 12q = 3 (3q 2 + 4q + 1)+1

Again, substitute, 3q 2 +4q+1 = m, to get,

x 2 = 3m + 1…………………………… (3)

Hence, from equation 1, 2 and 3, we can say that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get,

Now, by taking the cube of all the three above expressions, we get,

Case (i): When r = 0, then,

x 2 = (3q) 3 = 27q 3 = 9(3q 3 )= 9m; where m = 3q 3

Case (ii): When r = 1, then,

x 3 = (3q+1) 3 = (3q) 3 +1 3 +3×3q×1(3q+1) = 27q 3 +1+27q 2 +9q

Taking 9 as common factor, we get,

x 3 = 9(3q 3 +3q 2 +q)+1

Putting = m, we get,

Putting (3q 3 +3q 2+ q) = m, we get ,

Case (iii): When r = 2, then,

x 3 = (3q+2) 3 = (3q) 3 +2 3 +3×3q×2(3q+2) = 27q 3 +54q 2 +36q+8

x 3 =9(3q 3 +6q 2 +4q)+8

Putting (3q 3 +6q 2 +4q) = m, we get ,

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Exercise 1.2 Page: 11

1. Express each number as a product of its prime factors:

By taking the LCM of 140, we will get the product of its prime factor.

Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2 2 ×5×7

By Taking the LCM of 156, we will get the product of its prime factor.

Hence, 156 = 2 × 2 × 13 × 3 × 1 = 2 2 × 13 × 3

By taking the LCM of 3825, we will get the product of its prime factor.

Hence, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3 2 ×5 2 ×17

By Taking the LCM of 5005, we will get the product of its prime factor.

Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

By taking the LCM of 7429, we will get the product of its prime factor.

Hence, 7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Expressing 26 and 91 as product of its prime factors, we get,

26 = 2 × 13 × 1

91 = 7 × 13 × 1

Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182

And HCF (26, 91) = 13

Verification

Now, product of 26 and 91 = 26 × 91 = 2366

And product of LCM and HCF = 182 × 13 = 2366

Hence, LCM × HCF = product of the 26 and 91.

Expressing 510 and 92 as product of its prime factors, we get,

510 = 2 × 3 × 17 × 5 × 1

92 = 2 × 2 × 23 × 1

Therefore, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

And HCF (510, 92) = 2

Now, product of 510 and 92 = 510 × 92 = 46920

And Product of LCM and HCF = 23460 × 2 = 46920

Hence, LCM × HCF = product of the 510 and 92.

Expressing 336 and 54 as product of its prime factors, we get,

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

54 = 2 × 3 × 3 × 3 × 1

Therefore, LCM(336, 54) = = 3024

And HCF(336, 54) = 2×3 = 6

Now, product of 336 and 54 = 336 × 54 = 18,144

And product of LCM and HCF = 3024 × 6 = 18,144

Hence, LCM × HCF = product of the 336 and 54.

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Writing the product of prime factors for all the three numbers, we get,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

HCF(17,23,29) = 1

LCM(17,23,29) = 17 × 23 × 29 = 11339

HCF(8,9,25)=1

LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

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

Solution: As we know that,

HCF×LCM=Product of the two given numbers

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

Hence, LCM(306,657) = 22338

5. Check whether 6 n can end with the digit 0 for any natural number n.

Solution: If the number 6 n ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.

Prime factorization of 6 n = (2×3) n

Therefore, the prime factorization of 6 n doesn’t contain prime number 5.

Hence, it is clear that for any natural number n, 6 n is not divisible by 5, and thus it proves that 6 n cannot end with the digit 0 for any natural number n.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution: By the definition of composite number, we know, if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression;

7 × 11 × 13 + 13

Taking 13 as common factor, we get,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take the other number,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 as a common factor, we get,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution: Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.

Therefore, LCM(18,12) = 2×3×3×2×1=36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

Exercise 1.3 Page: 14

1. Prove that √ 5 is irrational.

Solutions: Let us assume, that √ 5 is rational number.

i.e.  √ 5 = x/y (where, x and y are co-primes)

Squaring both the sides, we get,

(y √ 5) 2 = x 2

⇒5y 2 = x 2 ……………………………….. (1)

Thus, x 2 is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y 2 = (5k) 2

⇒y 2 = 5k 2

is divisible by 5 it means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about  √ 5 is rational is incorrect.

Hence,  √ 5 is an irrational number.

2. Prove that 3 + 2√5 + is irrational.

Solutions: Let us assume 3 + 2 √ 5 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y

Rearranging, we get,

Since, x and y are integers, thus,

Therefore, √ 5 is also a rational number. But this contradicts the fact that √ 5 is irrational.

So, we conclude that 3 + 2 √ 5 is irrational.

3. Prove that the following are irrationals:

(iii) 6 + √ 2

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

Hence, we can conclude that 1/√2 is irrational.

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Hence, we can conclude that 7√5 is irrational.

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

√2 = (x/y) – 6

Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.

Exercise 1.4 Page: 17

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2 3 5 2 ) (vii) 129/(2 2 5 7 7 5 ) (viii) 6/15 (ix) 35/50 (x) 77/210

Note: If the denominator has only factors of 2 and 5 or in the form of 2 m ×5 n then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

(i) 13/3125

Factorizing the denominator, we get,

3125 = 5 × 5 × 5 × 5 × 5 = 5 5

Since, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion.

8 = 2×2×2 = 2 3

Since, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion.

(iii) 64/455

455 = 5×7×13

Since, the denominator is not in the form of 2 m × 5 n , thus 64/455 has a non-terminating decimal expansion.

(iv) 15/ 1600

1600 = 2 6 ×5 2

Since, the denominator is in the form of 2 m × 5 n , thus 15/1600 has a terminating decimal expansion.

343 = 7×7×7 = 7 3 Since, the denominator is not in the form of 2 m × 5 n thus 29/343 has a non-terminating decimal expansion.

(vi)23/(2 3 5 2 )

Clearly, the denominator is in the form of 2 m × 5 n .

Hence, 23/ (2 3 5 2 ) has a terminating decimal expansion.

(vii) 129/(2 2 5 7 7 5 )

As you can see, the denominator is not in the form of 2 m × 5 n .

Hence, 129/ (2 2 5 7 7 5 ) has a non-terminating decimal expansion.

(viii) 6/15

Since, the denominator has only 5 as its factor, thus, 6/15 has a terminating decimal expansion.

35/50 = 7/10

Factorising the denominator, we get,

Since, the denominator is in the form of 2 m × 5 n thus, 35/50 has a terminating decimal expansion.

77/210 = (7× 11)/ (30 × 7) = 11/30

30 = 2 × 3 × 5

As you can see, the denominator is not in the form of 2 m × 5 n .Hence, 77/210 has a non-terminating decimal expansion.

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

13/3125 = 0.00416

17/8 = 2.125

(iii) 64/455 has a non terminating decimal expansion

(iv)15/ 1600

15/1600 = 0.009375

(v) 29/ 343 has a non terminating decimal expansion

(vi)23/ (2 3 5 2 ) = 23/(8×25)= 23/200

23/ (2 3 5 2 ) = 0.115

(vii) 129/ (2 2 5 7 7 5 ) has a non terminating decimal expansion

(viii) 6/15 = 2/5

(ix) 35/50 = 7/10

35/50 = 0.7

(x) 77/210 has a non-terminating decimal expansion.

3. 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, p q what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000. . .

Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.

Since it has non-terminating but repeating decimal expansion, it is a rational number in the form of p/q and q has factors other than 2 and 5.

NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

Real Number is one of the important topics in Maths, and it has a weightage of 6 marks in Class 10 (Unit – Number Systems) Maths board exams. The average number of questions asked in this chapter is usually 3. Three questions were asked from this chapter in the previous year board examination (2018).

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

This chapter talks about

• Euclid’s Division Algorithm
• The Fundamental Theorem of Arithmetic
• Revisiting Rational & Irrational Numbers
• Decimal Expansions

List of Exercises in Class 10 Maths Chapter 1: Exercise 1.1 Solutions 5 Questions ( 4 long, 1 short) Exercise 1.2 Solutions 7 Questions ( 4 long, 3 short) Exercise 1.3 Solutions 3 Questions ( 3 short) Exercise 1.4 Solutions 3 Questions ( 3 short)

Real Numbers is introduced in Class 9, and this is discussed in further detail in Class 10. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers provides the answers to the questions present in this chapter. The chapter discusses real numbers and their applications. The divisibility of integers using Euclid’s division algorithm says that any positive integer a can be divided by another positive integer b such that the remainder will be smaller than b. On the other hand, The Fundamental Theorem of Arithmetic works on the multiplication of positive integers.

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

• Euclid’s Division Algorithm – It includes 5 questions based on Theorem 1.1 – Euclid’s Division Lemma.
• The Fundamental Theorem of Arithmetic – Explore the applications of this topic which talks about the multiplication of positive integers, through solutions of the 7 problems in Exercise 1.2.

Next, it discusses the following topics, which were introduced in Class 9.

• Revisiting Rational & Irrational Numbers – In this, the solutions for 3 problems in Exercise 1.3 are given, which also use the topic in the last Exercise 1.2.
• Decimal Expansions – It explores when the decimal expansion of a rational number is terminating and when it is recurring. It includes a total of 3 problems with sub-parts in Exercise 1.4

Key Features of NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers

• These NCERT Solutions help you solve and revise the updated CBSE syllabus of Class 10 for 2023-24.
• After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
• It follows NCERT guidelines which help in preparing the students competently.
• It contains all the important questions from the examination point of view.

Disclaimer –

Dropped Topics – 

1.2 Euclid’s division lemma 1.5 Revisiting rational numbers and their decimal Expansions

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Chapter 1 Class 10 Real Numbers

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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. 

You can check the concepts by clicking on a link below Concept Wise.

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Class 10 Maths Chapter 1 Case Based Questions - Real Numbers

Case Study – 1

Q1: What is the minimum distance each should walk so that they can cover the distance in complete steps? (a) 120 m 40 cm (b) 122 m 40 cm (c) 12 m 4 cm (d) None of these Ans: (b) Explanation:  The process of solving this problem involves finding the least common multiple (LCM) of the three given measurements. The LCM of a set of numbers is the smallest number that is a multiple of each of the numbers in the set. Here are the steps to find the LCM of 80 cm, 85 cm, and 90 cm: Step 1:  Prime factorization The first step is to find the prime factors of each number.

• For 80, the prime factors are 2, 2, 2, 2, and 5 (or 2⁴ х 5)
• For 85, the prime factors are 5 and 17
• For 90, the prime factors are 2, 3, 3, and 5 (or 2 х 3² х 5)

Step 2:  Find the LCM Now, we find the LCM by taking the highest power of each prime factor from all the numbers.

• The highest power of 2 is 2⁴ from 80
• The highest power of 3 is 3² from 90
• The highest power of 5 is 5 from 80 or 90
• The highest power of 17 is 17 from 85

So, the LCM is 2⁴ х 3² х 5 х 17 = 12240 Step 3:  Convert cm to m Finally, we convert the LCM from centimeters to meters. Since 1 meter is 100 cm, we divide 12240 by 100 to get 122.4 meters, which can also be written as 122 meters and 40 cm. Therefore, the minimum distance each should walk so that they can cover the distance in complete steps is 122 meters and 40 cm. This corresponds to option (b). Q2: What is the minimum number of steps taken by any of the three friends, when they meet again? (a) 120 (b) 125 (c) 130 (d) 136 Ans: (d) Explanation:   To solve this problem, we need to find the least common multiple (LCM) of the step sizes of the three friends: 80 cm, 85 cm, and 90 cm. The LCM will give us the smallest distance that all three friends can walk together, taking whole steps. Here is the step-by-step process: Step 1: Prime Factorization We start by breaking down each number into its prime factors.

• 80 = 2 x 2 x 2 x 2 x 5 = 2⁴ x 5
• 85 = 5 x 17
• 90 = 2 x 3 x 3 x 5 = 2 x 3² x 5

Step 2:  Find the LCM The LCM is found by taking the highest power of all the prime numbers that appear in the prime factorization of any of the numbers. LCM = 2⁴ x 3² x 5 x 17 = 12240 cm This means that the three friends will meet again after walking a distance of 12240 cm. Step 3: Determine the Minimum Steps Among the three friends, Angelina has the longest step size (90 cm). Therefore, she will take the smallest number of steps to cover the distance of 12240 cm. Number of steps taken by Angelina = Total distance / Step size = 12240 cm / 90 cm = 136 steps Hence, the correct answer is (d) 136 steps. Q3: The HCF of 80, 85, and 90 is (a) 5 (b) 10 (c) 12 (d) 18 Ans: (a) Explanation: The Highest Common Factor (HCF) of a set of numbers is the largest number that divides evenly into all the numbers in the set. In this case, we are looking for the HCF of 80, 85, and 90. The first step is to determine the prime factors of each of the numbers. Prime factors are the factors of a number that are prime numbers. 1. For 80, the prime factors are 2 and 5. We obtain this by dividing 80 by the smallest prime number (2) as many times as possible until we are left with a prime number. This gives us 2 x 2 x 2 x 2 x 5 = 2⁴ x 5. 2. For 85, the prime factors are 5 and 17. We obtain this by dividing 85 by the smallest prime number (2) as many times as possible until we are left with a prime number. This gives us 5 x 17. 3. For 90, the prime factors are 2, 3, and 5. We obtain this by dividing 90 by the smallest prime number (2) as many times as possible, then doing the same with the next smallest prime number (3) until we are left with a prime number. This gives us 2 x 3 x 3 x 5 = 2 x 3² x 5. Now that we have the prime factors of each number, we can determine the HCF by finding the largest number that is a factor of all three numbers. In this case, the only common factor among 80, 85, and 90 is 5. Therefore, the HCF of 80, 85, and 90 is 5, which corresponds to answer choice (a). Q4:  The product of HCF and LCM of 80, 85, and 90 is (a) 60400 (b) 61000 (c) 61200 (d) 65500 Ans:  (c) Explanation:   The problem requires us to find the product of the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the numbers 80, 85, and 90. Step 1:  To find the HCF and LCM, we first need to find the prime factors of the three numbers. For 80, the prime factors are 2 x 2 x 2 x 2 x 5 (or 2 4 x 5). For 85, the prime factors are 5 x 17. For 90, the prime factors are 2 x 3 x 3 x 5 (or 2 x 3 2 x 5). Step 2:  To find the HCF, we look for common prime factors. The only common factor among all three numbers is 5. So, HCF = 5. Step 3:  For the LCM, we take the highest power of all the prime numbers in the factorization of each number. So, LCM = 2 2  x 3 2 x 5 x 17 = 12240. Step 4:  Finally, we need to find the product of the HCF and LCM. This is done by multiplying the HCF (5) with the LCM (12240), which gives us 61200. So, the product of the HCF and LCM of 80, 85, and 90 is 61200. Therefore, the correct answer is option (C).   Q5: 90 can be expressed as a product of its primes as (a) 2 х 3² х 5² (b) 2 х 3³ х 5 (c) 2² х 3² х 5 (d) 2 х 3² х 5 Ans: (d) Explanation:  The question asks us to express 90 as a product of its prime factors. Prime factors are the factors of a number that are prime numbers. A prime number is a number that only has two factors: 1 and itself. Here are the steps to find the prime factors of 90: Step 1:  Start by dividing the number 90 with the smallest prime number, which is 2. 90 is divisible by 2. So, divide 90 by 2. You get 45. Step 2: Now, try dividing 45 by 2. It can't be divided evenly. So, we move to the next prime number, which is 3. 45 divided by 3 gives 15. Step 3:  Try dividing 15 by 3. It can't be divided evenly. So, we move to the next prime number, which is 5. 15 divided by 5 gives 3. Step 4:  Now, we are left with 3. 3 is a prime number itself, so we stop here. So, the prime factors of 90 are 2, 3, 3, and 5. We can write this as 2 x 3² x 5, which matches option (d). Therefore, the correct answer is (d).  

Case Study – 2

Q1: What will be the value of x? (a) 15005 (b) 13915 (c) 56920 (d) 17429 Ans:  (b) Explanation:  The factor tree is a method used to break down any given number into its prime factors. In this case, we don't have the factor tree visually, but the question suggests that 'x' can be obtained by multiplying the numbers 5 and 2783. Step-by-step process: Step 1: Identify the numbers given. Here, we have 5 and 2783. Step 2:  Multiply the given numbers. In this case, x = 5 * 2783 Step 3: Perform the multiplication. 5 * 2783 = 13915 So, by using these steps, we find that the value of 'x' is 13915. Therefore, the correct option is (b) 13915. Q2: What will be the value of y? (a) 23 (b) 22 (c) 11 (d) 19 Ans: (c) Explanation: The given factor tree shows how a number is broken down into its prime factors. The number at the top of the tree is the original number and the numbers at the bottom are all prime factors. In the question, we are not given the specific factor tree, but we are asked to find the value of 'y' given that Y = 2783/253. To solve this, we need to perform the division operation: 2783 divided by 253 equals to 11. Hence, the correct answer is option (c), i.e., y = 11.   Q3: What will be the value of z? (a) 22 (b) 23 (c) 17 (d) 19 Ans:  (b) Explanation:  The given factor tree is not explicitly provided here, but from the available solution, we can assume that the number 253 is divided by 11 on the factor tree to obtain the value of z. The process for solving the problem is as follows: Step 1: Identify the numbers given in the factor tree. Here, it's 253 divided by 11 to get 'z'. Step 2: Divide the larger number (253) by the smaller number (11). 253 ÷ 11 = 23 So, z = 23. Therefore, the correct answer is (b) 23. In conclusion, a factor tree is a tool that breaks down any number into its prime factors. In this case, it helped to find the value of z by dividing 253 by 11.   Q4: According to Fundamental Theorem of Arithmetic 13915 is a (a) Composite number (b) Prime number (c) Neither prime nor composite (d) Even number Ans:  (a) Explanation:  The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime number, or can be represented as a unique product of prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. For example, the first six prime numbers are 2, 3, 5, 7, 11, and 13. A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words, a composite number is any positive integer greater than one that is not a prime number. Now, let's consider the number 13915. We are given that 13915 can be written as the product of primes: 13915 = 5 х 11 х 11 х 23 = 5 х 11² х 23. Here, we can see that 13915 has more divisors than just 1 and itself (which are 5, 11 and 23). This means that 13915 is not a prime number. Also, as 13915 can be expressed as a product of prime numbers, it is not a number that falls into the category of 'neither prime nor composite'. As for being an even number, we know that an even number is any integer that can be divided by 2. In the case of 13915, it is not divisible by 2, so it is not an even number. Therefore, by process of elimination and based on the definitions, we can conclude that 13915 is a composite number (option a).   Q5:  The prime factorisation of 13915 is (a) 5 х 11³ х 13² (b) 5 х 11³ х 23² (c) 5 х 11² х 23 (d) 5 х 11² х 13² Ans: (c) Explanation:  The prime factorisation of a number is the representation of that number as the product of its prime factors. Here's how you would calculate the prime factorisation of 13915 step-by-step:

• First, find the smallest prime number that divides 13915. This will be 5, because 13915 is not divisible by 2 (it's not an even number), nor by 3 (the sum of its digits is not divisible by 3). So, you can start with 5.
• Divide 13915 by 5, which gives you 2783.
• Now, repeat the process with 2783. The smallest prime number that divides 2783 is 11. Divide 2783 by 11 to get 253.
• Repeat the process with 253. It's not divisible by 2, 3, 5, or 7, but it is divisible by 11. Dividing by 11 gives you 23.
• 23 is a prime number itself, so that's the end of the process.

Therefore, the prime factorisation of 13915 is 5 x 11 x 11 x 23, or 5 x 11² x 23, which matches option (c).  

Case Study – 3

Q1: What is the maximum capacity of a container which can measure the petrol of either tanker in exact number of time? (a) 150 litres (b) 160 litres (c) 170 litres (d) 180 litres Ans: (c) Explanation: The question is asking for the highest common factor (HCF) of 850 and 680. The HCF is the largest number that can evenly divide both numbers. Step 1:  Find the prime factors of both numbers. Prime factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Prime factors of 680 = 2 х 2 х 2 х 5 х 17 = 2³ х 5 х 17 Step 2:  Identify the common prime factors. The common prime factors of 850 and 680 are 2, 5, and 17. Step 3: Multiply the common prime factors to get the HCF. HCF of 850 and 680 = 2 х 5 х 17 = 170 Therefore, the maximum capacity of a container that can measure the petrol of either tanker an exact number of times is 170 litres, which corresponds to option (c). Q2: 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 (a) 3100 (b) 3200 (c) 3300 (d) 3400 Ans: (d) Explanation:  The question is asking for the least common multiple (LCM) of 850 and 680. The LCM is the smallest number that is a multiple of both numbers. Step 1:  We already have the prime factors of both numbers from the previous question, and the HCF. Prime factors of 850 = 2 х 5 х 5 х 17 = 2 х 5² х 17 Prime 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 Step 2:  Use the formula for finding the LCM when the HCF is known. LCM (850, 680) = (850 х 680) / HCF Step 3:  Substitute the values into the formula. LCM (850, 680) = (850 х 680) / 170 = 3400 Therefore, the LCM of 850 and 680 is 3400, which corresponds to option (d).   Q3: 680 can be expressed as a product of its primes as (a) 2² х 5 х 17 (b) 2¹ х 5 х 17 (c) 2³ х 5 х 17 (d) 2³ х 5 х 17⁰ Ans: (c) Explanation:  To solve this problem, you need to understand what prime factorization is. Prime factorization is the process of breaking down a number into its smallest prime factors. Let's try to factorize the number 680. First, we need to find a prime number that can divide 680. The smallest prime number is 2, and it can divide 680, so we use it as our first factor. 680 ÷ 2 = 340 Now we continue the process with 340. Again, it can be divided by 2, so we use 2 as our next factor. 340 ÷ 2 = 170 We repeat the process with 170. It can be divided by 2, so we use 2 as our next factor. 170 ÷ 2 = 85 Now, 85 cannot be divided by 2, so we move to the next prime number, which is 3. However, 85 cannot be divided by 3 either. We continue this process until we find a prime number that can divide 85, which is 5. 85 ÷ 5 = 17 Finally, we have 17, which is a prime number itself, so our factorization process stops here. Therefore, the prime factorization of 680 is 2 × 2 × 2 × 5 × 17, or in the exponential form, it is 2³ × 5 × 17. Hence, option (c) is correct.   Q4: 2 х 3 х 5 х 11 х 17 + 11 is a (a) Prime number (b) Composite number (d) Neither prime nor composite (d) None of the above Ans:  (b) Explanation: The provided answer appears to be incorrect. The number 11 is indeed a prime number, not a composite number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. In other words, if a number is prime, it can only be divided without a remainder by 1 and itself. The number 11 meets this criteria, as it can only be divided evenly by 1 and 11. On the other hand, a composite number is a positive integer that has at least one positive divisor other than one or itself. In other words, it has more than two distinct divisors. Here, the number 11 does not have more than two distinct divisors. Thus, the number 11 is a prime number. Therefore, the correct answer is (a) Prime number.   Q5: 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 (a) ab (b) a³b³ (c) a³b⁵ (d) a⁵b³ Ans:  (b) Explanation:  To find the least common multiple (LCM) of two numbers, we need to consider the highest powers of all the factors in the numbers. In this case, we are given that p = a³b² and q = a²b³, where a and b are prime numbers. The factors of p are a and b, with a having a power of 3 and b having a power of 2. The factors of q are also a and b, but here a has a power of 2 and b has a power of 3. When finding the LCM, we need to take the highest powers of these common factors. So, we take a to the power of 3 (since 3 is higher than 2) and b to the power of 3 (since 3 is higher than 2). Hence, the LCM of p and q is a³b³. Therefore, the correct option is (b) a³b³.  

Case Study – 4

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.

Q1: 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 (a) 14 (b) 12 (c) 16 (d) 18 Ans:  (b) Explanation: In order to find the maximum number of participants that can be accommodated in each room, we need to find the Highest Common Factor (HCF) of the number of participants in each subject. The HCF of a set of numbers is the largest number that divides each of them without leaving a remainder. It can be found by listing all the factors of each number and finding the largest one that they have in common. Here are the factors of each number:

• Factors of 60: 2 x 2 x 3 x 5 = 2² x 3 x 5
• Factors of 84: 2 x 2 x 3 x 7 = 2² x 3 x 7
• Factors of 108: 2 x 2 x 3 x 3 x 3 = 2² x 3³

The HCF of 60, 84, and 108 is 2² x 3 = 12. Therefore, the maximum number of participants that can be accommodated in each room is 12, which corresponds to the option (b).   Q2: What is the minimum number of rooms required during the event? (a) 11 (b) 31 (c) 41 (d) 21 Ans:  (d) Explanation:  The question requires us to calculate the minimum number of rooms required for the seminar. This can be done by finding the highest common factor (HCF) of the number of participants in each subject. The HCF tells us the maximum number of participants that can be accommodated in each room such that all rooms have the same number of participants. Let's start by finding the prime factorization of the numbers. For 60, the prime factors are 2, 2, 3, and 5 (2² х 3 х 5). For 84, the prime factors are 2, 2, 3, and 7 (2² х 3 х 7). For 108, the prime factors are 2, 2, 3, 3, and 3 (2² х 3³). Now, the HCF is found by multiplying the lowest power of the common prime factors. In this case, the common prime factors are 2 and 3. The lowest power of 2 is 2 (as in 2²), and the lowest power of 3 is 1 (as in 3). So, the HCF is 2² х 3 = 12. Now, to find the number of rooms required for each subject, we divide the number of participants by the HCF. For Hindi, we need 60/12 = 5 rooms. For English, we need 84/12 = 7 rooms. For Mathematics, we need 108/12 = 9 rooms. Adding these together, the total number of rooms required is 5 + 7 + 9 = 21 rooms. Therefore, the answer is (d) 21.   Q3: The LCM of 60, 84, and 108 is (a) 3780 (b) 3680 (c) 4780 (d) 4680 Ans: (a) Explanation: The problem revolves around finding the Least Common Multiple (LCM) of three numbers: 60, 84, and 108. To find the LCM of these numbers, we first need to find their prime factors. Here's how:

• The prime factors of 60 are 2, 2, 3, and 5 (since 2*2*3*5 = 60). We can write it as 2² * 3 * 5.
• The prime factors of 84 are 2, 2, 3, and 7 (since 2*2*3*7 = 84). We can write it as 2² * 3 * 7.
• The prime factors of 108 are 2, 2, 3, 3, and 3 (since 2*2*3*3*3 = 108). We can write it as 2² * 3³.

Now, to find the LCM, we take the highest power of all the prime factors obtained from these numbers. If a prime factor is not present in one number but is present in another, we take the factor from the number where it is present.

• We have the factor 2 in all three numbers, and the highest power is 2². So, we take 2².
• We have the factor 3 in all three numbers, and the highest power is 3³. So, we take 3³.
• We have the factor 5 only in 60. So, we take 5.
• We have the factor 7 only in 84. So, we take 7.  

Q4: The product of HCF and LCM of 60, 84, and 108 is (a) 55360 (b) 35360 (c) 45500 (d) 45360 Ans: (d) Explanation: The first step to solving this problem is understanding what HCF (Highest Common Factor) and LCM (Least Common Multiple) are. The HCF is the highest number that can divide two or more numbers without leaving a remainder. The LCM is the smallest number that is a multiple of two or more numbers. To find the HCF and LCM, we first need to find the prime factors of each number. For 60, the prime factors are 2, 2, 3, and 5 (or 2², 3, 5). For 84, the prime factors are 2, 2, 3, and 7 (or 2², 3, 7). For 108, the prime factors are 2, 2, 3, 3, and 3 (or 2², 3³). The HCF of these three numbers is found by taking the highest common factor of all three numbers, which is 2² (or 4) and 3. Multiplying these together gives us an HCF of 12. The LCM is found by taking the highest power of all the prime factors present in the numbers. This gives us 2², 3³, 5, and 7. Multiplying these together gives us an LCM of 3780. Finally, to find the product of the HCF and LCM, we multiply 12 and 3780 together, which gives us 45360. Hence, the correct answer is (d) 45360.   Q5: 108 can be expressed as a product of its primes as (a) 2³ х 3² (b) 2³ х 3³ (c) 2² х 3² (d) 2² х 3³ Ans:  (d) Explanation: The process of finding the answer is called prime factorization. Step 1: Start with the smallest prime number, which is 2. Check if 108 is divisible by 2. If it is, then write down 2 as a factor and divide 108 by 2. Step 2:  You get 54 as the quotient. Now, repeat the process with 54. Is it divisible by 2? Yes, it is. So, write down 2 as a factor again and divide 54 by 2. Step 3: You now have a quotient of 27. Repeat the process. Is 27 divisible by 2? No, it's not. So, move on to the next prime number, which is 3. Step 4: Is 27 divisible by 3? Yes, it is. So, write down 3 as a factor and divide 27 by 3. Step 5:  You get a quotient of 9. Repeat the process. Is 9 divisible by 3? Yes, it is. So, write down 3 as a factor again and divide 9 by 3. Step 6: You now have a quotient of 3. Repeat the process. Is 3 divisible by 3? Yes, it is. So, write down 3 as a factor again and divide 3 by 3. Step 7:  You now have a quotient of 1. When you reach 1, you can stop the process. Step 8: Now, count the number of times each prime number appears in your list of factors. You have two 2s and three 3s. Step 9: Write down your answer as the product of the prime numbers, each raised to the power of its count. So, 108 = 2² х 3³. That's how you get the answer (d) 2² х 3³.  

Case Study – 5

Q1: 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 Ans:  (c) Explanation:  The question requires finding the minimum number of books that can be equally distributed among the students of either section A or section B. This is essentially finding the least common multiple (LCM) of the number of students in both sections. Step 1:  Find the prime factors of both 32 and 36. Prime factors of 32 are 2 x 2 x 2 x 2 x 2 = 2^5 Prime factors of 36 are 2 x 2 x 3 x 3 = 2^2 x 3^2 Step 2: Find the LCM of 32 and 36. For finding the LCM, we take the highest power of each prime factor that appears in the factorization of either 32 or 36. Here, for 2, the highest power is 5 (from 32) and for 3, it is 2 (from 36). So, LCM of 32 and 36 = 2^5 x 3^2 = 32 x 9 = 288. Hence, you would need a minimum of 288 books so that they can be equally distributed among the students of either section A or section B. So, the correct option is (C).   Q2: 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 Ans:  (b) Explanation:  To arrive at the solution, we need to understand two key concepts - Highest Common Factor (HCF) and Lowest Common Multiple (LCM). The HCF of two numbers is the highest number that can divide both of them without leaving a remainder. On the other hand, LCM of two numbers is the smallest number that can be divided by both of them without leaving a remainder. Let's break down the problem into steps: Step 1:  Find the factors of the given numbers 32 and 36. Factors of 32: 2*2*2*2*2 = 2⁵ Factors of 36: 2*2*3*3 = 2²*3² Step 2:  Find the LCM of 32 and 36. The LCM is found by multiplying the highest power of all the factors that appear in either number. Here we have 2⁵ from 32 and 2²*3² from 36. The higher power of 2 is 2⁵ from 32 and the higher power of 3 is 3² from 36. So, LCM = 2⁵*3² = 32*9 = 288 Step 3: Find the HCF of 32 and 36. The product of two integers (32 and 36) is equal to the product of their HCF and LCM. So, we can find the HCF by dividing the product of the two numbers by their LCM. HCF = (32*36) / LCM = (32*36) / 288 = 4 Therefore, the HCF of 32 and 36 is 4. Hence, the correct option is (b) 4. Q3:  36 can be expressed as a product of its primes as (a) 2² х 3² (b) 2¹ х 3³ (c) 2³ х 3¹ (d) 2⁰ х 3⁰ Ans:  (a) Explanation: Prime factorization is the process of breaking down a number into its smallest prime factors. A prime number is a number that has only two distinct positive divisors: 1 and itself. For example, the first six prime numbers are 2, 3, 5, 7, 11, and 13. To express 36 as a product of its primes, we follow these steps: Step 1:  Begin by dividing the number 36 with the smallest prime number, i.e., 2. 36 divided by 2 is 18. Step 2: Now divide 18 by 2 to get 9. Step 3:  As 9 cannot be divided by 2, we move to the next prime number, which is 3. 9 divided by 3 is 3. Step 4:  Finally, divide 3 by 3 to get 1. Now we stop because we have reached 1. The prime factors of 36 are therefore 2, 2, 3, and 3. We write this as 2² х 3². So, the answer is (a) 2² х 3².   Q4: 7 х 11 х 13 х 15 + 15 is a (a) Prime number (b) Composite number (d) Neither prime nor composite (d) None of the above Ans:  (b) Explanation: The question is asking to determine the type of number 15 is, when it's part of the given multiplication equation. Step 1: Let's understand the definitions first. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The first few prime numbers are 2, 3, 5, 7, 11, and 13. A composite number is a positive integer that has at least one positive divisor other than one or itself. In other words, a composite number is any positive integer greater than one that is not a prime number. Step 2:  Now, consider the number 15. We can see that 15 is not a prime number because it has more than two factors, which are 1, 3, 5, and 15. Step 3:  Therefore, 15 is a composite number. Hence, the correct option is (B).   Q5: 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 (a) ab (b) a²b² (c) a³b² (d) a³b³ Ans: (b) Explanation:  Let's break down the solution to understand it better. Firstly, we are given two positive integers, p and q which are represented as p = ab² and q = a²b, where a and b are prime numbers. The LCM (Least Common Multiple) is the smallest number that is a multiple of both numbers. In other words, it is the smallest number that both numbers can divide into evenly. When finding the LCM of two numbers represented as the product of prime numbers raised to some powers (as in this case), the LCM is simply the product of these primes each raised to the highest power that appears in either number. In this case, the prime number 'a' is raised to the first power in p and to the second power in q. Thus, in the LCM, 'a' is raised to the highest power of these, which is 2. Similarly, the prime number 'b' is raised to the second power in p and to the first power in q. In the LCM, 'b' is raised to the highest power of these, which is 2. Hence, the LCM of p and q is a²b² which corresponds to option (b). So, the correct answer is option (b) a²b².  

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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.

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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?

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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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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Chapter 1 Real Numbers NCERT Solutions for Class 10 Maths

Ncert solutions for class 10 maths chapter 1 real numbers.

• Exercise 1.1

1. Use Euclid's division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

(i) 225 > 135 we always divide greater number with smaller one.

Divide 225 by 135 we get 1 quotient and 90 as remainder so that, 225= 135 × 1 + 90

Divide 135 by 90 we get 1 quotient and 45 as remainder so that, 135= 90 × 1 + 45

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as 90 = 2 × 45+ 0

As there are no remainder so divisor 45 is our HCF.

(ii) 38220 > 196 we always divide greater number with smaller one.

Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as 38220 = 196 × 195 + 0

As there is no remainder so divisor 196 is our HCF.

(iii) 867 > 255 we always divide greater number with smaller one.

Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as 867 = 255 × 3 + 102

Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as 255 = 102 × 2 + 51

Divide 102 by 51 we get quotient 2 and no remainder so we can write it as 102 = 51 × 2 + 0

As there is no remainder so divisor 51 is our HCF.

2. Show that any positive odd integer is of the form 6 q  + 1, or 6 q  + 3, or 6 q  + 5, where  q is some integer.

Let take  a  as any positive integer and  b  = 6.

Then using Euclid’s algorithm we get a = 6 q  +  r  here  r  is remainder and value of  q  is more than or equal to 0 and  r  = 0, 1, 2, 3, 4, 5 because 0 ≤  r  < b and the value of  b  is 6 

So, total possible forms will 6 q  + 0 , 6 q  + 1 , 6 q  + 2,6 q  + 3, 6 q  + 4, 6 q  + 5

6 q  + 0 6 is divisible by 2 so it is a even number 

6 q  + 1 6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number

6 q  + 2 6 is divisible by 2 and 2 is also divisible by 2 so it is a even number

6 q + 3 6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number 

6 q  + 4 6 is divisible by 2 and 4 is also divisible by 2 it is a even number

6 q  + 5 6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So, odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

HCF (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid's algorithm to find the HCF. 616 = 32 × 19 + 8 32 = 8 × 4 + 0 The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.

4. Use Euclid's division lemma to show that the square of any positive integer is either of form 3 m  or 3 m  + 1 for some integer m.

[Hint: Let  x  be any positive integer then it is of the form 3 q , 3 q  + 1 or 3 q  + 2. Now square each of these and show that they can be rewritten in the form 3 m  or 3 m  + 1.]

Let a be any positive integer and  b  = 3.

Then a = 3 q  +  r  for some integer  q  ≥ 0 And  r  = 0, 1, 2 because 0 ≤  r  < 3 Therefore,  a  = 3 q  or 3 q  + 1 or 3 q  + 2 Or, a 2  = (3 q ) 2  or (3 q  + 1) 2  or (3 q  + 2) 2 a 2  = (9 q ) 2  or 9 q 2  + 6 q  + 1 or 9 q 2  + 12 q  + 4 = 3 × (3 q 2 ) or 3(3 q 2  + 2 q ) + 1 or 3(3 q 2  + 4 q  + 1) + 1 = 3 k 1  or 3 k 2  + 1 or 3 k 3  + 1

Where  k 1 ,  k 2 , and  k 3  are some positive integers Hence, it can be said that the square of any positive integer is either of the form 3 m  or 3 m  + 1.

5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9 m , 9 m  + 1 or 9 m + 8.

Let a be any positive integer and b = 3

a = 3 q + r , where q ≥ 0 and 0 ≤ r < 3

∴ a = 3q or 3 q  + 1 or 3 q  + 2

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3 q , a 3 = (3 q ) 3 = 27 q 3 = 9(3 q ) 3 = 9 m , where,  m is an integer such that m = 3 q 3

Case 2: When a = 3q + 1, a 3 = (3 q +1) 3 a 3 = 27 q 3 + 27 q 2 + 9 q + 1 a 3 = 9(3 q 3 + 3 q 2 + q ) + 1 a 3 = 9 m + 1 where,  m is an integer such that m = (3 q 3 + 3 q 2 + q )

Case 3: When a = 3 q + 2, a 3 = (3 q +2) 3 a 3 = 27 q 3 + 54 q 2 + 36 q + 8 a 3 = 9(3 q 3 + 6 q 2 + 4q) + 8 a 3 = 9 m + 8 where m is an integer such that m = (3 q 3 + 6 q 2 + 4 q )

Therefore, the cube of any positive integer is of the form 9 m , 9 m + 1, or 9 m + 8.

Page No: 11

• Exercise 1.2

1. Express each number as product of its prime factors:

(i) 140 = 2 × 2 × 5 × 7 = 2 2 × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2 2 × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3 2 × 5 2 × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92 

(iii) 336 and 54

(i) 26 = 2 × 13

91 =7 × 13

LCM =2 × 7 × 13 =182

Product of two numbers 26 × 91 = 2366

Product of HCF and LCM 13 × 182 = 2366

Hence, product of two numbers = product of HCF × LCM

(ii) 510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of two numbers 510 × 92 = 46920

Product of HCF and LCM 2 × 23460 = 46920

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 = 6

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024

Product of two numbers 336 × 54 =18144

Product of HCF and LCM 6 × 3024 = 18144

Hence, product of two numbers = product of HCF × LCM.

3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21 

(ii) 17, 23 and 29 

(iii) 8, 9 and 25

(i) 12 = 2 × 2 × 3

15 = 3 × 5

21 =3 × 7

LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

LCM = 1 × 17 × 19 × 23 = 11339

(iii) 8 =1 × 2 × 2 × 2

9 =1 × 3 × 3

25 =1 × 5 × 5

LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

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

We have the formula that,

Product of LCM and HCF = product of number

LCM × 9 = 306 × 657

Divide both side by 9 we get,

⇒ LCM = 34 × 657 = 22338

5. Check whether 6 n can end with the digit 0 for any natural number n .

If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.

So, value 6 n should be divisible by 2 and 5 both.

6 n is divisible by 2 but not divisible by 5 as the prime factors of 6 are 2 and 3.

So, it can not end with 0.

6. Explain why 7×11×13 + 13 and 7×6×5×4×3×2×1 + 5 are composite numbers.

7 × 11 × 13 + 13

Taking 13 common, we get

13 (7×11 +1 )

⇒ 13(77 + 1 )

⇒ 13 (78)

It is product of two numbers and both numbers are more than 1 so it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 common, we get

5(7 × 6 × 4 × 3 × 2 × 1 +1)

⇒ 5(1008 + 1)

⇒ 5(1009)

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point.

They will be meet again after LCM of both values at the starting point.

18 = 2 × 3 × 3

12 = 2 × 2 × 3

LCM = 2 × 2 × 3 × 3 = 36

Therefore, they will meet together at the starting point after 36 minutes.

Page No: 14

• Exercise 1.3

1. Prove that √5 is irrational.

Let us take √5 as rational number

If a and b are two co prime number and b is not equal to 0.

We can write √5 = a/b

Multiply by b both side we get,

b√5 = a

To remove root, Squaring on both sides, we get

5 b 2 = a 2 … (i)

Therefore, 5 divides a 2 and according to theorem of rational number, for any prime number p which is divides a 2 then it will divide a also.

That means 5 will divide a . So we can write,

Putting value of a in equation (i) we get

5 b 2 = (5 c ) 2

⇒ 5 b 2 = 25 c 2

Divide by 25 we get,

b 2 /5 = c 2

Similarly, we get that b will divide by 5 and we have already get that a is divide by 5 but a and b are co prime number. So it contradicts.

Hence, √5 is not a rational number, it is irrational.

2. Prove that 3 + 2√5 is irrational.

Let take that 3 + 2√5 is a rational number.

So we can write this number as,

3 + 2√5 = a / b

Here, a and b are two co prime number and b is not equal to 0

Subtract 3 both sides we get

2√5 = a/b – 3

⇒ 2√5 = (a-3b)/b

Now, divide by 2, we get

√5 = (a-3b)/2b

Here,  a and b are integer so ( a -3 b ) / 2 b is a rational number so √5 should be a rational number. But √5 is a irrational number, so it contradicts.

Hence, 3 + 2√ 5  is a irrational number.

3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

(i) Let take that 1/√2 is a rational number.

So, we can write this number as

Here,  a and b are two co prime number and b is not equal to 0

Multiply by √2 both sides we get,

1 = ( a √ 2 )/ b

Now, multiply by b

b = a √ 2

Divide by a we get

Here,  a and b are integer, so b/a is a rational number. So, √2 should be a rational number

But √2 is a irrational number so it contradicts.

Hence, 1/√2 is a irrational number

So we can write this number as

7√5 = a / b

Divide by 7 we get

√5 =  a /(7 b )

Here, a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts.

Hence, 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number.

6 + √2 = a / b

Subtract 6 both side we get

√2 = a / b – 6

⇒ √2 = ( a  - 6 b )/ b

Here,  a and b are integer so (a-6 b )/ b is a rational number, So, √2 should be a rational number.

Hence, 6 + √ 2  is a irrational number.

Page No: 17

• Exercise 1.4

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125

(ii) 17/8 

(iii) 64/455 

(iv) 15/1600 

(v) 29/343 

(vi) 23/2 3  × 5 2  

(vii) 129/2 2  × 5 7  × 7 5  

(viii) 6/15 

(ix) 35/50 

Factorize the denominator we get

3125 =5 × 5 × 5 × 5 × 5 = 5 5

So, denominator is in form of 5 m so it is terminating.

Factorize the denominator we get,

8 =2 × 2 × 2 = 2 3

So, denominator is in form of 2 m  so it is terminating.

(iii) 64 / 455

455 = 5 × 7 × 13

There are 7 and 13 also in denominator so denominator is not in form of 2 m  × 5 n . So, it is not terminating.

(iv) 15 / 1600

1600 = 2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 2 6  × 5 2

So, denominator is in form of 2 m  × 5 n

Hence, it is terminating.

(v) 29 / 343

343 = 7 × 7 × 7 = 7 3

There are 7 also in denominator so denominator is not in form of 2 m  × 5 n

Hence, it is non-terminating.

(vi) 23 / (2 3  × 5 2 )

Denominator is in form of 2 m  × 5 n

Hence, it is terminating.

(vii) 129 / (2 2  × 5 7  × 7 5 )

Denominator has 7 in denominator so denominator is not in form of 2 m  × 5 n

Hence, it is none terminating.

(viii) 6 / 15

Divide nominator and denominator both by 3 we get 2 / 5

Denominator is in form of 5 m so it is terminating.

(ix) 35 / 50 divide denominator and nominator both by 5 we get 7 / 10

10 = 2 × 5

So, denominator is in form of 2 m  × 5 n  so it is terminating.

(x) 77 / 210

Simplify it by dividing nominator and denominator both by 7 we get, 11 / 30

30 = 2 × 3 × 5

Denominator has 3 also in denominator so denominator is not in form of 2 m  × 5 n

Page No: 18

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

(i) 13 / 3125

(ii) 17 / 8

(vi) 23 / 2 3 5 2

Dividing numerator and denominator by 3.

(ix) 35 / 50

Dividing numerator and denominator by 5.

3. 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 p , q you say about the prime factors of q ?

(i) 43.123456789

(ii) 0.120120012000120000...

(iii) 43. 123456789

(i) Since this number has a terminating decimal expansion, it is a rational number of the form p / q, and q is of the form 2 m  × 5 n .

(ii) The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii) Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form  p / q, and  q  is not of the form 2 m  × 5 n .

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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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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.

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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.

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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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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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2.9 billion records, including Social Security numbers, stolen in data hack: What to know

An enormous amount of sensitive information including Social Security numbers for millions of people could be in the hands of a hacking group after a data breach and may have been released on an online marketplace, The Los Angeles Times reported this week.

The hacking group USDoD claimed it had allegedly stolen personal records of 2.9 billion people from National Public Data, according to a class-action lawsuit filed in U.S. District Court in Fort Lauderdale, Florida, reported by Bloomberg Law. The breach was believed to have happened in or around April, according to the lawsuit.

Here's what to know about the alleged data breach.

Social security hack: National Public Data confirms massive data breach included Social Security numbers

What information is included in the data breach?

The class-action law firm Schubert, Jonckheer & Kolbe said in a news release that the stolen file includes 277.1 gigabytes of data , and includes names, address histories, relatives and Social Security numbers dating back at least three decades.

According to a post from a cybersecurity expert on X, formerly Twitter, USDoD claims to be selling the 2.9 billion records for citizens of the U.S., U.K. and Canada on the dark web for $3.5 million.

Since the information was posted for sale in April, others have released different copies of the data, according to the cybersecurity and technology news site Bleeping Computer.

A hacker known as " Fenice " leaked the most complete version of the data for free on a forum in August, Bleeping Computer reported.

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2025 COLA: Estimate dips with inflation, but high daily expenses still burn seniors

What is National Public Data?

National Public Data is a Florida-based background check company operated by Jerico Pictures, Inc. USA TODAY has reached out to National Public Data for comment.

The company has not publicly confirmed a data breach, but The Los Angeles Times reported that it has been telling people who contacted via email that "we are aware of certain third-party claims about consumer data and are investigating these issues."

What to do if you suspect your information has been stolen

If you believe your information has been stolen or has appeared on the dark web, there are a few steps you can take to prevent fraud or identity theft.

Money.com recommends taking the following steps:

• Make sure your antivirus is up to date and perform security scans on all your devices. If you find malware, most antivirus programs should be able to remove it, but in some cases you may need professional help.
• Update your passwords for bank accounts, email accounts and other services you use, and make sure they are strong and different for every account. Include uppercase and lowercase letters, numbers and punctuation marks, and never use personal information that a hacker could guess.
• Use multifactor authentication for any accounts or services that offer it to ensure you are the person logging in.
• Check your credit report, and report any unauthorized use of of your credit cards. If you notice any suspicious activity, you can ask credit bureaus to freeze your credit.
• Be careful with your email and social media accounts, and beware of phishing, an attempt to get your personal information by misrepresenting who a message or email is from.