case study based on 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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CBSE 10th Standard Maths Subject Real Number Case Study Questions With Solution 2021

CBSE 10th Standard Maths Subject Real Number Case Study Questions With Solution 2021 QB365 - Question Bank Software May-21 , 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

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Case Study Questions

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

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Case Study Class 10 Maths Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

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.

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

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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
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• Chapter 5: Arithmetic Progressions Case Study Questions

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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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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.

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Important Questions for CBSE Class 10 Maths Chapter 1 - Real Numbers 2024-25

• Class 10 Important Question
• Chapter 1: Real Numbers

Boost Your Performance in CBSE Class 10 Maths Exam Chapter 1 with These Important Questions

Important questions for Class 10 Maths Chapter 1, Real Numbers is prepared by experts of Vedantu with a purpose of creating important questions for the chapter is to enable the students to pivotal concepts of the chapter that has been introduced in Maths NCERT Solutions Class 10 . The important questions pdf version is prepared to give a better conceptual understanding to the students and help them to perceive what questions they can expect from this chapter in exams. There are pdf versions of important questions  for other subjects also. You can download them anytime on any device and practice them at your convenient time. The important questions will surely give an insight about what questions can be expected in exams. 

Chapter 1, Real numbers for class 10 , continues with the real number operations that you studied in earlier grades and also introduces two important properties of positive integers namely Euclid's Division Lemma and algorithm and the fundamental theorem of arithmetic. The important questions are based on the topics that are discussed in this chapter. Let us have a quick glance at the summary of the chapter so that you can solve the important questions for Maths Chapter 1, Real Numbers for Class 10. 

At Vedantu you will find reliable and accurate NCERT Solutions for all classes and subjects. You can also Download NCERT Solutions for Class 10 Science to help you to revise complete Syllabus and score more marks in your examinations.

Download CBSE Class 10 Maths Important Questions 2024-25 PDF

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Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 1 Real Numbers

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Study Important Questions for Class 10 Mathematics Chapter 1 - Real Numbers

Very Short Answer Questions                                                                        (1 Mark)

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

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

Ans: Let $a$ be any positive integer, then we can write $a=3q+r$ for some integer $q\ge 0$  …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.    ..…(2)

Therefore from (1) and (2) we can get, $a=3q$ or $3q+1$ or $3q+2$.    ……(3)

Squaring both sides of equation (3) we get,

${{a}^{2}}={{(3q)}^{2}}$ or ${{(3q+1)}^{2}}$ or ${{(3q+2)}^{2}}$ 

$\Rightarrow {{a}^{2}}=9{{q}^{2}}$ or $9{{q}^{2}}+6q+1$ or $9{{q}^{2}}+12q+4$  …..(4)

Taking $3$ common from LHS of equation (4) we get,

${{a}^{2}}=3\times (3{{q}^{2}})$ or $3(3{{q}^{2}}+2q)+1$ or $3(3{{q}^{2}}+4q+1)+1$.   …..(5)

Equation (5) could be written as 

${{a}^{2}}=3{{k}_{1}}$ or $3{{k}_{2}}+1$ or $(3{{k}_{3}}+1)$ for some positive integers ${{k}_{1}},{{k}_{2}}$ and ${{k}_{3}}$.

Hence, it can be said that the square of any positive integer is either of the form $3m$ or $3m+1$.

2. Express each number as product of its prime factors.

(i). $140$ 

Ans: The product of prime factors of number $140$ is \[140=2\cdot 2\cdot 5\cdot 7={{2}^{2}}\cdot 5\cdot 7\].

(ii). $156$ 

Ans: The product of prime factors of number $156$ is $156=2\cdot 2\cdot 3\cdot 13={{2}^{2}}\cdot 3\cdot 13$.

(iii). \[\mathbf{382}5\] 

Ans: The product of prime factors of number $3825$ is $3825=3\cdot 3\cdot 5\cdot 5\cdot 17={{3}^{2}}\cdot {{5}^{2}}\cdot 17$.

(iv). \[5\mathbf{005}\] 

Ans: The product of prime factors of number $5005$ is $5005=5\cdot 7\cdot 11\cdot 13$.

(v). \[\mathbf{7429}\] 

Ans: The product of prime factors of number $7429$ is $7429=17\cdot 19\cdot 23$.

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

Ans : It is given that the HCF of the two numbers $306,657$ is $9$. We have to find its LCM.

We know that, LCM$\times $HCF = Product of two numbers. Therefore,

LCM$\times $HCF $=306\times 657$ 

$\Rightarrow LCM=\dfrac{306\times 657}{HCF}$

$\Rightarrow LCM=\dfrac{306\times 657}{9}$

$\therefore LCM=22338$

4. Check whether ${{6}^{n}}$can end with the digit $0$ for any natural number $n$.

Ans: If any number ends with the digit $0$, it should be divisible by $10$.

Let us check this for natural number $n=2$. 

Clearly, ${{6}^{2}}=36$, which is not divisible by $10$. Therefore, ${{6}^{n}}$ cannot end with the digit $0$ for any natural number $n$.

5. Prove that $\left( 3+2\sqrt{5} \right)$ is irrational.    

Ans : We will prove this by contradiction. To solve this problem, suppose that $\left( 3+2\sqrt{5} \right)$ is rational i.e., $\exists $ two co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{a}{b}=3+2\sqrt{5}$     …..(1)

Subtracting $3$ from both sides of the equation (1) and simplifying it further we get,

$\dfrac{a}{b}-3=2\sqrt{5}$

$\Rightarrow \dfrac{a-3b}{b}=2\sqrt{5}$ 

$\Rightarrow \dfrac{a-3b}{2b}=\sqrt{5}$        …..(2)

Since we know that $\sqrt{5}$ is irrational. Therefore, from (2) $\dfrac{a-3b}{2b}$ should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $(3+2\sqrt{5})$ cannot be rational.

Hence, $(3+2\sqrt{5})$ is irrational.

6. The number $7\times 11\times 13\times 15+15$ is a

Composite Number

Whole Number

Prime Number

None of these

Ans: (a) and (b) both

7. For what least value of $'n'$ a natural number, ${{(24)}^{n}}$ is divisible by $8$?

 $0$ 

No value of $'n'$ is possible

8. The sum of a rational and an irrational number is

Both (a) & (b)

Either (a) & (b)

Ans: (b) Irrational

9. HCF of two numbers is $113,$ their LCM is $56952$. If one number is $904$, then other number is:

Ans: (b) $7119$

10. A lemma is an axiom used for providing

other statement

no statement

contradictory statement

none of these

Ans: (a) other statement

11. If HCF of two numbers is $1$, the two numbers are called relatively ______ or _______.

prime, co-prime

composite, prime

Both (a) and (b)

Ans: (a) prime, co-prime

12. The number $2.\overline{35}$ is

a terminating decimal number

a rational number 

an irrational number

Ans: (b) a rational number

13. The number $2.13113111311113......$ is        

a rational number

a non-terminating decimal number

Both (a) & (c)

Ans: (b) a non-terminating decimal number

14. The smallest composite number is      

Ans: (d) $4$ 

15. The number $1.23\overline{48}$ is

None if these

Ans: (c) a rational number

16. The number $\pi $ is

neither rational nor irrational

Ans : (b) an irrational number

17. The number $(2+\sqrt{5})$ is

not real number

Ans: (b) an irrational number

Short Answer Questions                                                                               (2 Marks)

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

Ans: Let $a$ be any odd positive integer, then by Euclid’s algorithm we can write $a=6q+r$ for some integer $q\ge 0$.        …..(1)

Clearly, in the expression (1) we are dividing $a$ by $6$ with quotient $q$ and remainder $r$, 

$r=1,3,5$ because $0\le r  < 6$ and $a$ is odd.    …..(2)

Therefore from (1) and (2) we can get, 

$a=6q+1$ or $6q+3$ or $6q+5$ 

And therefore, any odd integer can be expressed in the form $6q+1,6q+3,$ or $6q+5$.

2. An army contingent of $616$ members are to march behind a 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?                   

Ans: To solve this problem, we have to find the HCF of two numbers $(616,32)$.

To find the HCF, we can use Euclid’s algorithm which states that if there are any two integers \[a\] and \[b\], then there exists \[q\] and \[r\] such that it satisfies the condition \[a=bq+r\] where \[0\le r  < b\].

For $a=616$, $616=32\times 19+8$       …..(1)

For $a=32$, $32=8\times 4+0$       …..(2) 

Therefore, from (1) and (2), the HCF $(616,32)$ is $8$ and hence they can march in $8$ columns each.

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

Ans: Let $a$ be any positive integer, then by Euclid’s algorithm we can write $a=3q+r,$ for some integer $q\ge 0$.    …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.     …..(2)

$a=3q$ or $3q+1$ or $3q+2$    …..(3)

Cubing both sides of equation (3) we get,

Case 1: When $a=3q$,

${{a}^{3}}=27{{q}^{3}}$

$\Rightarrow {{a}^{3}}=9(3{{q}^{3}})$ 

$\Rightarrow {{a}^{3}}=9m$ 

Where $m$ is an integer such that $m=3{{q}^{3}}$

Case 2: When $a=3q+1$

${{a}^{3}}={{(3q+1)}^{3}}$

$\Rightarrow {{a}^{3}}=9{{q}^{3}}+9{{q}^{2}}+9q+1$ 

$\begin{align} & \Rightarrow {{a}^{3}}=9\left( {{q}^{3}}+{{q}^{2}}+q \right)+1 \\ & \Rightarrow {{a}^{3}}=9m+1 \\ \end{align}$

Where $m$ is an integer such that $m=(3{{q}^{3}}+3{{q}^{2}}+q)$

Case 3: When $a=3q+2$

${{a}^{3}}={{(3q+2)}^{3}}$

$\Rightarrow {{a}^{3}}=27{{q}_{3}}+54{{q}^{2}}+36q+8$ 

$\Rightarrow {{a}^{3}}=9(3{{q}^{3}}+6{{q}^{2}}+4q)+8$ 

$\Rightarrow {{a}^{3}}=9m+8$ 

Where $m$ is an integer such that $m=(3{{q}^{3}}+6{{q}_{2}}+4q)$

Therefore, from case 1,2 and 3 we conclude that the cube of any positive integer is of the form $9m,9m+1$ or $9m+8$.

4. Find the LCM and HCF of the following pairs of integers and verify that $LCM\times HCF=$product of two numbers.                       

(i). $26\text{ and }91$

Ans: Let us first find the HCF of the two given numbers.

Writing the prime factorization of both of the numbers we get, 

$26=2\times 13$ and $91=7\times 13$.       …..(1) 

Since $13$ is common in the prime factorization of both of the numbers we get $\text{HCF}=13$.      …..(2)

From (1), $\text{LCM = 2}\times \text{7}\times \text{13=182}$     …..(3) 

Product of two numbers, $26\times 91=2366$   …..(4)

From (2) and (3), $\text{HCF}\times \text{LCM =13}\times \text{182=2366}$.    …..(5) 

Hence from (4) and (5), product of two numbers \[=HCF\times LCM\]. 

(ii). $510\text{ and }92$

$510=2\times 3\times 5\times 17$ and $92=2\times 2\times 23$.   …..(1) 

Since $2$ is common in the prime factorization of both of the numbers we get $HCF=2$.         …..(2)

From (1),$LCM=2\times 2\times 3\times 5\times 17\times 23=23460$    …..(3) 

Product of two numbers, $510\times 92=46920$    …..(4)

From (2) and (3),$HCF\times LCM=2\times 23460=46920$.  …..(5) 

Hence from (4) and (5), product of two numbers \[=HCF\times LCM\].

(iii). $336\text{ and }54$

$336=2\times 2\times 2\times 2\times 3\times 7={{2}^{4}}\times 3\times 7$and$54=2\times 3\times 3\times 3=2\times {{3}^{3}}$.     …..(1) 

Since $2,3$ are common in the prime factorization of the numbers, $HCF=2\times 3=6$.      .....(2)

From (1),$LCM={{2}^{4}}\times {{3}^{3}}\times 7=3024$      …..(3) 

Product of two numbers $=336\times 54=18144$      …..(4)

From (2) and (3), $HCF\times LCM=6\times 3024=18144$.      …..(5) 

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

(i). $12,15\text{ and }221$

Writing the prime factorization of all the numbers we get, 

$12={{2}^{2}}\times 3$, $15=3\times 5$ and$21=3\times 7$.    …..(1) 

Since $3$ is common in the prime factorization of all the numbers we get $HCF=3$.                                   

From (1), $LCM={{2}^{2}}\times 3\times 5\times 7=420$. 

(ii). $17,23\text{ and }29$

$17=1\times 17$, $23=1\times 23$ and $29=1\times 29$.   …..(1) 

Since $1$ is common in the prime factorization of all the numbers we get $HCF=1$.                                   

From (1), $LCM=17\times 23\times 29=11339$. 

(iii). $8,9\text{ and }25$

$8=2\times 2\times 2={{2}^{3}}$, $9=3\times 3={{3}^{2}}$ and $25=5\times 5={{5}^{2}}$.    …..(1) 

From (1), $LCM={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}}=1800$. 

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

Ans: Numbers are of two types – prime and composite.                  

Prime numbers can be divided by $1$ and only itself, whereas composite numbers have factors other than $1$ and itself.

It can be observed that $7\times 11\times 13+13=13\times (7\times 11+1)$

The given expression has $13$ as one of the factors. Therefore, it is a composite number.

Also, $7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times (7\times 6\times 4\times 3\times 2\times 1+1)$.

The given expression has $5$ as one of the factors. Therefore, it 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?                                                                     

Ans: It can be observed that Ravi takes lesser time than Sonia for completing one round of the circular path. As they are going in the same direction, they will meet again at the time that will be the $LCM$ of time taken by Sonia and Ravi for completing $1$ round of circular path respectively i.e., $LCM$ of $18$ minutes and $12$ minutes.

Let us now calculate $LCM$ of $18$ minutes and $12$ minutes.

$18=2\times 3\times 3$ and $12=2\times 2\times 3$    …..(1)

$LCM\left( 12\text{,}18 \right)=2\times 2\times 3\times 3=36$

Therefore, Ravi and Sonia will meet together at the starting point after $36$ minutes.

8. Prove that $\sqrt{5}$ is irrational.

Ans: We will prove this by contradiction.

Let us suppose that $\sqrt{5}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow b\sqrt{5}=a$   …..(1)

Squaring both sides of (1), we get $5b={{a}^{2}}$    …..(2)

From (2) we can conclude that $5$ is factor of ${{a}^{2}}$, this means that $5$ is also factor of $a$.     …..(3)

From (3) we can write $a=5c$ for some integer $c$.   …..(4)

Substituting value of (4) in (2) we get,

$\begin{align} & 5{{b}^{2}}=25{{c}^{2}} \\ & \Rightarrow {{b}^{2}}=5{{c}^{2}} \\ \end{align}$

It means that $5$ is factor of ${{b}^{2}}$, this means that $5$ is also factor of $b$.   …..(5)

From $(3)\text{,}(5)$, we can say that $5$ is factor of both $a\text{,}b$, which is a contradiction to our assumption. Therefore, our assumption was wrong and $\sqrt{5}$ is irrational.

9. Write down the decimal expansion of those rational numbers in Question 1 which have terminated decimal expansion.        

Ans:  

(i). The decimal expansion of $\dfrac{13}{3125}$ is $\dfrac{13}{{{5}^{5}}}=\dfrac{13\times {{2}^{5}}}{{{5}^{5}}\times {{2}^{5}}}=\dfrac{13\times {{2}^{5}}}{{{10}^{5}}}=\dfrac{416}{{{10}^{5}}}=0.00416$ 

(ii) The decimal expansion of $\dfrac{17}{8}$ is $\dfrac{17}{{{2}^{3}}}=\dfrac{17\times {{5}^{3}}}{{{2}^{3}}\times {{5}^{3}}}=\dfrac{17\times {{5}^{3}}}{{{10}^{3}}}=\dfrac{2125}{{{10}^{3}}}=2.215$ 

(iii) The decimal expansion of $\dfrac{15}{1600}$ is $\dfrac{15}{{{2}^{6}}\times {{5}^{2}}}=\dfrac{15\times {{5}^{4}}}{2\times {{5}^{2}}\times {{5}^{4}}}=\dfrac{15\times {{5}^{4}}}{{{10}^{6}}}=\dfrac{9375}{{{10}^{6}}}=0.009375$ 

(iv) The decimal expansion of $\dfrac{23}{{{2}^{3}}\times {{5}^{2}}}$ is $\dfrac{23\times {{5}^{1}}}{{{2}^{3}}\times {{5}^{2}}\times {{5}^{1}}}=\dfrac{23\times {{5}^{1}}}{{{10}^{3}}}=\dfrac{115}{{{10}^{3}}}=0.115$ 

(v) The decimal expansion of $\dfrac{6}{15}$ is $\dfrac{2}{5}=\dfrac{2\times 2}{5\times 2}=\dfrac{4}{10}=0.4$ 

(vi) The decimal expansion of $\dfrac{35}{50}$ is $\dfrac{7}{10}=0.7$

10. 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 form $\dfrac{p}{q}$, what can you say about the prime factors of $q$?                                                                               

(i). Number - $43.123456789$

Ans: Since the decimal expansion is terminating so it is a rational number and therefore it can be expressed in $\dfrac{p}{q}$ form.

 \[43.123456789=\dfrac{43123456789}{{{10}^{9}}}\]

Hence, the factors of $q$ are of the form ${{2}^{n}}\times {{5}^{m}}$ where $n\text{ and }m$ are non-negative integers.

(ii). Number - $0.1201120012000120000...$

Ans: Since the decimal expansion is neither terminating nor non-terminating repeating so it is an irrational number.

(iii). Number - $43.\overline{123456789}$

Ans: Since the decimal expansion is non-terminating repeating so it is a rational number and therefore it can be expressed in $\dfrac{p}{q}$ form.

\[43.\overline{123456789}=43\cdot \left( \dfrac{123456789}{{{10}^{9}}} \right)\cdot \left( \dfrac{123456789}{{{10}^{9}}} \right)\cdot \left( \dfrac{123456789}{{{10}^{9}}} \right)\cdot ...\]

11. Show that every positive integer is of the form $2q$ and that every positive odd integer is of the form $2q+1$ for some integer $q$.                               

Ans: Let $a$ be any positive integer, then by Euclid’s algorithm we can write $a=2q+r,$ for some integer $q\ge 0$.   …..(1)

Clearly, in the expression (1) we are dividing $a$ by $2$ with quotient $q$ and remainder $r$, $r=0,1$ because $0\le r  < 2$.  …..(2)

$a=2q$ or $2q+1$   …..(3)

If $a=2q$ (which is even)

If $a=2q+1$ (which is odd)

So, every positive even integer is of the form $2q$ and odd integer is of the form $2q+1$.

12. Show that any number of the form ${{4}^{n}}$, $n\in N$ can never end with the digit $0$.  

Ans : The prime factorization of $10$ is $2\times 5$. Therefore, for a number to end with $0$ it should have both 2 and 5 as a factor.

For the given number of the form ${{4}^{n}}={{[{{2}^{2}}]}^{n}}={{2}^{2n}}$, we can observe that it has only $2$ as a factor and not $5$. So ${{4}^{n}},n\in N$ can never end with the digit $0.$

13. Use Euclid’s Division Algorithm to find the $HCF$ of $4052\text{ and }12576$.

Ans: Step 1: Since$12576 >4052$ , we will apply Euclid’s division lemma such that $12576=4052\times 3+420$   …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[4052\] and \[420\], such that $4052=420\times 9+272$   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $420$ and $272$, such that $420=272\times 1+148$.   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to $148$ and $272$, such that $272=148\times 1+124$.   …..(4) 

Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s division lemma to $148$ and $124$, such that $148=124\times 1+24$.  …..(5) 

Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s division lemma to $24$ and $124$, such that $124=24\times 5+4$.  …..(6) 

Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s division lemma to $24$ and $4$, such that $24=4\times 6+0$.  …..(7) 

Now the remainder in equation (7) is zero. Therefore, $HCF$ of $12576$ and $4052$ is $'4'.$

14. Given that $HCF$ of two numbers is $23$ and there $LCM$ is $1449$. If one of the numbers is $161$, find the other.                             

Ans : It is given that the HCF of the two numbers is $23$ and their LCM is $1449$. We have to find the other number if one of the numbers is $161$. Let the other number be $a$. 

$1449\times 23=a\times 161$ 

$\Rightarrow a=\dfrac{1449\times 23}{161}$

$\therefore a=207$

15. Show that every positive odd integer is of the form $(4q+1)$ or $(4q+3)$ for some integer $q$.

Ans: Let $a$ be any positive integer, then by Euclid’s algorithm we can write $a=4q+r,$ for some integer $q\ge 0$. …..(1)

Clearly, in the expression (1) we are dividing $a$ by $4$ with quotient $q$ and remainder $r$, $r=0,1,2,3$ because $0\le r  < 4$.     …..(2)

$a=4q$ or $4q+1$ or $4q+2$ or $4q+3$     …..(3)

If $a=4q=2\left( 2q \right)$ (which is even)

If $a=4q+1=2\left( 2q \right)+1$ (which is odd)

If $a=4q+2=2(2q+1)$ (which is even)

If $a=4q+3=2(2q+1)+1$ (which is odd) 

So, every positive odd integer is of the form $(4q+1)$ or $(4q+3)$.

16. Show that any number of the form ${{6}^{x}},x\in N$ can never end with the digit $0.$       

For the given number of the form ${{6}^{n}}={{(2\times 3)}^{n}}={{2}^{n}}\times 3{}^{n}$ . , we can observe that it has only $2$ as a factor and not $5$. So ${{6}^{x}},x\in N$ can never end with the digit $0.$

17. Find $HCF$ and $LCM$ of $18$ and $24$ by the prime factorization method.   

Ans: The prime factorization of $18$ is $18=2\times 3\times 3=2\times {{3}^{2}}$  …..(1)

The prime factorization of $24$ is $24=2\times 2\times 2\times 3={{2}^{3}}\times 3$   …..(2)

From (1) and (2) we can see that $18,24$ has prime factors $2,3$ in common. 

$\therefore HCF=2\times 3=6$

From (1) and (2) we can see that $3$ appears twice in prime factorization of $18$ and $2$ appears thrice in prime factorization of $24$.

 $\therefore LCM=\left( {{3}^{2}} \right)\times \left( {{2}^{3}} \right)=72$

18. The $HCF$ of two numbers is $23$ and their $LCM$ is $1449$. If one of the numbers is $161,$ find the other.

Ans: It is given that the HCF of the two numbers is $23$ and their LCM is $1449$. We have to find the other number if one of the numbers is $161$. Let the other number be $a$. 

19. Prove that the square of any positive integer of the form $5g+1$ is of the same form.

Ans: Let $a$ be any positive integer, of the form $a=5g+1$      …..(1)

Squaring both sides of equation (1) we get,

\[{{a}^{2}}={{(5g+1)}^{2}}\]

\[\Rightarrow {{a}^{2}}=25{{g}^{2}}+10g+1\] 

\[\Rightarrow {{a}^{2}}=5(5{{g}^{2}}+2g)+1\]

\[\Rightarrow {{a}^{2}}=5m+1\] 

Hence, the square of any positive integer of the form $5g+1$ is of the same form.

20. Use Euclid’s Division Algorithm to find the $HCF$ of $4052$ and $12576$.      

Ans: Step 1 : Since$12576 >4052$ , we will apply Euclid’s division lemma such that $12576=4052\times 3+420$    …..(1)

Step 3 : Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $420$ and $272$, such that $420=272\times 1+148$.    …..(3) 

Step 5: Again, the remainder in (4) is not zero so we have to apply the Euclid’s division lemma to $148$ and $124$, such that $148=124\times 1+24$.     …..(5) 

Step 6: Again, the remainder in (5) is not zero so we have to apply the Euclid’s division lemma to $24$ and $124$, such that $124=24\times 5+4$.    …..(6) 

Step 7: Again, the remainder in (6) is not zero so we have to apply the Euclid’s division lemma to $24$ and $4$, such that $24=4\times 6+0$.   …..(7) 

Now the remainder equation (7) is zero. Therefore, $HCF$ of $12576$ and $4052$ is $'4'.$

21. Find the largest number which divides $245$ and $1029$ leaving remainder $5$ in each case.

Ans. Since we want the remainder to be $5$, therefore the required number is the $HCF$ of $(245-5)$ and $(1029-5)$ i.e., $240$ and $1024$.

Step 1: Since$1024 >240$ , we will apply Euclid’s division lemma such that $1024=240\times 4+64$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[240\] and \[64\], such that $240=64\times 3+48$  …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $64$ and $48$, such that$64=48\times 1+16$.   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to $48$ and $16$, such that$48=16\times 3+0$.    …..(4) 

Now the remainder in equation (4) is zero. Therefore, $HCF$ of $1024$ and $240$ is $'16'$.

The largest number which divides $245$ and $1029$ leaving remainder $5$ in each case is $16$. 

22. A shopkeeper has $120$ litres of petrol, $180$ litres of diesel and $240$ litres of kerosene. He wants to sell oil by filling the three kinds of oils in tins of equal capacity. What should be the greatest capacity of such a tin?                                                                         

Ans: The required greatest capacity is the $HCF$ of $120,180$ and $240$. First let us find the HCF of $240,180$ by Euclid’s division algorithm.

Step 1: Since$240 >180$ , we will apply Euclid’s division lemma such that $240=180\times 1+60$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[180\] and \[60\], such that $180=60\times 3+0$     …..(2)

Now the remainder in equation (2) is zero. Therefore, $HCF$ of $180$ and $240$ is $60$.

Let us now find the HCF of $60,120$. Clearly, $120=60\times 2+0$.

\[\therefore HCF\] of $120,180$ and $240$ is $60$ and hence the required capacity is $60$ litres.

Short Answer Questions                                                                               (3 Marks)

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

(i). $135$ and $225$

Ans: Step 1: Since $225 >135$, we will apply Euclid’s division lemma such that $225=135\times 1+90$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to \[135\] and \[90\], such that $135=90\times 1+45$ …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $90$ and $45$, such that $90=2\times 45+0$.   …..(3) 

Now the remainder in equation (3) is zero. Therefore, the $HCF$ of $135$ and $225$ is $45$.

(ii). $196$ and $38220$

Ans: Since $38220 >196$, we will apply Euclid’s division lemma such that $38220=196\times 195+0$                     …..(1)

Since, the remainder in (1) is zero. Therefore, the $HCF$ of $38220$ and $196$ is $196$.

(iii). $867$ and $255$

Ans: Step 1: Since $867 >255,$, we will apply Euclid’s division lemma such that $867=255\times 3+102$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to $255$ and $102$, such that $255=102\times 2+51$   …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $102$ and $51$, such that $102=51\times 2+0$.   …..(3) 

Now the remainder is zero. Therefore, the $HCF$ of $867$ and $255$ is $51$.

2. Find the greatest number of $6$ digits exactly divisible by $24,15$ and $36.$

Ans: To find the greatest number of $6$ digits exactly divisible by $24,15$ and $36$, first we have to find the LCM of $24,15$ and $36$.

The prime factorization of $24,15$ and $36$ are:

$24=2\times 2\times 2\times 3$    …..(1)

$15=5\times 3$    …..(2)

$36=2\times 2\times 3\times 3$   …..(3)

Hence, from (1), (2) and (3), $LCM\left( 15,24,36 \right)=2\times 2\times 2\times 5\times 3\times 3=360$.  .….(4)

Now, we know that the greatest number of $6$ digits is $999999.$  …..(5)

From (4) and (5), using Euclid’s division lemma we get, $999999=360\times 2777+279$.

$\Rightarrow 999999-279=360\times 2777$ 

$\Rightarrow 999720=360\times 2777$

Hence, the greatest number of $6$ digits exactly divisible by $24,15$ and $36$ is $999720$.  

3. Prove that the square of any positive integer is of the form $4m$ or $4m+1$ for some integer $m$.

Ans: Let $a$ be any positive integer, then by Euclid’s algorithm we can write $a=4q+r,$ for some integer $q\ge 0$.      …..(1)

Clearly, in the expression (1) we are dividing $a$ by $4$ with quotient $q$ and remainder $r$, $r=0,1,2,3$ because $0\le r  < 4$.   …..(2)

$a=4q$ or $4q+1$ or $4q+2$ or $4q+3$   …..(3)

Case 1: If $a=4q$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}$

$\Rightarrow {{a}^{2}}=4\left( 4{{q}^{2}} \right)$ …..(4)

Equation (4) could be written as \[{{a}^{2}}=4m\] for some integer $m$ ….. (5)

Case 2: If $a=4q+1$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}+8q+1$

$\Rightarrow {{a}^{2}}=4\left( 4{{q}^{2}}+2q \right)+1$ …..(6)

Equation (6) could be written as \[{{a}^{2}}=4m+1\] for some integer $m$ ….. (7)

Case 3: If $a=4q+2$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}+16q+4$

$\Rightarrow {{a}^{2}}=4\left( 4{{q}^{2}}+4q+1 \right)$ …..(8)

Equation (8) could be written as \[{{a}^{2}}=4m\] for some integer $m$ ….. (9)

Case 4: If $a=4q+3$ then, squaring both sides we get,

${{a}^{2}}=16{{q}^{2}}+24q+9$

${{a}^{2}}=4\left( 4{{q}^{2}}+6q+2 \right)+1$ …..(10)

Equation (10) could be written as \[{{a}^{2}}=4m+1\] for some integer $m$ ….. (11)

Therefore, from equations (5), (7), (9) and (11) it can be said that the square of any positive integer is either of the form $4m$ or $4m+1$ for some integer $m$.

4. There are $144$ cartons of coke can and $90$ cartons of Pepsi can to be stacked in a canteen. If each stack is of the same height and is to contain cartoons of the same drink. What would be the greater number of cartons each stack would have?     

Ans: Given $144$ cartons of coke can and $90$ cartons of Pepsi can are to be stacked in a canteen. For stack to be of the same height and is to contain cartoons of the same drink, we have to find the $HCF$ of $144$ and $90$.

Using Euclid’s division algorithm,

Step 1: Since $144 >90,$, we will apply Euclid’s division lemma such that $144=90\times 1+54$                     …..(1)

Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to $90$ and $54$, such that $90=54\times 1+36$  …..(2)

Step 3: Again, the remainder in (2) is not zero so we have to apply the Euclid’s division lemma to $54$ and $36$, such that $54=36\times 1+18$   …..(3) 

Step 4: Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to $18$ and $36$, such that$36=18\times 2+0$.      …..(4)

Now the remainder in (4) is zero. Therefore, the $HCF$ of $144$ and $90$ is $18$.

So, greatest number of cartoons is $18.$

5. Prove that product of three consecutive positive integers is divisible by $6$.   

Ans: Let us consider three consecutive positive integers $x,(x+1)$ and $(x+2)$.

Using Euclid’s lemma, let us write $x=6q+r$   …..(1)

Clearly, in the expression (1) we are dividing $a$ by $6$ with quotient $q$ and remainder $r$, $r=0,1,2,3,4,5$ because $0\le r  < 6$.     …..(2)

Therefore from (1) and (2) we can get,                                                                                       

$x=6q,6q+1,6q+2,6q+3,6q+4,6q+5$

Case 1: If $x=6q$ then product 

$x(x+1)(x+2)=6q(6q+1)(6q+2)$, which is divisible by $6$.   …..(3)

Case 2: If $x=6q+1$ then product 

$x(x+1)(x+2)=(6q+1)(6q+2)(6q+3)$

$\Rightarrow x(x+1)(x+2)=2(3q+1).3(2q+1)(6q+1)$ 

$\Rightarrow x(x+1)(x+2)=6(3q+1).(2q+1)(6q+1)$ 

 which is divisible by $6$.    …..(4)

Case 3: If $x=6q+2$ then product 

$x(x+1)(x+2)=(6q+2)(6q+3)(6q+4)$

$\Rightarrow x(x+1)(x+2)=3(2q+1).2(3q+1)(6q+4)$ 

$\Rightarrow x(x+1)(x+2)=6(2q+1)(3q+1)(6q+4)$ 

 which is divisible by $6$.  …..(5)

Case 4: If $x=6q+3$ then product 

$x(x+1)(x+2)=(6q+3)(6q+4)(6q+5)$

$\Rightarrow x(x+1)(x+2)=6(2q+1)(3q+2)(6q+5)$ 

 which is divisible by $6$.   …..(6)

Case 5: If $x=6q+4$ then product 

$x(x+1)(x+2)=(6q+4)(6q+5)(6q+6)$

$\Rightarrow x(x+1)(x+2)=6(6q+4)(6q+5)(q+1)$ 

 which is divisible by $6$.     …..(7)

Case 6: If $x=6q+5$ then product 

$x(x+1)(x+2)=(6q+5)(6q+6)(6q+7)$

$\Rightarrow x(x+1)(x+2)=6(6q+5)(q+1)(6q+7)$ 

 which is divisible by $6$.   …..(8)

From equations (3) to (8) we can conclude that the product of three consecutive positive integers is divisible by $6$.

6. Prove that $\left( 3-\sqrt{5} \right)$ is an irrational number.

Ans: We will prove this by contradiction. To solve this problem, suppose that $\left( 3-\sqrt{5} \right)$ is rational i.e., $\exists $ two co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{a}{b}=3-\sqrt{5}$      …..(1)

$\dfrac{a}{b}-3=-\sqrt{5}$

$\Rightarrow \dfrac{3b-a}{b}=\sqrt{5}$      …..(2)

Since we know that $\sqrt{5}$ is irrational. Therefore, from (2) $\dfrac{3b-a}{b}$should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $\left( 3-\sqrt{5} \right)$ cannot be rational.

7. Prove that if $x$ and $y$ are odd positive integers, then ${{x}^{2}}+{{y}^{2}}$ is even but not divisible by $4$.

Ans: Let $x=2p+1$ and $y=2q+1$ be two odd positive integers $\forall p,q\in \mathbb{N}$.       …..(1)

Squaring $x,y$ from (1) and adding them we get, 

\[{{x}^{2}}+{{y}^{2}}={{(2p+1)}^{2}}+{{(2q+1)}^{2}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=\left( 4{{p}^{2}}+4p+1 \right)+\left( 4{{q}^{2}}+4q+1 \right)\]

$\Rightarrow {{x}^{2}}+{{y}^{2}}=2(2{{p}^{2}}+2{{q}^{2}}+2p+2q+1)$

$\Rightarrow {{x}^{2}}+{{y}^{2}}=2m$    …..(2)

Where \[m=(2{{p}^{2}}+2{{q}^{2}}+2p+2q+1)\] 

From (2), it is clear that ${{x}^{2}}+{{y}^{2}}$ is even but not divisible by $4$.

8. Show that one and only one out of $n,(n+2)$ or $(n+4)$ is divisible by $3$, where $n\in N.$

Ans: Let $n$ be a number such that $n=3q+r$.   …..(1)

Clearly, in the expression (1) we are dividing $n$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.    …..(2)

Therefore from (1) and (2) we can get, $n=3q,3q+1,3q+2$

Case 1: If $n=3q$ then, numbers are $3q,(3q+2)(3q+4)$ out of these $3q$ is divisible by $3$.     …..(1)

Case 2: If $n=3q+1$ then, numbers are $(3q+1),(3q+3),(3q+5)$ out of these $(3q+3)$ is divisible by $3$.     …..(2)

Case 3: If $n=3q+2$ then, numbers are $(3q+2),(3q+4),(3q+6)$ out of these $(3q+6)$ is divisible by $3$.     …..(3)

$\therefore $ from (1), (2) and (3) we can conclude that out of $n,(n+2)$ and $(n+4)$ only one is divisible by $3$.

9. Use Euclid’s Division Lemma to show that the square of any positive integer of the form $3m$ or $(3m+1)$ for some integer $q$.

Ans: Let $a$ be any positive integer, then we can write $a=3q+r$ for some integer $q\ge 0$    …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r  < 3$.     ..…(2)

Therefore from (1) and (2) we can get, $a=3q$ or $3q+1$ or $3q+2$.     ……(3)

$\Rightarrow {{a}^{2}}=9{{q}^{2}}$ or $9{{q}^{2}}+6q+1$ or $9{{q}^{2}}+12q+4$    …..(4)

${{a}^{2}}=3\times (3{{q}^{2}})$ or $3(3{{q}^{2}}+2q)+1$ or $3(3{{q}^{2}}+4q+1)+1$.    …..(5)

10. Prove that if $\sqrt{n}$ is not a rational number, if $n$ is not a perfect square. 

Let us suppose that $\sqrt{n}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\sqrt{n}=\dfrac{a}{b}$

$\Rightarrow b\sqrt{n}=a$   …..(1)

Squaring both sides of (1), we get $n{{b}^{2}}={{a}^{2}}$   …..(2)

From (2) we can conclude that $n$ is factor of ${{a}^{2}}$, this means that $n$ is also factor of $a$.    …..(3)

From (3) we can write $a=nc$ for some integer $c$.   …..(4)

$\begin{align} & n{{b}^{2}}={{n}^{2}}{{c}^{2}} \\ & \Rightarrow {{b}^{2}}=n{{c}^{2}} \\ \end{align}$

It means that $n$ is factor of ${{b}^{2}}$, this means that $n$ is also factor of $b$.    …..(5)

From $(3)\text{,}(5)$, we can say that $n$ is factor of both $a\text{,}b$, which is a contradiction to our assumption. Therefore, our assumption was wrong and $\sqrt{n}$ is irrational.

11. Prove that the difference and quotient of $(3+2\sqrt{3})$ and $(3-2\sqrt{3})$ are irrational. 

Ans: Difference of $(3+2\sqrt{3})$ and $(3-2\sqrt{3})$ is $\left( 3+2\sqrt{3} \right)-\left( 3-2\sqrt{3} \right)=4\sqrt{3}$.

We will prove this by contradiction.

Let us suppose that $4\sqrt{3}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $4\sqrt{3}=\dfrac{a}{b}$

$\Rightarrow 4b\sqrt{3}=a$   …..(1)

Squaring both sides of (1), we get $48{{b}^{2}}={{a}^{2}}$  …..(2)

From (2) we can conclude that $48$ is factor of ${{a}^{2}}$, this means that $48$ is also factor of $a$.   …..(3)

From (3) we can write $a=48c$ for some integer $c$.   …..(4)

$\begin{align} & 4\sqrt{3}{{b}^{2}}=48{{c}^{2}} \\ & \Rightarrow {{b}^{2}}=4\sqrt{3}{{c}^{2}} \\ \end{align}$

It means that $48$ is factor of ${{b}^{2}}$, this means that $48$ is also factor of $b$.   …..(5)

From $(3)\text{,}(5)$, we can say that $48$ is factor of both $a\text{,}b$, which is a contradiction to our assumption. Therefore, our assumption was wrong and $4\sqrt{3}$ is irrational.

Now, the quotient of $\left( 3+2\sqrt{3} \right),\left( 3-2\sqrt{3} \right)$ is

$\dfrac{3+2\sqrt{3}}{3-2\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3-2\sqrt{3}}\times \dfrac{3+2\sqrt{3}}{3+2\sqrt{3}}$

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

Similarly, using the approach used above, we can prove that \[-\left( 5+4\sqrt{3} \right)\] is also irrational. 

12. Show that $({{n}^{2}}-1)$ is divisible by $8$, if $n$ is an odd positive integer.

Ans: Let $n=4q+1$ be an odd positive integer.   …..(1)

Then from (1),

${{n}^{2}}-1={{(4q+1)}^{2}}-1$

$\Rightarrow {{n}^{2}}-1=16{{q}^{2}}+8q$    …..(2)

Taking $8$ common from RHS of equation (2) we get,

${{n}^{2}}-1=8\left( 2{{q}^{2}}+q \right)$ 

\[\Rightarrow {{n}^{2}}-1=8m\] where $m=2{{q}^{2}}+q$.   …..(3)

Hence, from (3) we can conclude that $({{n}^{2}}-1)$ is divisible by $8$, if $n$ is an odd positive integer.

13. Use Euclid Division Lemma to show that cube of any positive integer is either of the form $9m,(9m+1)$ or $(9m+8)$.

Ans : Let $a$ be any positive integer, then we can write $a=3q+r$ for some integer $q\ge 0$    …..(1)

Clearly, in the expression (1) we are dividing $a$ by $3$ with quotient $q$ and remainder $r$, $r=0,1,2$ because $0\le r <3$.    ..…(2)

${{a}^{3}}={{(3q)}^{3}}$ or ${{(3q+1)}^{3}}$ or ${{(3q+2)}^{3}}$ 

$\Rightarrow {{a}^{3}}=27{{q}^{3}}$ or $27{{q}^{3}}+27{{q}^{2}}+9q+1$ or $27{{q}^{3}}+54{{q}^{2}}+36q+8$    …..(4)

Taking $9$ common from RHS of equation (4) we get,

${{a}^{3}}=9\left( 3{{q}^{3}} \right)$ or $9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$ or $9\left( 3{{q}^{3}}+6{{q}^{2}}+4q \right)+8$    …..(5)

${{a}^{3}}=9{{k}_{1}}$ or $9{{k}_{2}}+1$ or $(9{{k}_{3}}+1)$ for some positive integers ${{k}_{1}},{{k}_{2}}$ and ${{k}_{3}}$.

Hence, it can be said that the square of any positive integer is either of the form $9m,9m+1$ or $3m+8$.

Long Answer Questions                                                                                (4 Marks)

1. Prove that the following are irrationals.       

(i). $\dfrac{1}{\sqrt{2}}$

Ans: (i) We will prove this by contradiction. Let us suppose that $\dfrac{1}{\sqrt{2}}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$

$\Rightarrow \dfrac{b}{a}=\sqrt{2}$      …..(1)

Since we know that $\sqrt{2}$ is irrational. Therefore, from (1) $\dfrac{b}{a}$ should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $\dfrac{1}{\sqrt{2}}$ cannot be rational.

(ii). $7\sqrt{5}$

Ans: We will prove this by contradiction. Let us suppose that $7\sqrt{5}$ is rational. It means that we have co-prime integers $a$ and $b$ $(b\ne 0)$ such that $7\sqrt{5}=\dfrac{a}{b}$  

$\Rightarrow \sqrt{5}=\dfrac{a}{7b}$      …..(1)

Since we know that $\sqrt{5}$ is irrational. Therefore, from (1) $\dfrac{a}{7b}$ should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $7\sqrt{5}$ cannot be rational.

(iii). $6+\sqrt{2}$

Ans: We will prove this by contradiction. To solve this problem, suppose that $6+\sqrt{2}$ is rational i.e., $\exists $ two co-prime integers $a$ and $b$ $(b\ne 0)$ such that $\dfrac{a}{b}=6+\sqrt{2}$        …..(1)

Subtracting $6$ from both sides of the equation (1) and simplifying it further we get,

$\dfrac{a}{b}-6=\sqrt{2}$

$\Rightarrow \dfrac{a-6b}{b}=\sqrt{2}$       …..(2)

Since we know that $\sqrt{2}$ is irrational. Therefore, from (2) $\dfrac{a-6b}{b}$should be irrational but it could be written in $\dfrac{p}{q}$ form which means it is rational. It is not possible and hence our assumption is wrong and $6+\sqrt{2}$ cannot be rational.

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

(i). $\dfrac{13}{3125}$

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{13}{3125}$, the prime factorization of $q$ is $q=3125=5\times 5\times 5\times 5\times 5={{5}^{5}}$.                       …..(2)

From (2) we can see that the denominator $q=3125$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=5$ and $n=0$.

Therefore, it follows from theorem (1), the rational number $\dfrac{13}{3125}$ has a terminating decimal expansion.

(ii). $\dfrac{17}{8}$

For rational number $\dfrac{p}{q}=\dfrac{17}{8}$, the prime factorization of $q$ is 

$q=8=2\times 2\times 2={{2}^{3}}$.    …..(2)

From (2) we can see that the denominator $q=8$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where \[m=0\] and $n=3$.

Therefore, it follows from theorem (1), the rational number $\dfrac{17}{8}$ has a terminating decimal expansion.

(iii). $\dfrac{64}{455}$

For rational number $\dfrac{p}{q}=\dfrac{64}{455}$, the prime factorization of $q$ is 

$q=455=5\times 91$.   …..(2)

From (2) we can see that the denominator $q=455$ is not of the form ${{2}^{n}}\times {{5}^{m}}$.

Therefore, it follows from theorem (1), the rational number $\dfrac{64}{455}$ has a non-terminating repeating decimal expansion.

(iv). $\dfrac{15}{1600}$

For rational number $\dfrac{p}{q}=\dfrac{15}{1600}$, the prime factorization of $q$ is 

$q=320=2\times 2\times 2\times 2\times 2\times 2\times 5={{2}^{6}}\times 5$.    …..(2)

From (2) we can see that the denominator $q=8$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=1$ and $n=6$.

Therefore, it follows from theorem (1), the rational number $\dfrac{15}{1600}$ has a terminating decimal expansion.

(v). $\dfrac{29}{343}$

For rational number $\dfrac{p}{q}=\dfrac{29}{343}$, the prime factorization of $q$ is 

$q=343=7\times 7\times 7$.   …..(2)

From (2) we can see that the denominator $q=343$ is not of the form ${{2}^{n}}\times {{5}^{m}}$.

Therefore, it follows from theorem (1), the rational number $\dfrac{29}{343}$ has a non-terminating repeating decimal expansion.

(vi). $\dfrac{23}{{{2}^{3}}\times {{5}^{2}}}$

For rational number $\dfrac{23}{{{2}^{3}}\times {{5}^{2}}}$, the prime factorization of $q$ is 

$q={{2}^{3}}\times {{5}^{2}}$.   …..(2)

From (2) we can see that the denominator $q={{2}^{3}}\times {{5}^{2}}$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=2$ and $n=3$.

Therefore, it follows from theorem (1), the rational number $\dfrac{23}{{{2}^{3}}\times {{5}^{2}}}$ has a terminating decimal expansion.

(vii). $\dfrac{129}{{{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}}$

For rational number $\dfrac{129}{{{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}}$, the prime factorization of $q$ is 

$q={{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}$.    …..(2)

From (2) we can see that the denominator $q={{2}^{3}}\times {{5}^{2}}$ is not of the form ${{2}^{n}}\times {{5}^{m}}$.

Therefore, it follows from theorem (1), the rational number $\dfrac{129}{{{2}^{2}}\times {{5}^{7}}\times {{7}^{5}}}$ has a non-terminating repeating decimal expansion.

(viii). $\dfrac{6}{15}=\dfrac{2}{5}$

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative  integers. Then $x$ has a decimal expansion which terminates.            …..(1)

For rational number $\dfrac{p}{q}=\dfrac{6}{15}=\dfrac{2}{5}$, the prime factorization of $q$ is $q=5={{5}^{1}}$.    …..(2)

From (2) we can see that the denominator $q=5$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=1$ and $n=0$. Therefore, it follows from theorem (1), the rational number $\dfrac{6}{15}=\dfrac{2}{5}$ has a terminating decimal expansion.

(ix). $\dfrac{35}{50}=\dfrac{7}{10}$

Ans: Theorem: Let $x$ be a rational number expressed in the form , $\dfrac{p}{q}$ where $p$ and $q$are coprime, and the prime factorisation of $q$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m$ and $n$ are non-negative integers. Then $x$ has a decimal expansion which terminates.   …..(1)

For rational number $\dfrac{35}{50}=\dfrac{7}{10}$, the prime factorization of $q$ is $q=10=2\times 5={{2}^{1}}\times {{5}^{1}}$.  …..(2)

From (2) we can see that the denominator $q=5$ is of the form ${{2}^{n}}\times {{5}^{m}}$, where $m=1$ and $n=1$. Therefore, it follows from theorem (1), the rational number $\dfrac{35}{50}$ has a terminating decimal expansion.

(x). $\dfrac{77}{210}=\dfrac{11}{30}$

For rational number $\dfrac{77}{210}=\dfrac{11}{30}$, the prime factorization of $q$ is 

$q=30=2\times 3\times 5$.  …..(2)

Therefore, it follows from theorem (1), the rational number $\dfrac{77}{210}=\dfrac{11}{30}$ has a non-terminating repeating decimal expansion.

Maths Chapter 1- Real Numbers

Introduction.

All numbers can be classified into two categories -- Real and Imaginary. A real number can be actually perceived and can be represented in a number line. It includes all integers -- positive and negative, all natural numbers (1, 2, 3, ….., ) all whole numbers (0, 1, 2, 3, ….., ), fractions, rational and irrational numbers.

An imaginary number on the other hand cannot be perceived. Square root of all negative numbers is imaginary.

A real number can be represented as an infinite decimal expansion. Ex: 5 = 5.0000….., ⅓ = 0.3333……

A real number can be rational or irrational.

All rational numbers and all irrational numbers together make a group of real numbers.

(i) A group of rational numbers is denoted by Q.

(ii) A group of irrational numbers is denoted by Q’.

(iii) A group of real numbers is denoted by R.

A rational number an be written in the form pq, both p and q being integers (q 0) while an irrational number can not be impressed in such a way.

The decimal expansion of a rational number say x can terminate after a finite number of digits.

Ex: ⅝ = 0.625 or eventually begins to repeat the same finite sequence of digits continuously. 

Ex: 56/99 = 0.565656……….

On the other hand in case of an irrational number, the decimal expansion continues without repeating. Ex: 1.7320508….,  𝜋= 3.141592….)

Important Questions For CBSE Class 10 Math :

Important Questions For CBSE Class 10 Math Chapter 2 Polynomials

Important Questions For CBSE Class 10 Math Chapter 3 Pair of Linear Equations In Two Variables

Important Questions For CBSE Class 10 Math Chapter 4 Quadratic Equations

Important Questions For CBSE Class 10 Math Chapter 5 Arithmetic Progression

Important Questions For CBSE Class 10 Math Chapter 6 Triangles

Important Questions For CBSE Class 10 Math Chapter 7 Coordinate Geometry

Important Questions For CBSE Class 10 Math Chapter 8 Introduction To Trigonometry

Important Questions For CBSE Class 10 Math Chapter 9 Some Application To Trigonometry

Important Questions For CBSE Class 10 Math Chapter 10 Circles

Important Questions For CBSE Class 10 Math Chapter 11 Constructions

Important Questions For CBSE Class 10 Math Chapter 12 Area Related To Circles

Important Questions For CBSE Class 10 Math Chapter 13 Surface Area And Volume

Important Questions For CBSE Class 10 Math Chapter 14 Statistics

Important Questions For CBSE Class 10 Math Chapter 15 Probability

Euclid’s Division Lemma and Algorithm

First, let us understand the meaning of Lemma and Algorithms.

A Lemma is a minor proven statement which is used to prove other statements. The Euclid’s division Lemma is actually another statement of the common long division process.

For example, on dividing 36 by 5, we get quotient = 7 and remainder = 1 which is less than 5.

If there are positive integers a and b, then there are unique integers existing q and r that satisfy a = bq + r, 0 rb.

Example: If a =47 and b = 9, then we can write 

  47 = 9 x 5 + 2, where q = 4 and r = 15 b

Note: 

(i) The unique characteristics of qand r are nothing but the quotient and remainder respectively.

(ii) Although Euclid’s division algorithm is expressed for only positive integers but it can be extended for all integers except 0.

(iii) When aand bare positive integers, then qand rcan take values only from whole numbers.

An algorithm means a series of well defined steps, which provides  a procedure of calculations repeatedly successfully on the results of earlier steps. With the help of defined procedure the desired result can be obtained. 

Euclid’s Division Algorithm

The Euclid’s Division Algorithm is also an algorithm to determine the H.C.F (Highest Common Factor) of the given positive integers. 

Ordinarily, if we want to find the HCF of two positive integers, we first find the factors of the two integers. 

For ex: For two integers 24 and 42, the common factors are 1, 2, 3 and 6. As 6 is the highest of other numbers, 6 is the HCF of 24 and 42. 

In Euclid’s Division Algorithm, the largest integer is divided by the smaller one and a remainder is obtained. 

Next the smaller integer is divided by the above remainder and the process is repeated till zero is obtained. The remainder in the last but one division is the required HCF.

The Fundamental Theorem of Arithmetic

According to this theorem every positive integer is either a prime number or it can be expressed as the product of primes. It is also known as Unique Factorization Theorem. Thus, any integer that is greater than 1 can be either a prime number or can be expressed as a product of prime factors.

Example: 8 = 2 x 2 x 2, where 2 is a prime factor. 

(i) When a number has no factors other than 1 and the number itself, the number is called a prime number.

(ii) 1 is neither prime nor composite.

(iii) A number is a composite number if it has at least one factor other than 1 and the number itself.

(iv) 2 is the smallest prime number and an even number. It is the only number that is both prime and even.

(v) Two numbers are co-prime if they have no common factors other than 1, i.e, their HCF = 1.

This theorem is termed as ‘Fundamental’ because of its importance in the development of number theory. 

Factor Tree

A chain of factors that is demonstrated in the form of a tree, is called a factor tree. 

HCF and LCM using Prime Factorization

HCF is the product of the smallest power of each common prime factor present in the numbers.

The product of the largest power of each prime factor present in the numbers is the LCM. 

Important Note: 

(i) LCM of two or more numbers is the smallest number which is the smallest number and divisible by all the given numbers.

(ii) If there is no common prime factor, then the HCF of the given number is 1.

Relationship Between Numbers and Their HCF and LCM

For given positive integers aand b, the relation between these numbers and their HCF and LCM is 

HCF (a, b) = \[\frac{a \times b}{LCM(a,b)}\] or LCM (a, b) = \[\frac{a          \times b}{HCF(a,b)}\]

For given three positive integers a, b and c, the relation between these numbers and their HCF and LCM is  HCF (a, b, c) = \[\frac{a \times b \times c \times LCM(a, b, c) }{LCM(a,b) \times LCM(b, c) \times LCM(c, a)}\] or LCM (a, b,c) = \[\frac{a \times b \times c \times HCF(a, b, c) }{HCF(a,b) \times HCF(b, c) \times HCF(c, a)}\]

Method of Proving Irrationality of Numbers

We use the method of contradiction in order to prove irrationality of the numbers. i.e., first we assume that the given number is rational and after reaching a contradiction we prove that the given number is irrational. 

If a prime number  p divides a 2 , then p divides a, where ais a positive integer.

Let us understand how to write the decimal expansion of those rational numbers with terminating decimal expansion

Let a rational number in the lowest form be  p/q such that the prime factorization of q is of the form 2 m x 5 n , where m, n are non-negative integers. To write decimal expansion of p/q, convert p/q to an equivalent rational number of the form c/d, where dis a power of 10. 

The above notes will be a huge benefit for solving the Important Questions Of Chapter 1 of Maths for Class 10, Real Numbers . You can now master the Euclid’s Division Lemma and Algorithm. The important questions for this chapter is basically a compilation of higher and more advanced techniques a student can expect in Class 10 examinations. The questions of the chapter and the notes related to the topic provided by Vedantu will not only help you to understand the concept better but also solve the questions successfully. If you harbour any doubts then you can get answers to all your queries with our experienced teachers by registering with Vedantu and gain expertise in the subject. 

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Important Related Links for CBSE Class 10 Maths

FAQs on Important Questions for CBSE Class 10 Maths Chapter 1 - Real Numbers 2024-25

1. What do you mean by HCF and LCM?

Ans: HCF stands for "Highest Common Factor," also known as the Greatest Common Divisor (GCD). It refers to the largest number that divides two or more integers without leaving a remainder. In other words, the HCF is the largest common factor shared by a set of numbers.

LCM stands for "Least Common Multiple." It refers to the smallest positive integer that is divisible by two or more integers without leaving a remainder. In other words, the LCM is the smallest common multiple shared by a set of numbers.

2. How will Important Questions for Chapter 1 for Class 10 help in exams?

Ans: Important Questions for Chapter 1 in Class 10 will help in exams by focusing on the key concepts, topics, and problem-solving techniques. These questions are carefully selected to cover the important aspects of the chapter, allowing students to practice and gain confidence in their understanding. They serve as a valuable tool for exam preparation and improving overall performance.

3. What does Chapter 1, real number for Class 10 discuss?

Ans: Chapter 1, "Real Numbers," for Class 10 discusses the concept of real numbers and their properties. It covers topics such as rational numbers, irrational numbers, decimal representation of rational numbers, fundamental theorem of arithmetic, Euclid's division algorithm, and the concept of prime and composite numbers. The chapter also introduces the concept of HCF (Highest Common Factor) and LCM (Least Common Multiple) of numbers. It provides a foundation for understanding number systems and lays the groundwork for further mathematical concepts in higher grades.

4. Which is the most important chapter in maths class 10?

Ans: The most important chapter in maths are:

Chapter 3 - Linear Equations

Chapter 5 - Arithmetic Progression

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometry

Chapter 8 - Introduction to Trigonometry

Chapter 9 - Applications of Trigonometry

Chapter 13 - Surface Areas & Volumes

Chapter 14 - Statistics

5. What is the weightage of marks in class 10 maths 2024-25?

Ans: CBSE Class 10 Mathematics in the academic year 2024-25 includes a theory paper worth 80 marks and internal assessment carrying 20 marks. The theory paper consists of 30 questions divided into four sections, with different marks assigned to each question. The total duration of the examination is three hours.

CBSE Class 10 Maths Important Questions

Cbse study materials.

• NCERT Solutions
• NCERT Solutions for Class 10
• NCERT Solutions for Class 10 Maths
• 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

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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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MCQ Questions for Class 10 Maths Real Numbers with Answers

October 18, 2019 by Veerendra

Free PDF Download of CBSE Class 10 Maths Chapter 1 Real Numbers Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Real Numbers MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 1 Real Numbers

1. The decimal form of \(\frac{129}{2^{2} 5^{7} 7^{5}}\) is (a) terminating (b) non-termining (c) non-terminating non-repeating (d) none of the above

2. HCF of 8, 9, 25 is (a) 8 (b) 9 (c) 25 (d) 1

3. Which of the following is not irrational? (a) (2 – √3)2 (b) (√2 + √3)2 (c) (√2 -√3)(√2 + √3) (d)\(\frac{2 \sqrt{7}}{7}\)

4. The product of a rational and irrational number is (a) rational (b) irrational (c) both of above (d) none of above

5. The sum of a rational and irrational number is (a) rational (b) irrational (c) both of above (d) none of above

6. The product of two different irrational numbers is always (a) rational (b) irrational (c) both of above (d) none of above

7. The sum of two irrational numbers is always (a) irrational (b) rational (c) rational or irrational (d) one

8. If b = 3, then any integer can be expressed as a = (a) 3q, 3q+ 1, 3q + 2 (b) 3q (c) none of the above (d) 3q+ 1

9. The product of three consecutive positive integers is divisible by (a) 4 (b) 6 (c) no common factor (d) only 1

10. The set A = {0,1, 2, 3, 4, …} represents the set of (a) whole numbers (b) integers (c) natural numbers (d) even numbers

11. Which number is divisible by 11? (a) 1516 (b) 1452 (c) 1011 (d) 1121

12. LCM of the given number ‘x’ and ‘y’ where y is a multiple of ‘x’ is given by (a) x (b) y (c) xy (d) \(\frac{x}{y}\)

13. The largest number that will divide 398,436 and 542 leaving remainders 7,11 and 15 respectively is (a) 17 (b) 11 (c) 34 (d) 45

Answer: a Explaination:(a); [Hint. Algorithm 398 – 7 – 391; 436 – 11 = 425; 542 – 15 = 527; HCF of 391, 425, 527 = 17]

14. There are 312, 260 and 156 students in class X, XI and XII respectively. Buses are to be hired to take these students to a picnic. Find the maximum number of students who can sit in a bus if each bus takes equal number of students (a) 52 (b) 56 (c) 48 (d) 63

Answer: a Explaination:(a); [Hint. HCF of 312, 260, 156 = 52]

15. There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field. Harish takes 12 minutes. 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 ? (a) 36 minutes (b) 18 minutes (c) 6 minutes (d) They will not meet

Answer: a Explaination:(a); [Hint. LCM of 18 and 12 = 36]

16. Express 98 as a product of its primes (a) 2² × 7 (b) 2² × 7² (c) 2 × 7² (d) 2 3 × 7

17. Three farmers have 490 kg, 588 kg and 882 kg of wheat respectively. Find the maximum capacity of a bag so that the wheat can be packed in exact number of bags. (a) 98 kg (b) 290 kg (c) 200 kg (d) 350 kg

Answer: a Explaination:(a); [Hint. HCF of 490, 588, 882 = 98 kg]

18. For some integer p, every even integer is of the form (a) 2p + 1 (b) 2p (c) p + 1 (d) p

19. For some integer p, every odd integer is of the form (a) 2p + 1 (b) 2p (c) p + 1 (d) p

20. m² – 1 is divisible by 8, if m is (a) an even integer (b) an odd integer (c) a natural number (d) a whole number

21. If two positive integers A and B can be ex-pressed as A = xy3 and B = xiy2z; x, y being prime numbers, the LCM (A, B) is (a) xy² (b) x 4 y²z (c) x 4 y 3 (d) x 4 y 3 z

22. The product of a non-zero rational and an irrational number is (a) always rational (b) rational or irrational (c) always irrational (d) zero

23. If two positive integers A and B can be expressed as A = xy3 and B = x4y2z; x, y being prime numbers then HCF (A, B) is (a) xy² (b) x 4 y²z (c) x 4 y 3 (d) x 4 y 3 z

24. The largest number which divides 60 and 75, leaving remainders 8 and 10 respectively, is (a) 260 (b) 75 (c) 65 (d) 13

25. The least number that is divisible by all the numbers from 1 to 5 (both inclusive) is (a) 5 (b) 60 (c) 20 (d) 100

Answer: b Explaination:(b); [Hint. LCM of 2, 3, 4, 5 = 60

26. The least number that is divisible by all the numbers from 1 to 8 (both inclusive) is (a) 840 (b) 2520 (c) 8 (d) 420

27. The decimal expansion of the rational number \(\frac{14587}{250}\) will terminate after: (a) one decimal place (b) two decimal places (c) three decimal places (d) four decimal places

28. The decimal expansion of the rational number \(\frac{97}{2 \times 5^{4}}\) will terminate after: (a) one decimal place (b) two decimal places (c) three decimal places (d) four decimal places

29. The product of two consecutive natural numbers is always: (a) prime number (b) even number (c) odd number (d) even or odd

30. If the HCF of 408 and 1032 is expressible in the form 1032 x 2 + 408 × p, then the value of p is (a) 5 (b) -5 (c) 4 (d) -4

Answer: b Explaination:(b); [Hint. HCF of 408 and 1032 is 24, .-. 1032 x 2 + 408 x (-5)]

31. The number in the form of 4p + 3, where p is a whole number, will always be (a) even (b) odd (c) even or odd (d) multiple of 3

32. When a number is divided by 7, its remainder is always: (a) greater than 7 (b) at least 7 (c) less than 7 (d) at most 7

33. (6 + 5 √3) – (4 – 3 √3) is (a) a rational number (b) an irrational number (c) a natural number (d) an integer

34. If HCF (16, y) = 8 and LCM (16, y) = 48, then the value of y is (a) 24 (b) 16 (c) 8 (d) 48

35. According to the fundamental theorem of arith-metic, if T (a prime number) divides b2, b > 0, then (a) T divides b (b) b divides T (c) T2 divides b2 (d) b2 divides T2

36. The number ‘π’ is (a) natural number (b) rational number (c) irrational number (d) rational or irrational

37. If LCM (77, 99) = 693, then HCF (77, 99) is (a) 11 (b) 7 (c) 9 (d) 22

38. Euclid’s division lemma states that for two positive integers a and b, there exist unique integer q and r such that a = bq + r, where r must satisfy (a) a < r < b (b) 0 < r ≤ b (c) 1 < r < b (d) 0 ≤ r < b

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