ncert class 10 maths chapter 1 case study questions

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CBSE Class 10 Study Material

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.

ncert class 10 maths chapter 1 case study questions

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

Answer: c) 288

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

Answer: b) 4

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

a) 2 2 × 3 2

b) 2 1 × 3 3

c) 2 3 × 3 1

d) 2 0 × 3 0

Answer: a) 2 2 × 3 2

4. 7 × 11 × 13 × 15 + 15 is a

a) Prime number

b) Composite number

c) Neither prime nor composite

d) None of the above

Answer: b) Composite number

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

Answer: b) a 2 b 2

CASE STUDY 2:

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

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

Answer: b) 12

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

Answer: d) 21

3. The LCM of 60, 84 and 108 is

Answer: a) 3780

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

Answer: d) 45360

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

a) 2 3 × 3 2

b) 2 3 × 3 3

c) 2 2 × 3 2

d) 2 2 × 3 3

Answer: d) 2 2 × 3 3

CASE STUDY 3:

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

Observe the following factor tree and answer the following:

1. What will be the value of x?

Answer: b) 13915

2. What will be the value of y?

Answer: c) 11

3. What will be the value of z?

Answer: b) 23

4. According to Fundamental Theorem of Arithmetic 13915 is a

a) Composite number

b) Prime number

d) Even number

Answer: a) Composite number

5. The prime factorisation of 13915 is

a) 5 × 11 3 × 13 2

b) 5 × 11 3 × 23 2

c) 5 × 11 2 × 23

d) 5 × 11 2 × 13 2

Answer: c) 5 × 11 2 × 23

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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Class 10 Maths Case Study Questions of Chapter 1 Real Numbers

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

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Real Numbers Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths Chapter 1 Real Numbers

Case Study/Passage-Based Questions

Case Study 1: Srikanth has made a project on real numbers, where he finely explained the applicability of exponential laws and divisibility conditions on real numbers. He also included some assessment questions at the end of his project as listed below. (i) For what value of n, 4 n ends in 0?

Answer: (d) no value of n

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

Answer: (c) for all n > 1

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

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

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

Answer: (a) always true

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

Answer: (d) 8

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

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

Answer: (b) LCM = pq2 x HCF

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

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

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

Answer: (d) 17

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

Answer: (b) 12599

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

Answer: (a) 3

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

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

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

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

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

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

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 1 Real Numbers with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Real Numbers Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Class 10 Maths Case Study

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

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

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

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

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

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

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

CBSE Class 10th Case study Questions

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

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

CBSE Class 10th Assertion Reasoning Questions

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

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

CBSE Class 10 Maths Case Based MCQ

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

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

Class 10th Mathematics Multiple Choice Questions

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

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

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

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

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

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CBSE Class 10 Maths: Case Study Questions of Chapter 1 Real Numbers PDF Download

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 case study-based questions for Class 10 Maths Chapter 1 Real Numbers

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

Question 1:

Srikanth has made a project on real numbers, where he finely explained the applicability of exponential laws and divisibility conditions on real numbers. He also included some assessment questions 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

Question 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.

Answer: (b) LCM = pq2 x HCF

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

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

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

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

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

Class 10 Maths has the following chapters.

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

Format of Maths Case-Based Questions

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

Maths Case Study Question Paper 2023

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

Case Study Question in Mathematics

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

Case Study Question – 1

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

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

Case Study Question – 2

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

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

Case Study Question – 3

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

Case Study Question – 4

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

Case Study Question – 5

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

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

Case Study Question – 6

$ncert class 10 maths chapter 1 case study questions$

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

$ncert class 10 maths chapter 1 case study questions$

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

How to Solve Case-Based Questions?

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

What are Class 10 competency-based questions?

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

Case Study Questions in Maths Question Paper

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

Case Studies on CBSE’s Official Website

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

10 Maths Case Studies in myCBSEguide App

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

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Class 10 Maths Chapter 1 MCQ

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

Case Study – 1

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

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

View Answer

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

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

36 can be expressed as a product of its primes as

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

7 х 11 х 13 х 15 + 15 is a

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

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

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

Case Study – 2

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

Class 10 Maths Chapter 1 Real Numbers MCQ

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

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

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

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

The LCM of 60, 84, and 108 is

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

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

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

108 can be expressed as a product of its primes as

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

Case Study – 3

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

Case study based MCQ for 10th Maths chapter 1

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

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

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

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

680 can be expressed as a product of its primes as

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

2 х 3 х 5 х 11 х 17 + 11 is a

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

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

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

Case Study – 4

Class 10 Maths Case study MCQ Factor tree

What will be the value of x?

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

What will be the value of y?

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

What will be the value of z?

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

According to Fundamental Theorem of Arithmetic 13915 is a

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

The prime factorisation of 13915 is

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

Case Study – 5

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

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

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

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

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

The HCF of 80, 85, and 90 is

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

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

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

90 can be expressed as a product of its primes as

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

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

Two tankers contain 850 litres & 680 litres of petrol respectively. Maximum capacity of a container which can measure the petrol of either tanker in exact number of times.

Maximum capacity of container means HCF of 850 & 680, applying Euclid’s algorithm we get the HCF of two numbers is 170. Clearly HCF of 850 & 680 is 170, hence capacity of the container must be 170litres.

m^2 -1 is divisible by 8, if m is

Square of an odd integer will be always odd and m^2 – 1 will be even and will be divisible by 8 for an odd integer.

The number in the form of 4p +3, where p is a whole number, will always be

Because 4p will be always an even number. Sum of an even number and an odd number will be always an odd number.

(6 + 5 √3) – (4 – 3 √3 ) is

After simplifying the expression: 6 +5 √3 – 4 + 3 √3 = 2 + 8 √3 is a irrational number

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

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

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

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

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

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

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

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

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

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

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

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Chapter 1 Real Numbers NCERT Solutions for Class 10 Maths

Ncert solutions for class 10 maths chapter 1 real numbers.

Exercise 1.1

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

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let a be any positive integer and b = 3.

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

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

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

Let a be any positive integer and b = 3

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

∴ a = 3q or 3 q + 1 or 3 q + 2

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

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

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

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

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

Page No: 11

Exercise 1.2

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

(i) 140 = 2 × 2 × 5 × 7 = 2 2 × 5 × 7

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

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

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

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

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

(i) 26 = 2 × 13

91 =7 × 13

LCM =2 × 7 × 13 =182

Product of two numbers 26 × 91 = 2366

Product of HCF and LCM 13 × 182 = 2366

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

(ii) 510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of two numbers 510 × 92 = 46920

Product of HCF and LCM 2 × 23460 = 46920

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 = 6

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

Product of two numbers 336 × 54 =18144

Product of HCF and LCM 6 × 3024 = 18144

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

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

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

(i) 12 = 2 × 2 × 3

15 = 3 × 5

21 =3 × 7

LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

LCM = 1 × 17 × 19 × 23 = 11339

(iii) 8 =1 × 2 × 2 × 2

9 =1 × 3 × 3

25 =1 × 5 × 5

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

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

We have the formula that,

Product of LCM and HCF = product of number

LCM × 9 = 306 × 657

Divide both side by 9 we get,

⇒ LCM = 34 × 657 = 22338

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

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

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

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

So, it can not end with 0.

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

7 × 11 × 13 + 13

Taking 13 common, we get

13 (7×11 +1 )

⇒ 13(77 + 1 )

⇒ 13 (78)

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

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 common, we get

5(7 × 6 × 4 × 3 × 2 × 1 +1)

⇒ 5(1008 + 1)

⇒ 5(1009)

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

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

18 = 2 × 3 × 3

12 = 2 × 2 × 3

LCM = 2 × 2 × 3 × 3 = 36

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

Page No: 14

Exercise 1.3

1. Prove that √5 is irrational.

Let us take √5 as rational number

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

We can write √5 = a/b

Multiply by b both side we get,

b√5 = a

To remove root, Squaring on both sides, we get

5 b 2 = a 2 … (i)

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

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

Putting value of a in equation (i) we get

5 b 2 = (5 c ) 2

⇒ 5 b 2 = 25 c 2

Divide by 25 we get,

b 2 /5 = c 2

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

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

2. Prove that 3 + 2√5 is irrational.

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

So we can write this number as,

3 + 2√5 = a / b

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

Subtract 3 both sides we get

2√5 = a/b – 3

⇒ 2√5 = (a-3b)/b

Now, divide by 2, we get

√5 = (a-3b)/2b

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

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

3. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

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

So, we can write this number as

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

Multiply by √2 both sides we get,

1 = ( a √ 2 )/ b

Now, multiply by b

b = a √ 2

Divide by a we get

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

But √2 is a irrational number so it contradicts.

Hence, 1/√2 is a irrational number

So we can write this number as

7√5 = a / b

Divide by 7 we get

√5 = a /(7 b )

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

Hence, 7√5 is a irrational number.

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

6 + √2 = a / b

Subtract 6 both side we get

√2 = a / b – 6

⇒ √2 = ( a - 6 b )/ b

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

Hence, 6 + √ 2 is a irrational number.

Page No: 17

Exercise 1.4

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

(i) 13/3125

(ii) 17/8

(iii) 64/455

(iv) 15/1600

(v) 29/343

(vi) 23/2 3 × 5 2

(vii) 129/2 2 × 5 7 × 7 5

(viii) 6/15

(ix) 35/50

Factorize the denominator we get

3125 =5 × 5 × 5 × 5 × 5 = 5 5

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

Factorize the denominator we get,

8 =2 × 2 × 2 = 2 3

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

(iii) 64 / 455

455 = 5 × 7 × 13

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

(iv) 15 / 1600

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

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

Hence, it is terminating.

(v) 29 / 343

343 = 7 × 7 × 7 = 7 3

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

Hence, it is non-terminating.

(vi) 23 / (2 3 × 5 2 )

Denominator is in form of 2 m × 5 n

Hence, it is terminating.

(vii) 129 / (2 2 × 5 7 × 7 5 )

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

Hence, it is none terminating.

(viii) 6 / 15

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

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

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

10 = 2 × 5

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

(x) 77 / 210

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

30 = 2 × 3 × 5

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

Page No: 18

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

(i) 13 / 3125

(ii) 17 / 8

(vi) 23 / 2 3 5 2

Dividing numerator and denominator by 3.

(ix) 35 / 50

Dividing numerator and denominator by 5.

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p , q you say about the prime factors of q ?

(i) 43.123456789

(ii) 0.120120012000120000...

(iii) 43. 123456789

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

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

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

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

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NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers - Exercise 1.2
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NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers is focused on providing the students with a reliable study tool. The Vedantu solutions for CBSE Class 10 Maths Chapter 1 Exercise 1.2 allow students to understand the basics that are included in the chapter and help understand and identify different questions in Class 10 Maths Chapter 1 Exercise 1.2. Our team of experts have developed the NCERT Solution Class 10 Maths Chapter 1 Exercise 1.2 to understand more around the basics and complex problems of Real Numbers. Moreover, these Class 10 Maths NCERT Solutions Chapter 1 Exercise 1.2 helps students understand the schematics around marking system as per the latest syllabus as well. Class 10 Science NCERT Solutions are also available on Vedantu so download it to prepare well for science exams.

Access NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers

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

(i) $140$

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 140=2\times 2\times 5\times 7$

$\therefore 140={{2}^{2}}\times 5\times 7$

Therefore, the prime factors of $140$ are $2,5,7$.

(ii) $156$

$\Rightarrow 156=2\times 2\times 3\times 13$

$\therefore 156={{2}^{2}}\times 3\times 13$

Therefore, the prime factors of $156$ are $2,3,13$.

(iii) $3825$

$\Rightarrow 3825=3\times 3\times 5\times 5\times 17$

$\therefore 3825={{3}^{2}}\times {{5}^{2}}\times 17$

Therefore, the prime factors of $3825$ are $3,5,17$.

(iv) $5005$

$\Rightarrow 5005=5\times 7\times 11\times 13$

$\therefore 5005=5\times 7\times 11\times 13$

Therefore, the prime factors of $5005$ are $5,7,11,13$.

(v) $7429$

$\Rightarrow 7429=17\times 19\times 23$

$\therefore 7429=17\times 19\times 23$

Therefore, the prime factors of $7429$ are $17,19,23$.

2. Find the LCM and HCF of the following pairs of integers and verify that $LCM\times HCF=\text{Product of two numbers}$.

(i) $26$ and $91$

Ans: First we write the prime factors of $26$ and $91$. We get

$ 26=2\times 13$ and

$91=7\times 13$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $26$ and $91$ is $13$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $26$ and $91$ will be

$ 2\times 7\times 13=182$

Therefore, the LCM of $26$ and $91$ is $182$.

Now, the product of two numbers is

$ 26\times 91=2366$

Product of LCM and HCF is

$ 13\times 182=2366$

We get $LCM\times HCF=\text{Product of two numbers}$ .

The desired result has been verified.

(ii) $510$ and $92$

Ans: First we write the prime factors of $510$ and $92$. We get

$510=2\times 3\times 5\times 17$ and

$92=2\times 2\times 23$

Therefore, the HCF of $510$ and $92$ is $2$.

Then the LCM of $510$ and $92$ will be

$2\times 2\times 3\times 5\times 17\times 23=23460$

Therefore, the LCM of $510$ and $92$ is $23460$.

$510\times 92=46920$

$2\times 23460=46920$

(iii) $336$ and $54$

Ans: First we write the prime factors of $336$ and $54$. We get

\[336=2\times 2\times 2\times 2\times 3\times 7\] and

$54=2\times 3\times 3\times 3$

Therefore, the HCF of $336$ and $54$ is $2\times 3=6$.

Then the LCM of $336$ and $54$ will be

$ 2\times 2\times 2\times 2\times 3\times 3\times 3\times 7=3024$

Therefore, the LCM of $336$ and $54$ is $3024$.

$336\times 54=18144$

$ 6\times 3024=18144$

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

(i) $12,15$ and $21$

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $12,15$ and $21$ are as follows:

\[12=2\times 2\times 3\]

\[15=3\times 5\] and

$21=3\times 7$

Therefore, the HCF of $12,15$ and $21$ is $3$.

Then the LCM of $12,15$ and $21$ will be

$2\times 2\times 3\times 5\times 7=420$

Therefore, the LCM of $12,15$ and $21$ is $420$.

(ii) $17,23$ and $29$

The prime factors of $17,23$ and $29$ are as follows:

\[17=17\times 1\]

\[23=23\times 1\] and

$29=29\times 1$

Therefore, the HCF of $17,23$ and $29$ is $1$.

Then the LCM of $17,23$ and $29$ will be

$\times 23\times 29=11339$

Therefore, the LCM of $17,23$ and $29$ is $11339$.

(iii) $8,9$ and $25$

The prime factors of $8,9$ and $25$ are as follows:

\[8=2\times 2\times 2\]

\[9=3\times 3\] and

$25=5\times 5$

Now, we know that HCF is the highest factor, among the common factors of two numbers. as there is no common factor.

Therefore, the HCF of $8,9$ and $25$ is $1$.

Then the LCM of $8,9$ and $25$ will be

$2\times 2\times 2\times 3\times 3\times 5\times 5=1800$

Therefore, the LCM of $8,9$ and $25$ is $1800$.

4. Given that HCF $\left( 306,657 \right)=9$, find LCM $\left( 306,657 \right)$.

Ans: We have been given the HCF of two numbers $\left( 306,657 \right)=9$.

We have to find the LCM of $\left( 306,657 \right)$.

Now, we know that $LCM\times HCF=\text{Product of two numbers}$

Substitute the values, we get

$LCM\times 9=306\times 657$

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

$\therefore LCM=22338$

Therefore, the LCM of $\left( 306,657 \right)=22338$.

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

Ans: We have to check whether ${{6}^{n}}$ can end with the digit $0$ for any natural number $n$.

By divisibility rule we know that if any number ends with the digit $0$, it is divisible by $2$ and $5$.

Thus, the prime factors of ${{6}^{n}}$ is

$ {{6}^{n}}={{\left( 2\times 3 \right)}^{n}}$

Now, we will observe that for any value of $n$, ${{6}^{n}}$ is not divisible by $5$.

Therefore, ${{6}^{n}}$ cannot end with the digit $0$ for any natural number $n$.

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: The given numbers are $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$.

We can rewrite the given numbers as

$7\times 11\times 13+13=13\times \left( 7\times 11+1 \right)$

$\Rightarrow 7\times 11\times 13+13=13\times \left( 77+1 \right)$

$\Rightarrow 7\times 11\times 13+13=13\times 78$

$\Rightarrow 7\times 11\times 13+13=13\times 13\times 6$

$7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times \left( 7\times 6\times 4\times 3\times 2\times 1+1 \right)$

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times \left( 1008+1 \right)$

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times 1009$

Here, we can observe that the given expression has its factors other than $1$ and the number itself.

A composite number has factors other than $1$ and the number itself.

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

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 less time than Sonia for completing the $1$ round of the circular path. Both are going in the same direction, they will meet again when Ravi will have completed $1$ round of that circular path with respect to Sonia.

The total time taken for completing this $1$ round of circular path will be the LCM of time taken by Sonia and Ravi for ending the $1$ round of circular path respectively, i.e., LCM of $18$ minutes and $12$ minutes.

The prime factors of $12$ and $18$ are as follows:

\[12=2\times 2\times 3\] and

$18=2\times 3\times 3$

Then the LCM of $12$ and $18$ will be

$2\times 2\times 3\times 3=36$

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

Important Topics under NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.2) Exercise 1.2

Real Numbers is the first chapter of the class 10 maths syllabus. It is a significant chapter in maths that is covered in class 10. The chapter on Real Numbers has 4 major parts that need to be read through and understood properly to get a good grasp on the topic.

The following is a list of the 4 important topics that are covered under the chapter Real Numbers. It is recommended that students go through these topics carefully to get a hold of the concepts without any confusion.

Introduction to Real Numbers

Fundamental Theorem of Arithmetic (H.C.F. and L.C.M.)

Euclid’s Division Lemma

Revisiting Irrational Numbers

Importance of NCERT Class 10 Maths Chapter 1 Real Numbers

Real numbers comprise both rational and irrational numbers. Rational numbers include integers, decimals, and fractions, while irrational numbers are numbers like root overs, pi (22/7), and so on. In short, real numbers are all numbers excluding imaginary numbers.

Real numbers are important due to their use in almost every sphere of mathematics and in real life. We encourage students to learn as much as they can from this chapter on real numbers to grow a good number sense and be able to solve all the problems that rely on the use of real numbers in their exams.

NCERT Solutions for Class 10 Maths All Chapters

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

Class 10 Maths Chapter 1 - Exercise 1.2

Importance of Preparing Real Numbers from NCERT Solutions for Class 10 Maths Exercise 1.2

The NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers covers the important aspects of prime numbers to be able to find the HCF & LCM of given pairs and the integers that will allow users to find the product of any two numbers. Further, when students prepare from NCERT Solutions Class 10 Maths Ch 1 Ex 1.2, they uncover the different aspects of numbers and how they can be used to solve the study of numbers. Vedantu’s analytical approach to Class 10 Maths Exercises 1.2 helps students prepare in a full-fledged manner for their upcoming exams and also understand the concept clearly. Along with Exercise 1.2 we also covered all the solutions for Exercise 1.1, Exercise 1.3 and Exercise 1.4

NCERT Solutions for Class 10 Maths Chapter 1 All Exercises

Practical problems from maths class 10 chapter 1 exercise 1.2.

Question 1: Prove that the positive odd integer is of the suggested form, 6q+1 or 6q+3, or 6q+5, making q the integer part.

Solution:

Let’s assume that a is any given positive integer and b=6. Now, we take the approach of Euclid’s algorithm a= 6q +r for some of the given integers and q ≥ 0, or r = 0, 1, 2, 3, 4, 5 because of 0 ≤ r < 6.

Therefore, a = 6q or 6q+1 or 6q+2, 6q+3, 6q+4, 6q+5. Moreover, 6q+1 = 2x 3q +1 = 2k 1 + 1, where k 1 is a positive integer.

6q+3 = (6q+2) + 1= 2 (3q + 1) + 1= 2 k 2 +1, in this situation, k 2 is an integer

6q+5 = (6q+4) + 1= 2 (3q + 2) + 1= 2 k 3 + 1, in this situation, k 3 is an integer

6q+1, 6q+3, 6q+5 are of form 2k+1, where the product k is an integer.

Hence, 6q+1, 6q+3, 6q+5 are not precisely divisible by 2.

Lastly, we can conclude that the given expressions are odd numbers and can be from 6q+1 or 6q+3, or 6q+5.

Question 2:

An army unit comprises 616 members, who are ordered to march behind a band of 32 parades. At the time of their marching, these 2 groups are marching at the same columns. You need to find out the maximum number of columns in which the parade can march.

Given, HCF = (616, 32) is the maximum number of columns in which the army can march, Now, we put into place Euclid's algorithm to find the suggested HCF.

By calculations,

616 = 32 x 9 8

32 = 8 x 4 + 0

Therefore, we can conclude that the HCF (616, 32) is 8.

Conclusion: The army can march into 8 maximum columns

Did You Know?

A real number can be any number that is generally denoted by the symbol R along the number line. It can be zero, negative, rational, irrational or positive integer. This chapter of Class 10 Maths Exercise 1.2 helps students to understand the fundamental and intricate concepts of real numbers.

FAQs on NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers - Exercise 1.2

1. Does Exercise 1.2 Class 10 Maths NCERT Solutions prove to be helpful for Class 10 Exams?

Preparing from Maths Ncert Solutions Class 10 Chapter 1 Exercise 1.2 helps students secure the best and effective results. It makes sure that they understand the concept of real numbers, from its core and retain the usefulness of this subject beyond exams. Our solutions for Exercise 1.2 Class 10 Maths help students uncover the best of the process with enhanced problem-solving skills. Further, Ex 1.2 Class 10 Maths Ncert Solutions come with problems for every level, intermediate and advanced, allowing students to get the best insight on the subject and be able to prepare well for their Class 10 Maths Ch 1 Ex 1.2.

2. What are the different aspects covered in Exercise 1.2 Class 10 Maths Solution?

The solutions of Class 10th Maths Chapter 1 Exercise 1.2 revolves around several questions that help students prepare the best for their exams.

The suggested topics for practice include:

Proving number as a product of its prime factors (Important Question For Class 10 Chapter 1 Maths Exercise 1.2)

Evaluating the HCF and LCM of two given numbers (a repeated question of Exercise 1.2 Class 10th Maths, over the years)

Finding LCM of two numbers when HCF is given. (Important Question From Class 10 Maths Chapter 1 Ex 1.2)

Evaluate HCF and LCM of integers using a method of prime factorization (an important question of Exercise 1.2 Class 10th Maths to be practised)

3. How many questions are there in Class 10 Maths Exercise 1.2?

The First Chapter given in the syllabus of Class 10 Mathematics is Real Numbers. Exercise 1.2 of this chapter includes seven questions for students to practice. These seven questions are based on the Fundamental Theorem of Arithmetic. Through this exercise, students can also understand the concept of the Highest Common Factor (HCF) and Least Common Multiple (LCM).

4. Which examples are important for solving Exercise 1.2 of Class 10 Mathematics?

All the examples provided in the NCERT Book for Class 10 Mathematics are equally important. Therefore, students must practice solving the examples as well. Four examples are based on exercise 1.2. These include examples five, six, seven, and eight. All of these examples hold equal importance. All of them should be studied to score well in the exam.

5. Is Class 10 Maths chapter 1 difficult?

No, it is not. Regular practice of this chapter helps students achieve a proper understanding of all concepts. Referring to NCERT solutions for Class 10 Maths chapter 1 on Vedantu can further assist them to get a better hold of this chapter. You can also refer to revision notes of this chapter to revise before the exam. These notes are prepared by experts with decades of experience and hence are 100% reliable.

6. Where can I find NCERT Solutions for Class 10 Maths Exercise 1.2?

Students can find the NCERT Solutions for Class 10 Maths Exercise 1.2 on Vedantu . NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2 helps the students to solve the exercise easily. These solutions have been prepared in a step-by-step manner by the subject-matter experts. With the help of these NCERT Solutions, students will be able to enhance their understanding, and they can solve any kind of difficulties that they might face while practising Class 10 Maths.

7. Do I need to practice all the questions provided in Class 10 Maths Chapter 1 NCERT Solutions?

Question papers in your Class 10 exams are designed on the basis of the NCERT books. Questions can be asked from any exercise of the chapter, and there is no particular way to find which of them can be asked. Ignoring any questions that are provided in the NCERT can lead to a loss of marks in the exam. Hence, it is best to practice all the questions provided in the NCERT Solutions for Class 10 Maths Chapter 1.

NCERT Solutions for Class 10

Cbse study materials for class 10, cbse study materials.

Maths Notes Class 10
NCERT Solutions Class 10
RD Sharma Solutions Class 10
Maths Formulas Class 10
Class 10 Syllabus
Class 10 Revision Notes
Physics Notes Class 10
Chemistry Notes Class 10
Biology Notes Class 10
History Notes class 10
political science class 10
Geography Notes class 10
Social science Notes class 10
Class 10 NCERT Solutions- Chapter 1 Real Numbers - Exercise 1.3
Class 10 NCERT Solutions- Chapter 1 Real Numbers - Exercise 1.2
Class 10 NCERT Solutions- Chapter 1 Real Numbers - Exercise 1.4
Class 8 NCERT Solutions - Chapter 1 Rational Numbers - Exercise 1.2
Class 8 NCERT Solutions- Chapter 1 Rational Numbers - Exercise 1.1
Class 9 NCERT Solutions- Chapter 1 Number System - Exercise 1.1
Class 11 NCERT Solutions - Chapter 1 Sets - Exercise 1.3
Class 11 NCERT Solutions - Chapter 1 Sets - Exercise 1.6
Class 11 NCERT Solutions - Chapter 1 Sets - Exercise 1.2
Class 11 NCERT Solutions - Chapter 1 Sets - Exercise 1.4
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Class 9 NCERT Solutions - Chapter 1 Number System - Exercise 1.2
Class 9 NCERT Solutions - Chapter 1 Number System - Exercise 1.6
Class 9 NCERT Solutions - Chapter 1 Number System - Exercise 1.4
Class 9 NCERT Solutions- Chapter 1 Number System - Exercise 1.5
Class 8 NCERT Solutions - Chapter 16 Playing with Numbers - Exercise 16.1
Class 9 NCERT Solutions- Chapter 1 Number System - Exercise 1.3
Class 10 RD Sharma Solutions- Chapter 1 Real Numbers - Exercise 1.5
Class 8 NCERT Solutions - Chapter 16 Playing with Numbers - Exercise 16.2

NCERT Solutions Class 10 – Chapter 1 Real Numbers – Exercise 1.1

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

(i) 140 Taking the LCM of 140, 140 = 2 × 2 × 5 × 7 × 1 = 2 2 × 5 × 7 Therefore, the product of the prime factors is 2 2 × 5 × 7 (ii) 156 Taking the LCM of 156, 156 = 2 × 2 × 3 × 13 × 1 = 2 2 × 3 × 13 Therefore, the product of the prime factors is 2 2 × 3 × 13 (iii) 3825 Taking the LCM of 3825, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3 2 × 5 2 × 17 Therefore, the product of the prime factors is 3 2 × 5 2 × 17 (iv) 5005 Taking the LCM of 5005, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13 Therefore, the product of the prime factors is 5 × 7 × 11 × 13 (v) 7429 Taking the LCM of 7429, 7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23 Therefore, the product of the prime factors is 17 × 19 × 23

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

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

(i) 26 and 91 Taking the LCM of 26, 26 = 2 × 13 × 1 Taking the LCM of 91, 91 = 7 × 13 × 1 Therefore, LCM of 26 and 91 together = 2 × 7 × 13 × 1 = 182 HCF of 26 and 91 = 13 Now, the product of 26 and 91 = 26 × 91 = 2366 And the product of LCM and HCF = 182 × 13 = 2366 Therefore, LCM × HCF = product of the 26 and 91. (ii) 510 and 92 Taking the LCM of 510, 510 = 2 × 3 × 17 × 5 × 1 Taking the LCM of 92, 92 = 2 × 2 × 23 × 1 Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460 HCF of 510 and 92 = 2 Now, the product of 510 and 92 = 510 × 92 = 46920 And the product of LCM and HCF = 23460 × 2 = 46920 Therefore, LCM × HCF = product of the 510 and 92. (iii) 336 and 54 Taking the LCM of 336, 336 = 2 × 2 × 2 × 2 × 7 × 3 × 1 Taking the LCM of 54, 54 = 2 × 3 × 3 × 3 × 1 Therefore, LCM of 336 and 54 = 3024 HCF of 336 and 54 = 2×3 = 6 Now, the product of 336 and 54 = 336 × 54 = 18144 And the product of LCM and HCF = 3024 × 6 = 18144 Therefore, LCM × HCF = product of the 336 and 54.

Question 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

(i) 12, 15 and 21 Taking the LCM of 12, 12=2×2×3 Taking the LCM of 15, 15=5×3 Taking the LCM of 21, 21=7×3 Therefore, HCF of 12, 15 and 21 = 3 LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420 (ii) 17, 23 and 29 Taking the LCM of 17, 17=17×1 Taking the LCM of 23, 23=23×1 Taking the LCM of 29, 29=29×1 Therefore, HCF of 17, 23 and 29 = 1 LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339 (iii) 8, 9 and 25 Taking the LCM of 8, 8=2×2×2×1 Taking the LCM of 9, 9=3×3×1 Taking the LCM of 25, 25=5×5×1 Therefore, HCF of 8, 9 and 25 =1 LCM of 8, 9 and 25 = 2×2×2×3×3×5×5 = 1800

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

Solution:

Given: HCF (306, 657) = 9 We know that, HCF×LCM=Product of the two given numbers Therefore, by substituting the value we get, 9 × LCM = 306 × 657 LCM = (306×657)/9 = 22338 LCM (306,657) = 22338 Therefore, the LCM (306,657) = 22338

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

If the given number 6 n ends with the digit 0, then it should be divisible by 5. We know that if any number with the unit place as 0 or 5 is divisible by 5. Therefore, By prime factorization of 6 n = (2×3) n Since the prime factorization of 6 n doesn’t contain prime number 5. Therefore, 6 n cannot end with the digit 0 for any natural number.

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

We know by the definition of a composite number, that, the composite number has factors other than 1 and itself. Therefore, in the given expression 7 × 11 × 13 + 13 By taking 13 as a common factor, we get, =13(7×11×1+1) = 13(77+1) = 13×78 [taking prime factors of 78] = 13×3×2×13 Therefore, 7 × 11 × 13 + 13 is a composite number. Now, for the 2 nd number 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 By taking 5 as a common factor, we get, =5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009 Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

Question 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?

Given: Both Sonia and Ravi move in the same direction and at the same time. Now, the time when they will be meeting again at the starting point can be calculated by finding the LCM of 18 and 12 Therefore, LCM (18, 12) = 2×3×3×2×1 = 36 Finally, they both will meet again at the starting point after 36 minutes.

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NCERT Solutions
NCERT Solutions for Class 10
NCERT Solutions for Class 10 Maths
Chapter 1: Real Numbers
Exercise 1.2

NCERT Solutions for Class 10 Maths Exercise 1.2 Chapter 1 Real Numbers

NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 are provided here. Our subject experts prepare these solutions with the aim of helping the students appearing for CBSE exams. They prepare NCERT Maths Solution for Class 10 – chapter-wise so that they help students solve the problems quickly.

The points kept in mind while preparing these solutions are ease of understanding and detailed explanation of the solutions provided. Exercise 1.2 is the second exercise of Chapter 1 -Real Numbers, and it deals with the fundamental theorem of Arithmetic. There are seven problems given in Exercise 1.2 Class 10 in NCERT, and solutions to all those can be found here. It is made sure that the solutions provided follow NCERT guidelines. These NCERT solutions cover all the topics of the chapter Real Numbers and help in scoring well in exams.

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 of Maths NCERT Class 10 Chapter 1 – Real Number Exercise 1.2

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 the 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 the 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 the 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,

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 the unit place as 0 or 5 is divisible by 5.

Prime factorisation of 6 n = (2×3) n

Therefore, the prime factorisation of 6 n doesn’t contain the 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 a 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 a 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: SincebBoth 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 1.2 of NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers is the second exercise of Chapter 1 of Class 10 Maths. Real Numbers is the first chapter students study in the Class 10 NCERT textbook. The fundamentals of Arithmetic is one of the exercise topics of this chapter. The factors discussed in this chapter are the Factorisation of composite numbers. It states every Composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

The Fundamental Theorem of Arithmetic – It includes 7 questions based on this theorem.

Key Features of NCERT Solutions for Class 10 Maths Chapter 1 – Real Number Exercise 1.2 Page number 14

These NCERT Class 10 Solutions help in solving and revising all questions of exercise 1.2 real numbers.
If students go through the step-wise solutions given here, they will be able to get more marks.
The solutions will help learners score well in Maths exams if they practise thoroughly.
They follow NCERT guidelines which help in preparing the students for the exam accordingly.
They contain all the important questions from the examination point of view.

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CBSE Class 10 Maths Case Study Questions for Class 10 Maths Chapter 1
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. By Gurmeet Kaur
Class 10 Maths Case Study Questions of Chapter 1 Real Numbers
Show Answer. (v) If A, B and C are three rational numbers such that 85C - 340A = 109, 425A + 85B = 146, then the sum of A, B and C is divisible by. (a) 3. (b) 6. (c) 7. (d) 9. Show Answer. Case Study 3: Real numbers are an essential concept in mathematics that encompasses both rational and irrational numbers.
PDF Real Numbers- Case Study Case Study 1
Let the no of questions whose answer is known to the student x and questions attempted by cheating be y x + y =120 x-1/4y =90 solving these two x=96 and y= 24 1. He answered 96 questions correctly. 2. He attempted 24 questions by guessing. 3. Marks = 80- ¼ 0f 40 =70 4. x - ¼ 0f (120-x) =95 5x=500, x = 100 CASE STUDY-2:
CBSE Class 10 Maths Case Study Questions PDF
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 ...
CBSE Class 10 Maths Case Study : Case Study With Solutions
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.
NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers
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.
CBSE Class 10 Maths: Case Study Questions of Chapter 1 ...
Here, we have provided case-based/passage-based questions for Class 10 Maths Chapter 1 Real Numbers. Case Study/Passage-Based Questions. Question 1: Srikanth has made a project on real numbers, where he finely explained the applicability of exponential laws and divisibility conditions on real numbers. He also included some assessment questions ...
Case Study Class 10 Maths Questions
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.
Case Study Questions for Class 10 Maths Chapter 1 Real Numbers
Case Study Questions for Class 10 Maths Chapter 1 Real Numbers. Question 1: HCF and LCM are widely used in number system especially in real numbers in finding relationship between different numbers and their general forms. Also, product of two positive integers is equal to the product of their HCF and LCM. Based on the above information answer ...
Class 10 Maths NCERT Solutions Chapter 1 Real Numbers
These NCERT Solutions for Class 10 of Maths subject includes detailed answers of all the questions in Chapter 1 - Class 10 Real Numbers provided in NCERT Book which is prescribed for class 10 in schools. Class 10 Real Numbers Ex 1.4 - 3 Questions Based in which you have to expand fractions into decimals and write decimals in their fraction ...
Important Questions for Class 10 Maths Chapter 1 Real Numbers
Real Numbers Class 10 Maths Important Questions Short Answer-II (3 Marks) 23. Prove that 3 + 2√3 is an irrational number. Solution. Let us assume to the contrary, that 3 + 2√3 is rational. So that we can find integers a and b (b ≠ 0). Such that 3 + 2√3 = ab, where a and b are coprime. Rearranging the equations, we get.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Class 10 Maths Chapter 1 Exercise 1.1 Question 4 Given that HCF (306, 657) = 9, find LCM (306, 657). The solution of this question is very simple and usually asked in MCQ type questions. For more revision, please write to us. Class 10 Maths Chapter 1 Exercise 1.1 Question 5 Check whether 6ⁿ can end with the digit 0 for any natural number n.
NCERT Exemplar Class 10 Maths Solutions Chapter 1 (Free PDF)
The NCERT Exemplar Class 10 Maths Chapter 1 Real Numbers is provided here for students to practise and prepare for the CBSE first- and second-term exams. The downloadable PDF is provided by our subject experts as per the CBSE guidelines (2023-2024). This material will help students to revise the syllabus of the Real Numbers chapter and score ...
Class 10 Maths Chapter 1 MCQ
Class 10 Maths Chapter 1 Real Numbers MCQ (Multiple Choice Objective Questions) with answers and complete explanation case study type questions for the first term examination 2024-25. The answers of 10th Maths Chapter 1 MCQ are given with explanation, so that students can understand easily.
Chapter 1 Real Numbers NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Here, we have providing Class 10 Maths NCERT Solutions for Chapter 1 Real Numbers which will be beneficial for students.These solutions are updated according to 2020-21 syllabus. As NCERT Solutions are prepared by Studyrankers experts, we have taken of every steps so you can understand the concepts without any difficulty.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Class 10 Maths NCERT Solutions Chapter 1: Real Numbers is one of the first and foremost important chapters in CBSE Class 10 NCERT Syllabus. Most of the students feel Maths difficult and to help them out we have listed detailed NCERT Solutions for Class 10 Maths Chapter 1 in this article. All the solutions prevailing here are prepared by experts and can be of great help during your board exam ...
NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers
NCERT Solutions. Ex 1.1 Class 10 Maths Question 1. Use Euclid's Division Algorithm to find the HCF of: (i) 135 and 225. (ii) 196 and 38220. (iii) 867 and 255. Solution: Ex 1.1 Class 10 Maths Question 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1
Access answers of Maths NCERT Class 10 Chapter 1 - Real Number Exercise 1.1 page number 7. 1. Use Euclid's division algorithm to find the HCF of: i. 135 and 225. ii. 196 and 38220. iii. 867 and 255. Solutions: i. 135 and 225. As you can see from the question, 225 is greater than 135.
NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers
Exercise 1.1 contains 5 questions. Class 10 Maths Chapter 1 is important for students to form a strong base of the subject as it will help them in their higher classes as well. It is important that you thoroughly practice all the exercises from the NCERT book. You can also refer to the NCERT Class 10 Maths solutions PDF for reference. There are ...
Important Questions Class 10 Maths Chapter 1 Real Numbers
Q.7: Give an example to show that the product of a rational number and an irrational number may be a rational number. Q.8: Prove that √3 - √2 and √3 + √5 are irrational. Q.9: Express 7/64, 12/125 and 451/13 in decimal form. Q.10: Find two irrational numbers lying between √2 and √3.
NCERT Solutions for Class 10 Maths Chapter 1: Real Numbers
NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers is focused on providing the students with a reliable study tool. The Vedantu solutions for CBSE Class 10 Maths Chapter 1 Exercise 1.2 allow students to understand the basics that are included in the chapter and help understand and identify different questions in Class 10 Maths Chapter 1 Exercise 1.2.
NCERT Solutions Class 10
Case Studies in Designing Systems; Complete System Design Tutorial; ... NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers; Class 9 NCERT Solutions - Chapter 1 Number System - Exercise 1.2; ... Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number. Question 7: There is a circular path around a sports field. ...
NCERT Solutions for Class 10 Maths Updated for 2023-24 Exams
NCERT Solutions for Class 10 Maths Updated for 2023-24 Session - Free PDF Download. NCERT Solutions for Class 10 Maths for all the exercises from Chapters 1 to 15 are provided here.These NCERT Solutions are curated by our expert faculty to help students in their exam preparations. Students looking for the NCERT Solutions of Class 10 Maths can download all chapter-wise PDFs to find a better ...
NCERT Solutions for Class 10 Maths Exercise 1.2 Chapter 1 Real Numbers
Exercise 1.2 is the second exercise of Chapter 1 -Real Numbers, and it deals with the fundamental theorem of Arithmetic. There are seven problems given in Exercise 1.2 Class 10 in NCERT, and solutions to all those can be found here. It is made sure that the solutions provided follow NCERT guidelines. These NCERT solutions cover all the topics ...

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

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Class 10 Maths Chapter 1 MCQ

Case Study – 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?

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

36 can be expressed as a product of its primes as

7 х 11 х 13 х 15 + 15 is a

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

Case Study – 2

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

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

The LCM of 60, 84, and 108 is

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

108 can be expressed as a product of its primes as

Case Study – 3

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

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

680 can be expressed as a product of its primes as

2 х 3 х 5 х 11 х 17 + 11 is a

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

Case Study – 4

What will be the value of x?

What will be the value of y?

What will be the value of z?

According to Fundamental Theorem of Arithmetic 13915 is a

The prime factorisation of 13915 is

Case Study – 5

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

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

The HCF of 80, 85, and 90 is

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

90 can be expressed as a product of its primes as

Two tankers contain 850 litres & 680 litres of petrol respectively. Maximum capacity of a container which can measure the petrol of either tanker in exact number of times.

m^2 -1 is divisible by 8, if m is

The number in the form of 4p +3, where p is a whole number, will always be

(6 + 5 √3) – (4 – 3 √3 ) is

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