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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:

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## CBSE 10th Standard Maths Subject Real Number Case Study Questions With Solution 2021

By QB365 on 21 May, 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

## QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cbse 10th standard maths subject real number case study questions with solution 2021 answer keys.

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

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

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

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

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

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## 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 the following questions.

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

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

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

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

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

Question 2:

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

(i) What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B? (a) 144 (b) 128 (c) 288 (d) 272

(ii) If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is (a) 2 (b) 4 (c) 6 (d) 8

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

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

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

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

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

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

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

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

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## Case Study on Real Numbers Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Real Numbers Class 10 Maths can use this page to download the PDF file.

The case study questions on Real Numbers are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Real Numbers case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

## Case Study on Real Numbers Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Real Numbers, therefore, they prepared a set of solutions along with the case study questions on Real Numbers.

The case study on Real Numbers Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Real Numbers case study questions are very easy to grasp from the PDF - download links are given on this page.

## Why Solve Real Numbers Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Real Numbers case study questions on Class 10 Maths - all those major reasons are discussed below:

- To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Real Numbers Case study questions as it will help better prepare for the Class 10 board exam preparation.
- Develop Problem-Solving Skills: Class 10 Maths Real Numbers case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
- Understand Real-Life Applications: Several Real Numbers Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Real Numbers as well as real-life implications of those learnings too.

## How to Answer Case Study Questions on Real Numbers?

Students can choose their own way to answer Case Study on Real Numbers Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Real Numbers Case Study questions.

- Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
- Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Real Numbers questions quickly.
- Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Real Numbers Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

## What to Know to Solve Case Study Questions on Class 10 Real Numbers?

A few essential things to know to solve Case Study Questions on Class 10 Real Numbers are -

- Basic Formulas of Real Numbers: One of the most important things to know to solve Case Study Questions on Class 10 Real Numbers is to learn about the basic formulas or revise them before solving the case-based questions on Real Numbers.
- To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Real Numbers case study questions.
- Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

## Where to Find Case Study on Real Numbers Class 10 Maths?

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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

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

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.

(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

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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## Unit 1: Real numbers

Real numbers 1.1.

- HCF visualized (Opens a modal)
- LCM visualized (Opens a modal)
- HCF and LCM product (Opens a modal)
- Least common multiple of three numbers (Opens a modal)
- Proof: √2 is irrational (Opens a modal)
- Proof: square roots of prime numbers are irrational (Opens a modal)
- Real numbers 1.1 Get 7 of 10 questions to level up!
- Finding HCF & LCM of large numbers Get 3 of 4 questions to level up!

## Euclid's Division Algorithm (Bonus)

- Intro to Euclid's division algorithm (Opens a modal)
- Euclid's division algorithm visualised (Opens a modal)
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- Finding HCF through Euclid's division algorithm Get 3 of 4 questions to level up!

## 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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## 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 2023-24. The answers of 10th Maths Chapter 1 MCQ are given with explanation, so that students can understand easily. This page of Class 10 Maths MCQ contains the questions released by CBSE as well as extra questions for practice.

## Case Study – 1

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

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

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

- View Answer

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

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

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

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

## 7 х 11 х 13 х 15 + 15 is a

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

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

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

## Case Study – 2

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

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

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

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

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

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

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

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

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

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

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

## Case Study – 3

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

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

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

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

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

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

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

## 2 х 3 х 5 х 11 х 17 + 11 is a

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

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

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

## Case Study – 4

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

## What will be the value of x?

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

## What will be the value of y?

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

## What will be the value of z?

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

## According to Fundamental Theorem of Arithmetic 13915 is a

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

## The prime factorisation of 13915 is

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

## Case Study – 5

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

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

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

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

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

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

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

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

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

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

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

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

## 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.

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

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

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

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

## 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

## The number π is

We know the value of π = 3.14285714….. Which is non-repeating non-terminating. So this is an irrational.

## 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 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

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

## NCERT Class 10 Mathematics Chapter 1: Complete Resource for Real Numbers - Free PDF

As you go into higher classes, juggling between so many subjects and assignments is not easy for a class 10 student. That is why we at Vedantu have come up with detailed NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers. The solutions follow the updated CBSE curriculum. The subject matter experts at Vedantu have done an extensive research to design NCERT Solutions for Class 10 Maths Chapter 1 so that it is easily understandable by you. By going through these Chapter 1 Maths Class 10 NCERT Solutions, you can clear your concepts on real numbers at the root level so that you can solve complex problems on your own.

If you are looking for answers to all the questions of Class 10 Maths Chapter 1, then download the NCERT Solutions for Class 10 Maths Chapter 1 PDF from the official website of Vedantu. You can save these solutions on your device to access them offline, without an internet connection. You can even print out Maths Chapter 1 Class 10 NCERT Solution to have another mode of doing a quick revision of essential formulas and concepts. If you are looking for NCERT Solutions for Class 10 Science you can find that on Vedantu.

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

## Related Chapters

## Practice and Master the Concepts of Real Numbers with NCERT Solutions Exercises for Class 10 Maths Chapter 1

NCERT Solutions for Class 10 Maths Chapter 1, "Real Numbers," is based on the concept of real numbers and their properties. The chapter consists of the following exercises:

Exercise 1.1: This exercise discusses the concept of Euclid's division lemma and the algorithm to find the HCF of two numbers. (Not available in the current syllabus)

Exercise 1.1: This exercise covers the concept of irrational numbers and their decimal expansions.

Exercise 1.2: This exercise explains the concept of real numbers and their properties such as closure, commutativity, associativity, distributivity, etc.

Exercise 1.3: This exercise deals with the square roots of positive integers and their properties.

## Mastering Class 10 Maths Chapter 1: Real Numbers - Notes, Practice and Exercise Problems, and Tips for Success

Exercise 1.1 (Not available in the current syllabus)

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

(i) $135$ and $225$

Ans: We have to find the HCF of $135$ and $225$ by using Euclid’s division algorithm.

According to Euclid’s division algorithm, the HCF of any two positive integers $a$ and $b$, where $a > b$ is found as :

First find the values of $q$ and $r$, where $a=bq+r$, $0\le r < b$.

If $r=0$, the HCF is $b$. If $r\ne 0$, apply Euclid’s lemma to $b$ and $r$.

Continue steps till the remainder is zero. When we get the remainder zero, divisor will be the HCF.

Let $a=225$ and $b=135$.

Since, $a > b$

Using division algorithm, we get

$\Rightarrow 225=135\times 1+90$

$\Rightarrow b=135$

$\Rightarrow q=1$

$\Rightarrow r=90$

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ (new divisor) and $r$ (new remainder). We get

$\Rightarrow 135=90\times 1+45$

Here,

$\Rightarrow b=90$

$\Rightarrow r=45$

Since $r\ne 0$, we apply the Euclid’s lemma to $b$ and $r$. We get

$\Rightarrow 90=2\times 45+0$

Now, we get $r=0$, thus we can stop at this stage.

When we get the remainder zero, divisor will be the HCF.

Therefore, the HCF of $135$ and $225$ is $45$.

(ii) $196$ and $38220$

Ans: We have to find the HCF of $196$ and $38220$ by using Euclid’s division algorithm.

Let $a=38220$ and $b=196$.

$\Rightarrow 38220=196\times 195+0$

$\Rightarrow b=196$

$\Rightarrow q=195$

$\Rightarrow r=0$

Since, we get $r=0$, thus we can stop at this stage.

Therefore, the HCF of $196$ and $38220$ is $196$.

(iii) $867$ and $255$

Ans: We have to find the HCF of $867$ and $255$ by using Euclid’s division algorithm.

Let $a=867$ and $b=255$.

\[\Rightarrow 867=255\times 3+102\]

$\Rightarrow b=255$

$\Rightarrow q=3$

$\Rightarrow r=102$

$\Rightarrow 255=102\times 2+51$

$\Rightarrow b=102$

$\Rightarrow q=2$

$\Rightarrow r=51$

$\Rightarrow 102=51\times 2+0$

Therefore, 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.

Ans: Let $a$ be any positive integer and $b=6$.

Then, by Euclid’s division algorithm we get $a=bq+r$, where, $0\le r < b$.

Here, $0\le r < 6$.

Substitute the values, we get

$\Rightarrow a=6q+r$

If $r=0$, we get

$\Rightarrow a=6q+0$

$\Rightarrow a=6q$

If $r=1$, we get

$\Rightarrow a=6q+1$

If $r=2$, we get

$\Rightarrow a=6q+2$ and so on

Therefore, $a=6q$ or $6q+1$ or $6q+2$ or $6q+3$ or $6q+4$ or $6q+5$.

We can write the obtained expressions as

\[6q+1=2\times 3q+1\]

$\Rightarrow 6q+1=2{{k}_{1}}+1$

Where, ${{k}_{1}}$ is an integer.

\[6q+3=6q+2+1\]

$\Rightarrow 6q+3=2\left( 3q+1 \right)+1$

$\Rightarrow 6q+3=2{{k}_{2}}+1$

Where, ${{k}_{2}}$ is an integer.

\[6q+5=6q+4+1\]

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

$\Rightarrow 6q+5=2{{k}_{3}}+1$

Where, ${{k}_{3}}$ is an integer.

Thus, $6q+1,6q+3,6q+5$ are of the form $2k+1$ and are not exactly divisible by $2$.

Also, all these expressions are of odd numbers.

Therefore, any positive odd integer can be expressed in the form, $6q+1$ or $6q+3$, or $6q+5$, 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?

Ans: Given that 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.

We have to find the maximum number of columns in which they can march.

We need to find the HCF of $616$ and $32$ to find the maximum number of columns.

We will use the Euclid’s division algorithm to find the HCF.

Let $a=616$ and $b=32$.

\[\Rightarrow 616=32\times 19+8\]

$\Rightarrow b=32$

$\Rightarrow q=19$

$\Rightarrow r=8$

$\Rightarrow 32=8\times 4+0$

Therefore, the HCF of $616$ and $32$ is $8$.

Therefore, in $8$ columns members can march.

4. 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 than 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 and $b=3$.

Here, $0\le r < 3$.

$\Rightarrow a=3q+r$

$\Rightarrow a=3q+0$

$\Rightarrow a=3q$

$\Rightarrow a=3q+1$

$\Rightarrow a=3q+2$

Therefore, $a=3q$ or $3q+1$ or $3q+2$.

Now, squaring each of these, we get

$\Rightarrow {{a}^{2}}={{\left( 3q \right)}^{2}}$ or ${{\left( 3q+1 \right)}^{2}}$ or ${{\left( 3q+2 \right)}^{2}}$

Now, applying the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get

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

Thus, we get

$\Rightarrow {{a}^{2}}=3\times 3{{q}^{2}}$

$\Rightarrow {{a}^{2}}=3m$, where, $m=3{{q}^{2}}$

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

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+2q \right)+1$

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+2q$

${{a}^{2}}=3\times 3{{q}^{2}}+6\times 2q+4$

$\Rightarrow {{a}^{2}}=3\times 3{{q}^{2}}+6\times 2q+3+1$

$\Rightarrow {{a}^{2}}=3\left( 3{{q}^{2}}+4q+1 \right)+1$

$\Rightarrow {{a}^{2}}=3m+1$, where, $m=3{{q}^{2}}+4q+1$

Therefore, we can say that the square of any positive integer is either of 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 $9m+8$.

Now, consider $a=3q$, we get

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

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

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

$\Rightarrow {{a}^{3}}=9m$, where, $m=3{{q}^{3}}$

Now, consider $a=3q+1$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+1 \right)}^{3}}$

Now, applying the identity ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$, we get

$\Rightarrow {{a}^{3}}=27{{q}^{3}}+27{{q}^{2}}+9q+1$

$\Rightarrow {{a}^{3}}=9\left( 3{{q}^{3}}+3{{q}^{2}}+q \right)+1$

$\Rightarrow {{a}^{3}}=9m+1$, where, \[m=3{{q}^{3}}+3{{q}^{2}}+q\]

Now, consider $a=3q+2$, we get

$\Rightarrow {{a}^{3}}={{\left( 3q+2 \right)}^{3}}$

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

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

$\Rightarrow {{a}^{3}}=9m+8$, where, \[m=3{{q}^{3}}+6{{q}^{2}}+4q\]

Therefore, we can say that cube of any positive integer is of the form $9m$, $9m+1$ or $9m+8$.

Exercise 1.1 (Page No. 11)

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

$\Rightarrow 26=2\times 13$ and

$\Rightarrow 91=7\times 13$

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 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

$\Rightarrow 2\times 7\times 13=182$

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

Now, the product of two numbers is

$\Rightarrow 26\times 91=2366$

Product of LCM and HCF is

$\Rightarrow 13\times 182=2366$

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

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

$\Rightarrow 92=2\times 2\times 23$

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

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

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

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

$\Rightarrow 510\times 92=46920$

$\Rightarrow 2\times 23460=46920$

(iii) $336$ and $54$

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

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

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

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

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

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

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

$\Rightarrow 336\times 54=18144$

$\Rightarrow 6\times 3024=18144$

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:

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

\[\Rightarrow 15=3\times 5\] and

$\Rightarrow 21=3\times 7$

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

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 $12,15$ and $21$ will be

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

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:

\[\Rightarrow 17=17\times 1\]

$\Rightarrow 23=23\times 1$ and

$\Rightarrow 29=29\times 1$

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

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

$\Rightarrow 17\times 23\times 29=11339$

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:

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

\[\Rightarrow 9=3\times 3\] and

$\Rightarrow 25=5\times 5$

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

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

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

$LCM\times 9=306\times 657$

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

$\therefore LCM=22338$

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

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

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

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

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

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

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

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

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

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

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

A composite number have 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 lesser time than Sonia for completing $1$ 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 $1$ round of circular path respectively i.e., LCM of $18$ minutes and $12$ minutes.

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

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

$\Rightarrow 18=2\times 3\times 3$

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

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

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

Exercise 1.2 (Page No. 14)

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

Ans: We have to prove that $\sqrt{5}$ is irrational.

We will use contradiction method to prove it.

Let $\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\sqrt{5}=\dfrac{a}{b}$

Now, squaring both sides, we get

${{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}}$

\[\Rightarrow 5=\dfrac{{{a}^{2}}}{{{b}^{2}}}\]

$\Rightarrow {{a}^{2}}=5{{b}^{2}}$ …….(1)

If ${{a}^{2}}$ is divisible by $5$ than $a$ is also divisible by $5$.

Let $a=5k$, where, $k$ is any integer.

Again squaring both sides, we get

\[\Rightarrow {{a}^{2}}={{\left( 5k \right)}^{2}}\]

Substitute the value in eq. (1), we get

\[\Rightarrow {{\left( 5k \right)}^{2}}=5{{b}^{2}}\]

$\Rightarrow {{b}^{2}}=5{{k}^{2}}$ …..(2)

If ${{b}^{2}}$ is divisible by $5$ than $b$ is also divisible by $5$.

From, eq. (1) and (2), we can conclude that $a$ and $b$ have $5$ as a common factor.

This contradicts our assumption.

Therefore, we can say that $\sqrt{5}$ is irrational.

Hence proved.

2. Prove that $3+2\sqrt{5}$ is irrational.

Ans: We have to prove that $3+2\sqrt{5}$ is irrational.

Let $3+2\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

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

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

$\Rightarrow \sqrt{5}=\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ ……..(1)

From eq. (1) we can say that $\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ is rational so $\sqrt{5}$ must be rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. Hence the assumption is false.

Therefore, we can say that $3+2\sqrt{5}$ is irrational.

3. Prove that following are irrationals:

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

Ans: We have to prove that $\dfrac{1}{\sqrt{2}}$ is irrational.

Let $\dfrac{1}{\sqrt{2}}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$

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

From eq. (1) we can say that $\dfrac{b}{a}$ is rational so $\sqrt{2}$ must be rational.

But this contradicts the fact that $\sqrt{2}$ is irrational. Hence the assumption is false.

Therefore, we can say that $\dfrac{1}{\sqrt{2}}$ is irrational.

(ii) $7\sqrt{5}$

Ans: We have to prove that $7\sqrt{5}$ is irrational.

Let $7\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $7\sqrt{5}=\dfrac{a}{b}$

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

From eq. (1) we can say that $\dfrac{a}{7b}$ is rational so $\sqrt{5}$ must be rational.

Therefore, we can say that $7\sqrt{5}$ is irrational.

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

Ans: We have to prove that $6+\sqrt{2}$ is irrational.

Let $6+\sqrt{2}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are co-prime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $6+\sqrt{2}=\dfrac{a}{b}$

$\Rightarrow \sqrt{2}=\dfrac{a}{b}-6$ ………..(1)

From eq. (1) we can say that $\dfrac{a}{b}-6$ is rational so $\sqrt{2}$ must be rational.

Therefore, we can say that $6+\sqrt{2}$ is irrational.

Exercise 1.3 (Page No. 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) $\dfrac{13}{3125}$

Ans: Given a rational number $\dfrac{13}{3125}$.

If the denominator of a rational number has prime factors of the form ${{2}^{n}}{{5}^{m}}$, where, $m$ and $n$ are positive integers. Then the rational number has terminating decimal expansion. If the denominator has factors other than $2$ and $5$, then it has non-terminating decimal expansion.

The denominator of the given number is $3125$.

Then, factors of $3125$ are

$\Rightarrow 3125=5\times 5\times 5\times 5\times 5$

$\Rightarrow 3125={{5}^{5}}$

Here, the factors of denominator are of the form ${{5}^{m}}$.

Therefore, $\dfrac{13}{3125}$ has terminating decimal expansion.

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

Ans: Given a rational number $\dfrac{17}{8}$.

The denominator of the given number is $8$.

Then, factors of $8$ are

$\Rightarrow 8=2\times 2\times 2$

$\Rightarrow 8={{2}^{3}}$

Here, the factors of denominator are of the form ${{2}^{n}}$.

Therefore, $\dfrac{17}{8}$ has terminating decimal expansion.

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

Ans: Given a rational number $\dfrac{64}{455}$.

The denominator of the given number is $455$.

Then, factors of $455$ are

$\Rightarrow 455=5\times 7\times 13$

Here, the factors of denominator are not in the form ${{2}^{n}}{{5}^{m}}$. The denominator has factors other than $2$ and $5$.

Therefore, $\dfrac{64}{455}$ has non-terminating repeating decimal expansion.

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

Ans: Given a rational number $\dfrac{15}{1600}$.

The denominator of the given number is $1600$.

Then, factors of $1600$ are

$\Rightarrow 1600=2\times 2\times 2\times 2\times 2\times 2\times 5\times 5$

$\Rightarrow 1600={{2}^{6}}\times {{5}^{5}}$

Here, the factors of denominator are of the form ${{2}^{n}}{{5}^{m}}$.

Therefore, $\dfrac{15}{1600}$ has terminating decimal expansion.

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

Ans: Given a rational number $\dfrac{29}{343}$.

The denominator of the given number is $343$.

Then, factors of $343$ are

$\Rightarrow 343=7\times 7\times 7$

$\Rightarrow 343={{7}^{3}}$

Therefore, $\dfrac{29}{343}$ has non-terminating repeating decimal expansion.

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

Ans: Given a rational number $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$.

The denominator of the given number is ${{2}^{3}}{{5}^{2}}$.

Here, the denominator is of the form ${{2}^{n}}{{5}^{m}}$.

Therefore, $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$ has terminating decimal expansion.

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

Ans: Given a rational number $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$.

The denominator of the given number is ${{2}^{2}}{{5}^{7}}{{7}^{5}}$.

Here, the denominator is of the form ${{2}^{n}}{{5}^{m}}$ but also has factors other than $2$ and $5$.

Therefore, $\dfrac{129}{{{2}^{2}}{{5}^{7}}{{7}^{5}}}$ has non-terminating repeating decimal expansion.

(viii) $\dfrac{6}{15}$

Ans: Given a rational number $\dfrac{6}{15}$.

The denominator of the given number is $15$.

$\Rightarrow 15=3\times 5$

But we can write the numerator of the given number as

$\dfrac{6}{15}=\dfrac{2\times 3}{3\times 5}=\dfrac{2}{5}$

Therefore, $\dfrac{6}{15}$ has terminating decimal expansion.

(ix) $\dfrac{35}{50}$

Ans: Given a rational number $\dfrac{35}{50}$.

The denominator of the given number is $50$.

Then, factors of $50$ are

$\Rightarrow 50=10\times 5$

$\dfrac{35}{50}=\dfrac{7\times 5}{10\times 5}=\dfrac{7}{10}$

$\Rightarrow 10=2\times 5$

Therefore, $\dfrac{35}{50}$ has terminating decimal expansion.

(x) $\dfrac{77}{210}$

Ans: Given a rational number $\dfrac{77}{210}$.

The denominator of the given number is $210$.

Then, factors of $210$ are

$\Rightarrow 210=2\times 3\times 5\times 7$

Here, the denominator has factors other than $2$ and $5$.

Therefore, $\dfrac{77}{210}$ has non-terminating repeating decimal expansion.

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

Ans: To find the decimal expansion of $\dfrac{13}{3125}$, we will divide the numerator of the number by denominator using long division method. We get

$3125\overset{0.00416}{\overline{\left){\begin{align} & 13.00000 \\ & 0 \\ & \overline{130\text{ }} \\ & \text{ 0} \\ & \overline{\text{13000}} \\ & \text{12500} \\ & \overline{\text{ 5000 }} \\ & \text{ 3125} \\ & \overline{\text{ 18750 }} \\ & \underline{\text{ 18750 }} \\ & \text{ 0} \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{13}{3125}$ is $0.00416$.

Ans: To find the decimal expansion of $\dfrac{17}{8}$, we will divide the numerator of the number by denominator using long division method. We get

$8\overset{2.125}{\overline{\left){\begin{align} & 17 \\ & 16 \\ & \overline{\text{ 10 }} \\ & \text{ 8} \\ & \overline{\text{ 20 }} \\ & \text{ 16} \\ & \overline{\text{ 40 }} \\ & \text{ }\underline{\text{40}} \\ & \text{ 0} \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{17}{8}$ is $2.125$.

(iii) $\dfrac{15}{1600}$

Ans: To find the decimal expansion of $\dfrac{15}{1600}$, we will divide the numerator of the number by denominator using long division method. We get

$1600\overset{0.009375}{\overline{\left){\begin{align} & 15.000000 \\ & 0 \\ & \overline{150\text{ }} \\ & \text{ 0} \\ & \overline{\text{1500}} \\ & \text{0} \\ & \overline{\text{15000 }} \\ & \text{14400} \\ & \overline{\text{ 6000 }} \\ & \underline{\text{ 4800 }} \\ & \text{ 12000} \\ & \underline{\text{ 11200}} \\ & \text{ 8000} \\& \text{ }\underline{\text{8000}} \\ & \text{ 0} \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{15}{1600}$ is $0.009375$.

(iv) $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$

Ans: To find the decimal expansion of $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$, we will divide the numerator of the number by denominator using long division method. We get

$\dfrac{23}{{{2}^{3}}{{5}^{2}}}=\dfrac{23}{200}$

$200\overset{00.115}{\overline{\left){\begin{align} & 23.000 \\ & 0 \\ & \overline{23\text{ }} \\ & \text{ 0} \\ & \overline{\text{230 }} \\ & 200 \\ & \overline{\text{ 300 }} \\ & \text{ 200} \\ & \overline{\text{ 1000 }} \\ & \underline{\text{ 1000 }} \\ & \text{ 0} \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{23}{{{2}^{3}}{{5}^{2}}}$ is $0.115$.

(v) $\dfrac{6}{15}$

Ans: To find the decimal expansion of $\dfrac{6}{15}$, we will divide the numerator of the number by denominator using long division method. We get

$5\overset{0.4}{\overline{\left){\begin{align} & 2.0 \\ & 0 \\ & \overline{20\text{ }} \\ & \underline{20\text{ }} \\ & 0 \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{6}{15}$ is $0.4$.

(vi) $\dfrac{35}{50}$

Ans: To find the decimal expansion of $\dfrac{35}{50}$, we will divide the numerator of the number by denominator using long division method. We get

$50\overset{0.7}{\overline{\left){\begin{align} & 35.0 \\ & 0 \\ & \overline{\text{350 }} \\ & \underline{350\text{ }} \\ & \text{ 0} \\ \end{align}}\right.}}$

Therefore, the decimal expansion of $\dfrac{35}{50}$ is $0.7$.

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

(i) $43.123456789$

Ans: Given a decimal expansion $43.123456789$.

The given number has terminating expansion, we can write the number as $\dfrac{43123456789}{1000000000}$, which is of the form $\dfrac{p}{q}$.

Therefore, the number $43.123456789$ is a rational number.

Since the number has terminating decimal expansion, the factors of $q$ must be of the form ${{2}^{n}}{{5}^{m}}$.

(ii) $0.120120012000120000......$

Ans: Given a decimal expansion $0.120120012000120000......$.

When we observe the given expansion we can say that the number has non-terminating and non-repeating decimal expansion. Hence we cannot express it in the form of $\dfrac{p}{q}$.

Therefore, the number is irrational.

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

Ans: Given the decimal expansion $43.\overline{123456789}$.

The given number has non-terminating but repeating decimal expansion. So the number will be of the form $\dfrac{p}{q}$.

Therefore, the number $43.\overline{123456789}$ is a rational number.

But the factors of denominator are not of the form ${{2}^{n}}{{5}^{m}}$. Denominator also has factors other than $2$ and $5$.

Chapter 1 - Real Numbers

You can opt for Chapter 1 - Real Numbers NCERT Solutions for Class 10 Maths PDF for upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

## NCERT Solutions for Class 10 Maths

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

## NCERT Solutions for Class 10 Maths Chapter 1 All Exercises

Important topics under ncert solutions for class 10 maths chapter 1 real numbers.

The first chapter of the class 10 maths syllabus includes Real Numbers, which is an important chapter in maths covered in class 10. The chapter on Real Numbers is divided into 4 major parts. The following is a list of the important topics covered under the chapter Real Numbers. It is recommended that students read through these topics carefully to be able to learn, use, and master this chapter and be able to solve problems based on Real Numbers.

Introduction to Real Numbers

- Euclid’s Division Lemma (Not available in the current syllabus)

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

Revisiting Irrational Numbers

## 1.1 Introduction

In the introduction part of ch 1 Maths Class 10 students will be reminded of what they learned in class IX about real numbers and irrational numbers. This section gives a glimpse of what students would learn about positive numbers in the later sections of Chapter 1 Maths Class 10, i.e. Euclid’s division algorithm and the fundamental theorem of arithmetic. The Fundamental Theorem of Arithmetic is based on the fact that a composite number can be expressed as a product of prime numbers, in distinct ways. This theorem has deep and significant applications in mathematics.

## 1.2 Euclid’s Division Lemma (Not available in the current syllabus)

You will learn about euclid’s division lemma and algorithm in this section of ncert solutions class 10 maths chapter 1. a lemma is a statement that is proven and acts as a stepping stone to prove other statements. euclid’s division lemma states the usual division system in mathematics in a formal way, i.e. for every pair of positive numbers x and y; there are two unique whole numbers a and b that satisfy the equation: x = ya + b where 0 <= b <= y and x is a dividend, y is a divisor. a is the quotient. b is the remainder. in other words: dividend = (quotient * divisor) + remainder. you would also learn euclid’s division algorithm in this portion of ch 1 maths class 10 ncert solutions which is based on the lemma. an algorithm is a set of well-defined steps that can procedurally solve a problem. euclid’s division algorithm is used in calculating the hcf(highest common factor) of two positive integers., 1.2 the fundamental theorem of arithmetic.

According to this theorem, every composite number can be factorized as a product of some prime numbers. It is a unique prime factorization of natural numbers as the order of the factors does not matter. We will understand this with an example that is based on the following fundamentals:

HCF - The highest common factor of two or more integers is the greatest integer that can exactly divide all the given integers. For example, HCF of 60 and 75 is 15.

LCM - The Least Common Multiple of two or more integers is the smallest integer that is exactly divisible by all the given integers. For example, LCm of 2, 4, and 5 is 20.

For two positive integers a and b; HCF(a,b) * LCM (a, b) = a * b

So, going by this theorem we can express any natural number as a multiplication of prime number, for example, 253 = 11 * 23, 4 = 2 * 2, etc.

## 1.3 Revisiting Irrational Numbers

In this section of NCERT Solutions for Class 10th Maths Chapter 1, you will remember the definition of Irrational numbers learned in earlier classes and then prove p is an irrational number, where p is a prime number.

If a number “n” can not be written in the form x/y, then it is called an irrational number. Here x and y are integers and n <> 0. Few examples of irrational numbers are 2,3, etc.

## Key Features of NCERT Solutions for Class 10 Maths Chapter 1

The Class 10 Maths NCERT Solutions Chapter 1 prepared by the scholars of Vedantu is one of the most reliable online resources. The key takeaways of these NCERT Solutions for Class 10 Maths ch 1 are:

Students will get answers to all the questions in Maths Class 10 NCERT Solutions Chapter 1 and no question is left out.

The Class 10 Maths Chapter 1 Solutions are also available for download in a PDF format which makes revisions very quick and easy during stressful exam times.

You will find that going through Class 10 Maths Chapter 1 NCERT Solutions is effortless as they are written in a simple and easily comprehensible manner which is apt for the understanding level of class 10 students.

## Importance of 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.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers

1. What are the Steps of Calculating the HCF of Two Positive Integers?

To calculate the HCF (Highest Common Factor) of two positive integers x and y (x is greater than y), we can follow the steps mentioned below:

Find out the quotient q and remainder r that satisfy Euclid’s Division Lemma x = (y * q) + r.

If r or remainder is 0, then the HCF of the two numbers is n.

If r is <> 0, then we need to apply Euclid’s division Lemma to y and r.

The above process has to be continued till we get r = 0. When we reach this stage, the divisor would be the HCF of x and y.

2. What is the Difference Between Whole Numbers, Integers, and Natural Numbers?

The definitions of these numbers are as below:

Natural numbers are all non-negative counting numbers, excluding 0, i.e. 5, 6, 7, 8, etc.

Whole numbers are similar to natural numbers except that they include 0, so whole numbers are 0, 1, 2, 3, etc.

Integers comprise all negative and positive numbers (including 0). So integers are -2, -1, 0, 1, 2, etc.

3. How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 1?

Chapter 1 in the Class 10 NCERT Maths textbook is Real Numbers. There are a total of 4 exercises in this chapter. If you want the solutions to these exercise problems, then you can download Vedantu's NCERT Solutions for Class 10 Maths Chapter 1 in PDF format. All the exercises have a different PDF. These solutions will allow you to finish your NCERT syllabus and be ready for the Class 10 Maths exam.

4. Where can I download CBSE Class 10 NCERT Mathematics chapter 1 PDF Solutions?

CBSE Class 10 NCERT Mathematics PDF solutions are your best tool for scoring well in your Class 10 exams. Everything in your question paper will be directly or indirectly from your NCERT textbooks. You need to go through all the exercises and chapters in order to be prepared for the exam. Here is how you can download Vedantu’s Class 10 NCERT Mathematics PDF Solutions:

Visit the page of ‘NCERT Solutions for Class 10 Maths’ on the official website of Vedantu.

Find the exercise for which you want a solution and click on the ‘Download PDF’ link.

Then enter your phone no or email ID to sign in.

And, the solutions will be downloaded into your system. You’ll also receive a message/ mail with a direct download button of your preferred solution.

5. What are the real numbers in Class 10 Maths?

In simple terms, real numbers are a combination of rational as well as irrational numbers present in the number system. The general rule is that these numbers can be used for all types of arithmetic operations and can be represented on the number line as well. On Vedantu’s website, you will find that all the concepts of real numbers are explained in detail. We also provide solutions to exercise problems from NCERT textbooks as well as practice questions. You can download NCERT Solutions for Class 10 Maths in PDF format.

6. Are NCERT Solutions for Class 10 Maths Chapter 1 important from an exam point of view?

Yes, Chapter 1 in NCERT Class 10 Maths Solutions is important from an exam point of view. There are important questions in your NCERT Class 10 Maths book. This chapter includes short-type answer questions and long-type answer questions. This chapter is a fundamental chapter and the given concepts will be used in the upcoming chapters. This chapter develops your problem-solving skills. You can practice these problems from Vedantu’s NCERT Solutions for Class 10 Maths which is prepared by experts and will clear all your doubts.

7. What are the main topics covered in the NCERT Solutions for Class 10 Maths Chapter 1?

The main topics covered in this chapter are as follows:

Euclid’s division algorithm

The fundamental theorem of arithmetic

Revisiting Rational and Irrational Numbers

Decimal Expansions

The experts at Vedantu have designed the solutions based on these important concepts. They have framed the solutions in a simple and easy-to-understand manner. They focus on helping students to score good marks in exams.

## NCERT Solutions for Class 10

Class 10 revision notes.

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

- Exercise 1.1
- Exercise 1.2
- Exercise 1.3
- Exercise 1.4

## NCERT Solutions for Class 10 Maths Chapters:

How many exercises in chapter 1 real numbers, what is lemma, what do you mean by fundamental theorem of arithmetic., what is algorithm, contact form.

- NCERT Exemplar
- Maths Exemplar Class 10
- Real Numbers

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

Ncert exemplar solutions class 10 maths chapter 1 – free pdf download.

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 good marks in the board exam.

In this chapter, students will learn to solve problems based on topics like

- Euclid’s Division Lemma
- The Fundamental Theorem of Arithmetic
- Revisiting Irrational Numbers
- Revisiting Rational Numbers and Their Decimal Expansions

The above-given subtopics, explained in the NCERT Exemplar Class 10 Maths problems of Chapter 1 Class 10, are important for students as they will be appearing for the board exam, and the result they obtain will determine which subject to choose for further studies. Thus, in Class 10, students should develop a good understanding of each and every concept, especially in Maths. They should also sharpen their problem-solving skills if they want to succeed in the exam. Click here to get exemplars for all chapters.

There will be a significant amount of pressure of exams on students, and in order to help them get better results, we are offering free NCERT Exemplar for Real Numbers here. The exemplar has been designed by our subject experts, based on the CBSE syllabus, to help students get a clear idea about two main topics: the fundamental theorem of arithmetic and Euclid’s Division Algorithm .

## Download the PDF of the NCERT Exemplar Class 10 Maths Chapter 1 Real Numbers

## Access Answers to the NCERT Exemplar Class 10 Maths Chapter 1 Real Numbers

Real numbers exercise 1.1.

Choose the correct answer from the given four options in the following questions:

1. For some integer m, every even integer is of the form:

(A) m (B) m +1

(C) 2m (D) 2m+1

Explanation:

Even integers are those integers which are divisible by 2.

Hence, we can say that every integer which is a multiple of 2 must be an even integer.

Therefore, let us conclude that,

for an integer ‘m’, every even integer must be of the form

2 × m = 2m.

Hence, option (C) is the correct answer.

2. For some integer q, every odd integer is of the form

(A) q (B) q +1

(C) 2g (D) 2q +1

Odd integers are those integers which are not divisible by 2.

Hence, we can say that every integer which is a multiple of 2 must be an even integer, while 1 added to every integer which is multiplied by 2 is an odd integer.

for an integer ‘q’, every odd integer must be of the form

(2 × q)+1 = 2q+1.

Hence, option (D) is the correct answer.

3. n 2 – 1 is divisible by 8, if n is

(A) an integer (B) a natural number

(C) an odd integer (D) an even integer

(C) an odd integer

Let x = n 2 – 1

In the above equation, n can be either even or odd.

Let us assume that n= even.

So, when n = even i.e., n = 2k, where k is an integer,

⇒ x = (2k) 2 -1

⇒ x = 4k 2 – 1

At k = -1, x = 4(-1) 2 – 1 = 4 – 1 = 3, is not divisible by 8.

At k = 0, x = 4(0) 2 – 1 = 0 – 1 = -1, is not divisible by 8

Let us assume that n= odd:

So, when n = odd i.e., n = 2k + 1, where k is an integer,

⇒ x = 2k + 1

⇒ x = (2k+1) 2 – 1

⇒ x = 4k 2 + 4k + 1 – 1

⇒ x = 4k 2 + 4k

⇒ x = 4k(k+1)

At k = -1, x = 4(-1)(-1+1) = 0 which is divisible by 8.

At k = 0, x = 4(0)(0+1) = 0 which is divisible by 8 .

At k = 1, x = 4(1)(1+1) = 8 which is divisible by 8.

From the above two observation, we can conclude that, if n is odd, n 2 -1 is divisible by 8.

4. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is

(A) 4 (B) 2

(C) 1 (D) 3

Let us find the HCF of 65 and 117,

117 = 1×65 + 52

65 = 1× 52 + 13

52 = 4 ×13 + 0

Hence, we get the HCF of 65 and 117 = 13.

According to the question,

65m – 117 = 13

65m = 117+13 = 130

∴ m =130/65 = 2

Hence, option (B) is the correct answer.

5. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

(A) 13 (B) 65

(C) 875 (D) 1750

We have to find the largest number which divides 70 and 125, leaving remainders 5 and 8.

This can be also written as,

To find the largest number which exactly divides (70 – 5), and (125 – 8)

The largest number that divides 65 and 117 is also the Highest Common Factor of 65 and 117

Therefore, the required number is the HCF of 65 and 117

Factors of 65 = 1, 5, 13, 65

Factors of 117 = 1, 3, 9, 13, 39, 117

Common Factors = 1, 13

Highest Common factor (HCF) = 13

i.e., the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively = 13

Hence, option (A) is the correct answer.

## Real Numbers Exercise 1.2

1. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer.

No, every positive integer cannot be of the form 4q + 2, where q is an integer.

Justification:

All the numbers of the form 4q + 2, where ‘q’ is an integer, are even numbers which are not divisible by ‘4’.

For example,

4q+2 = 4(1) + 2= 6.

4q+2 = 4(2) + 2= 10

4q+2 = 4(0) + 2= 2 and so on.

So, any number which is of the form 4q+2 will give only even numbers which are not multiples of 4.

Hence, every positive integer cannot be written in the form 4q+2

2. “The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons.

Yes, the statement “the product of two consecutive positive integers is divisible by 2” is true.

Let the two consecutive positive integers = a, a + 1

According to Euclid’s division lemma,

a = bq + r, where 0 ≤ r < b

For b = 2, we have a = 2q + r, where 0 ≤ r < 2 … (i)

Substituting r = 0 in equation (i),

a = 2q, is divisible by 2.

a + 1 = 2q + 1, is not divisible by 2.

Substituting r = 1 in equation (i),

a = 2q + 1, is not divisible by 2.

a + 1 = 2q + 1+1 = 2q + 2, is divisible by 2.

Thus, we can conclude that, for 0 ≤ r < 2, one out of every two consecutive integers is divisible by 2. So, the product of the two consecutive positive numbers will also be even.

Hence, the statement “product of two consecutive positive integers is divisible by 2” is true.

3. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.

Yes, the statement “the product of three consecutive positive integers is divisible by 6” is true.

Consider the 3 consecutive numbers 2, 3, 4

(2 × 3 × 4)/6 = 24/6 = 4

Now, consider another 3 consecutive numbers 4, 5, 6

(4 × 5 × 6)/6 = 120/6 = 20

Now, consider another 3 consecutive numbers 7, 8, 9

(7 × 8 × 9)/6 = 504/6 = 84

Hence, the statement “product of three consecutive positive integers is divisible by 6” is true.

4. Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.

No, the square of any positive integer cannot be written in the form 3m + 2 where m is a natural number

A positive integer ‘a’ can be written in the form of bq + r

a = bq + r, where b, q and r are any integers,

a = 3(q) + r, where, r can be an integers,

For r = 0, 1, 2, 3……….

3q + 0, 3q + 1, 3q + 2, 3q + 3……. are positive integers,

(3q) 2 = 9q² = 3(3q²) = 3m (where 3q² = m)

(3q+1) 2 = (3q+1)² = 9q²+1+6q = 3(3q²+2q) +1 = 3m + 1 (Where, m = 3q²+2q)

(3q+2) 2 = (3q+2)² = 9q²+4+12q = 3(3q²+4q) +4 = 3m + 4 (Where, m = 3q²+2q)

(3q+3) 2 = (3q+3)² = 9q²+9+18q = 3(3q²+6q) +9 = 3m + 9 (Where, m = 3q²+2q)

Hence, there is no positive integer whose square can be written in the form 3m + 2 where m is a natural number.

5. A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer.

Consider the positive integer 3q + 1, where q is a natural number.

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

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

= 3m + 1, (where m is an integer which is equal to 3q 2 + 2q.

Thus (3q + 1) 2 cannot be expressed in any other form apart from 3m + 1.

## Real Numbers Exercise 1.3

1. Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

According to Euclid’s division lemma,

When b = 4.

a = 4k + r, 0 < r < 4

When r = 0, we get, a = 4k

a 2 = 16k 2 = 4(4k 2 ) = 4q, where q = 4k 2

When r = 1, we get, a = 4k + 1

a 2 = (4k + 1) 2 = 16k 2 + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2)

When r = 2, we get, a = 4k + 2

a 2 = (4k + 2) 2 = 16k 2 + 4 + 16k = 4(4k 2 + 4k + 1) = 4q, where q = 4k 2 + 4k + 1

When r = 3, we get, a = 4k + 3

a 2 = (4k + 3) 2 = 16k 2 + 9 + 24k = 4(4k 2 + 6k + 2) + 1

= 4q + 1, where q = 4k 2 + 6k + 2

Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Hence Proved.

2. Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.

Let a be any positive integer and b = 4.

According to Euclid Division Lemma,

According to the question, the possible values of r are,

r = 0, r = 1, r = 2, r = 3

When r = 0,

Taking cubes on LHS and RHS,

a³ = 4 (16q³)

When r = 1,

a³ = (4q + 1)³

a³ = 64q³ + 1³ + 3 × 4q × 1 (4q + 1)

a³ = 64q³ + 1 + 48q² + 12q

a³ = 4 (16q³ + 12q² + 3q) + 1

When r = 2,

a³ = (4q + 2)³

a³ = 64q³ + 2³ + 3 × 4q × 2 (4q + 2)

a³ = 64q³ + 8 + 96q² + 48q

a³ = 4 (16q³ + 2 + 24q² + 12q)

When r = 3,

a³ = (4q + 3)³

a³ = 64q³ + 27 + 3 × 4q × 3 (4q + 3)

a³ = 64q³ + 24 + 3 + 144q² + 108q

a³ = 4 (16q³ + 36q² + 27q + 6) + 3

Hence, the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.

3. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Let the positive integer = a

According to the question, b = 5

So, r= 0, 1, 2, 3, 4

When r = 0, a = 5m.

When r = 1, a = 5m + 1.

When r = 2, a = 5m + 2.

When r = 3, a = 5m + 3.

When r = 4, a = 5m + 4.

When a = 5m

a 2 = (5m) 2 = 25m 2

a 2 = 5(5m 2 ) = 5q, where q = 5m 2

When a = 5m + 1

a 2 = (5m + 1) 2 = 25m 2 + 10 m + 1

a 2 = 5 (5m 2 + 2m) + 1 = 5q + 1, where q = 5m 2 + 2m

When a = 5m + 2

a 2 = (5m + 2) 2

a 2 = 25m 2 + 20m + 4

a 2 = 5 (5m 2 + 4m) + 4

a 2 = 5q + 4 where q = 5m 2 + 4m

When a = 5m + 3

a 2 = (5m + 3) 2 = 25m 2 + 30m + 9

a 2 = 5 (5m 2 + 6m + 1) + 4

a 2 = 5q + 4 where q = 5m 2 + 6m + 1

When a = 5m + 4

a 2 = (5m + 4) 2 = 25m 2 + 40m + 16

a 2 = 5 (5m 2 + 8m + 3) + 1

a 2 = 5q + 1 where q = 5m 2 + 8m + 3

Therefore, square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

4. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

According to Euclid’s division algorithm,

a = 6q + r, where 0 ≤ r < 6

a 2 = 6(6q 2 + 2qr) + r 2 …(i), where,0 ≤ r < 6

When r = 0, substituting r = 0 in Eq.(i), we get

a 2 = 6 (6q 2 ) = 6m, where, m = 6q 2 is an integer.

When r = 1, substituting r = 1 in Eq.(i), we get

a 2 = 6 (6q 2 + 2q) + 1 = 6m + 1, where, m = (6q 2 + 2q) is an integer.

When r = 2, substituting r = 2 in Eq(i), we get

a 2 = 6(6q 2 + 4q) + 4 = 6m + 4, where, m = (6q 2 + 4q) is an integer.

When r = 3, substituting r = 3 in Eq.(i), we get

a 2 = 6(6q 2 + 6q) + 9 = 6(6q 2 + 6a) + 6 + 3

a 2 = 6(6q 2 + 6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer.

When r = 4, substituting r = 4 in Eq.(i) we get

a 2 = 6(6q 2 + 8q) + 16

= 6(6q 2 + 8q) + 12 + 4

⇒ a 2 = 6(6q 2 + 8q + 2) + 4 = 6m + 4, where, m = (6q 2 + 8q + 2) is integer.

When r = 5, substituting r = 5 in Eq.(i), we get

a 2 = 6 (6q 2 + 10q) + 25 = 6(6q 2 + 10q) + 24 + 1

a 2 = 6(6q 2 + 10q + 4) + 1 = 6m + 1, where, m = (6q 2 + 10q + 4) is integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Hence Proved

5. Show that the square of any odd integer is of the form 4q + 1, for some integer q.

Let a be any odd integer and b = 4.

According to Euclid’s algorithm,

a = 4m + r for some integer m ≥ 0

And r = 0,1,2,3 because 0 ≤ r < 4.

So, we have that,

a = 4m or 4m + 1 or 4m + 2 or 4m + 3 So, a = 4m + 1 or 4m + 3

We know that, a cannot be 4m or 4m + 2, as they are divisible by 2.

(4m + 1) 2 = 16m 2 + 8m + 1

= 4(4m 2 + 2m) + 1

= 4q + 1, where q is some integer and q = 4m 2 + 2m.

(4m + 3) 2 = 16m 2 + 24m + 9

= 4(4m 2 + 6m + 2) + 1

= 4q + 1, where q is some integer and q = 4m 2 + 6m + 2

Therefore, Square of any odd integer is of the form 4q + 1, for some integer q.

6. If n is an odd integer, then show that n 2 – 1 is divisible by 8.

We know that any odd positive integer n can be written in form 4 q + 1 or 4 q + 3.

So, according to the question,

When n = 4 q + 1,

Then n 2 – 1 = (4 q + 1) 2 – 1 = 16 q 2 + 8 q + 1 – 1 = 8 q (2 q + 1), is divisible by 8.

When n = 4 q + 3,

Then n 2 – 1 = (4 q + 3) 2 – 1 = 16 q 2 + 24 q + 9 – 1 = 8(2 q 2 + 3 q + 1), is divisible by 8.

So, from the above equations, it is clear that, if n is an odd positive integer

n 2 – 1 is divisible by 8.

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

Let the two odd positive numbers x and y be 2k + 1 and 2p + 1, respectively

i.e., x 2 + y 2 = (2k + 1) 2 +(2p + 1) 2

= 4k 2 + 4k + 1 + 4p 2 + 4p + 1

= 4k 2 + 4p 2 + 4k + 4p + 2

= 4 (k 2 + p 2 + k + p) + 2

Thus, the sum of square is even the number is not divisible by 4

Therefore, if x and y are odd positive integer, then x 2 + y 2 is even but not divisible by four.

## Real Numbers Exercise 1.4

1. Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5

Then, the positive integers are of form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.

Taking cube on L.H.S and R.H.S,

(6q)³ = 216 q³ = 6(36q)³ + 0

= 6m + 0, (where m is an integer = (36q)³)

(6q+1)³ = 216q³ + 108q 2 + 18q + 1

= 6(36q³ + 18q 2 + 3q) + 1

= 6m + 1, (where m is an integer = 36q³ + 18q 2 + 3q)

(6q+2)³ = 216q³ + 216q 2 + 72q + 8

= 6(36q³ + 36q 2 + 12q + 1) +2

= 6m + 2, (where m is an integer = 36q³ + 36q 2 + 12q + 1)

(6q+3)³ = 216q³ + 324q 2 + 162q + 27

= 6(36q³ + 54q 2 + 27q + 4) + 3

= 6m + 3, (where m is an integer = 36q³ + 54q 2 + 27q + 4)

(6q+4)³ = 216q³ + 432q 2 + 288q + 64

= 6(36q³ + 72q 2 + 48q + 10) + 4

= 6m + 4, (where m is an integer = 36q³ + 72q 2 + 48q + 10)

(6q+5)³ = 216q³ + 540q 2 + 450q + 125

= 6(36q³ + 90q 2 + 75q + 20) + 5

= 6m + 5, (where m is an integer = 36q³ + 90q 2 + 75q + 20)

Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

2. Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.

According to Euclid’s division Lemma,

Let the positive integer = n

n =3q+r, where q is the quotient and r is the remainder

0<r<3 implies remainders may be 0, 1 and 2

Therefore, n may be in the form of 3q, 3q+1, 3q+2

Here n is only divisible by 3

When n = 3q+1

Here only n+2 is divisible by 3

When n=3q+2

n+4=3q+2+4=3q+6

Here only n+4 is divisible by 3

So, we can conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3.

The solved questions in Exemplar Class 10 for Maths will help students to prepare and score good marks in the board exam. Students are also provided with 10th Standard Maths NCERT Solutions , notes, question papers and other learning materials online on our website for free. Also, solve previous years’ question papers and sample papers of Class 10 Maths to understand the question pattern and marking scheme for the chapter Real Numbers.

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

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

## CBSE Class 10 Maths Chapter Real Numbers Solutions

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

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

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

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

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

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Case Study Questions for Class 10 Maths Chapter 1 - Real Numbers To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class...

CASE STUDY 1. To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- section A and section B of grade X. There are 32 students in section A and 36 students in section B.

CBSE 10th Standard Maths Subject Real Number Case Study Questions With Solution 2021 By QB365 on 21 May, 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 .

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

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

Question Bank Case Study Questions » Class 10 Maths CBSE Case Study Questions for Class 10 Maths Real Numbers Free PDF Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Real Numbers in order to fully complete your preparation. They are very very important from exam point of view.

TopperLearning provides a complete collection of case studies for CBSE Class 10 Maths Real Numbers chapter. Improve your understanding of biological concepts and develop problem-solving skills with expert advice.

The case study questions on Real Numbers are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Real Numbers case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination.

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.

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

In other words, all natural numbers can be represented in the form of the product of its prime factors. Prime factors are the numbers that cannot be divisible by other numbers and are only divisible by 1 . For example, the number 56 can be written in the form of its prime factors as: 56 = 2³ × 7. For the number 56, the prime factors are 2 and 7.

Class 10. 14 units · 43 skills. Unit 1. Real numbers. Unit 2. Polynomials. Unit 3. Pair of linear equations in two variables. Unit 4. Quadratic equations. ... Real numbers 1.1 Get 7 of 10 questions to level up! Finding HCF & LCM of large numbers Get 3 of 4 questions to level up! Euclid's Division Algorithm (Bonus)

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, 4n ends in 0? Show Answer

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 2023-24. The answers of 10th Maths Chapter 1 MCQ are given with explanation, so that students can understand easily.

In general, Real numbers constitute the union of all rational and irrational numbers. Any real number can be plotted on the number line. To know more about real numbers, visit here. Students can refer to the short notes and MCQ questions along with a separate solution pdf of this chapter for quick revision from the links below.

Introduction HCF using Euclid's Division Algorithm Euclid's Division Algorithm - Proving Prime Factorization LCM/HCF Decimal Expansion Irrational numbers What's in it? Updated for NCERT 2023-2024 Book. Answers to all exercise questions and examples are solved for Chapter 1 Class 10 Real numbers.

Class 10 Real Numbers Important Questions and Answers Q.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. Solution: Let x be any positive integer and y = 3. By Euclid's division algorithm; x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3) Therefore,

9. The least number that is divisible by all the numbers from 1-10 (both inclusive) is (A) 10 (B) 100 (C) 504 (D)2520. Ans: D. Since, the required number is divisible from all the numbers from 1 to 10. Therefore, we need to find the LCM. Factors of numbers from 1 to 10. $1 = 1$ $2 = 1 \times 2$ $3 = 1 \times 3$ $4 = 2 \times 2$ $5 = 1 \times 5$

Chapter 1 in the Class 10 NCERT Maths textbook is Real Numbers. There are a total of 4 exercises in this chapter. If you want the solutions to these exercise problems, then you can download Vedantu's NCERT Solutions for Class 10 Maths Chapter 1 in PDF format. All the exercises have a different PDF.

Answer. 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 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. Answer.

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 ...

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