NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
NCERT solutions for class 10 maths chapter 1 Real Numbers- In mathematics, there are two kinds of numbers -
- Real number
- Imaginary number
Real numbers are those numbers that can be represented on the number line. If anyone wants to go deeper into the number system then studying real numbers is the first step to master the number system. Real numbers include the in-depth classification of numbers, its applications, and properties related to various kinds of numbers. In this particular chapter, we are going to talk about real numbers. The NCERT solutions for class 10 maths chapter 1 real numbers carry detailed explanations for each and every question. These NCERT solutions can assist you while doing homework as well as in the preparation of the board exam. NCERT solutions for class 10 maths chapter 1 real numbers will give the conceptual clarity which is are very important as board exam point of view as well as for the competitive examinations. Apart from this, NCERT solutions for other subjects and classes can be downloaded by clicking on the above link. Here you will get NCERT solutions for class 10 also.
Types of questions asked from class 10 maths chapter 1 Real Numbers
CBSE Class 10 maths board exam will have the following types of questions from real numbers:
Euclid's division algorithm
Prime factorization
Highest common factor
Prime factorization
Prime & composite numbers
Rational & irrational numbers
NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.1
Q1 (1) Use Euclid’s division algorithm to find the HCF of 135 and 225
Answer:
225 > 135. Applying Euclid's Division algorithm we get
since remainder 0 we again apply the algorithm
since remainder 0 we again apply the algorithm
since remainder = 0 we conclude the HCF of 135 and 225 is 45.
Q1 (2) Use Euclid’s division algorithm to find the HCF of 196 and 38220
Answer:
38220 > 196. Applying Euclid's Division algorithm we get
since remainder = 0 we conclude the HCF of 38220 and 196 is 196.
Q1 (3) Use Euclid’s division algorithm to find the HCF of 867 and 255
Answer:
867 > 225. Applying Euclid's Division algorithm we get
since remainder 0 we apply the algorithm again.
since 255 > 102
since remainder 0 we apply the algorithm again.
since 102 > 51
since remainder = 0 we conclude the HCF of 867 and 255 is 51.
Answer:
Let p be any positive integer. It can be expressed as
p = 6q + r
where and
but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.
Answer:
The maximum number of columns in which they can march = HCF (32, 616)
Since 616 > 32, applying Euclid's Division Algorithm we have
Since remainder 0 we again apply Euclid's Division Algorithm
Since 32 > 8
Since remainder = 0 we conclude, 8 is the HCF of 616 and 32.
The maximum number of columns in which they can march is 8.
Answer:
Let x be any positive integer.
It can be written in the form 3q + r where and r = 0, 1 or 2
Case 1:
For r = 0 we have
x ^{ 2 } = (3q) ^{ 2 }
x ^{ 2 } = 9q ^{ 2 }
x ^{ 2 } = 3(3q ^{ 2 } )
x ^{ 2 } = 3m
Case 2:
For r = 1 we have
x ^{ 2 } = (3q+1) ^{ 2 }
x ^{ 2 } = 9q ^{ 2 } + 6q +1
x ^{ 2 } = 3(3q ^{ 2 } + 2q) + 1
x ^{ 2 } = 3m + 1
Case 3:
For r = 2 we have
x ^{ 2 } = (3q+2) ^{ 2 }
x ^{ 2 } = 9q ^{ 2 } + 12q +4
x ^{ 2 } = 3(3q ^{ 2 } + 4q + 1) + 1
x ^{ 2 } = 3m + 1
Hence proved.
Answer:
Let x be any positive integer.
It can be written in the form 3q + r where and r = 0, 1 or 2
Case 1:
For r = 0 we have
x ^{ 3 } = (3q) ^{ 3 }
x ^{ 3 } = 27q ^{ 3 }
x ^{ 3 } = 9(3q ^{ 3 } )
x ^{ 3 } = 9m
Case 2:
For r = 1 we have
x ^{ 3 } = (3q+1) ^{ 3 }
x ^{ 3 } = 27q ^{ 3 } + 27q ^{ 2 } + 9q + 1
x ^{ 3 } = 9(3q ^{ 3 } + 3q ^{ 2 } +q) + 1
x ^{ 3 } = 3m + 1
Case 3:
For r = 2 we have
x ^{ 3 } = (3q+2) ^{ 3 }
x ^{ 3 } = 27q ^{ 3 } + 54q ^{ 2 } + 36q + 8
x ^{ 3 } = 9(3q ^{ 3 } + 6q ^{ 2 } +4q) + 8
x ^{ 3 } = 3m + 8
Hence proved.
NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.2
Q1 (1) Express each number as a product of its prime factors: 140
Answer:
The number can be as a product of its prime factors as follows
Q 1 (2) Express each number as a product of its prime factors: 156
Answer:
The given number can be expressed as follows
Q1 (3) Express each number as a product of its prime factors: 3825
Answer:
The number is expressed as the product of the prime factors as follows
Q1 (4) Express each number as a product of its prime factors: 5005
Answer:
The given number can be expressed as the product of its prime factors as follows.
Q1 (5) Express each number as a product of its prime factors: 7429
Answer:
The given number can be expressed as the product of their prime factors as follows
Answer:
26 = 2 x 13
91 = 7 x 13
HCF(26,91) = 13
LCM(26,91) = 2 x 7 x 13 = 182
HCF x LCM = 13 x 182 = 2366
26 x 91 = 2366
26 x 91 = HCF x LCM
Hence Verified
Answer:
The number can be expressed as the product of prime factors as
510 = 2 x 3 x 5 x 17
92 = 2 ^{ 2 } x 23
HCF(510,92) = 2
LCM(510,92) = 2 ^{ 2 } x 3 x 5 x 17 x 23 = 23460
HCF x LCM = 2 x 23460 = 46920
510 x 92 = 46920
510 x 92 = HCF x LCM
Hence Verified
Answer:
336 is expressed as the product of its prime factor as
336 = 2 ^{ 4 } x 3 x 7
54 is expressed as the product of its prime factor as
54 = 2 x 3 ^{ 3 }
HCF(336,54) = 2 x 3 = 6
LCM(336,54) = 2 ^{ 4 } x 3 ^{ 3 } x 7 = 3024
HCF x LCM = 6 x 3024 = 18144
336 x 54 = 18144
336 x 54 = HCF x LCM
Hence Verified
Answer:
The numbers can be written as the product of their prime factors as follows
12 = 2 ^{ 2 } x 3
15 = 3 x 5
21 = 3 x 7
HCF = 3
LCM = 2 ^{ 2 } x 3 x 5 x 7 = 420
Answer:
The given numbers are written as the product of their prime factors as follows
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
HCF = 1
LCM = 17 x 23 x 29 = 11339
Answer:
The given numbers are written as the product of their prime factors as follows
8 = 2 ^{ 3 }
9 = 3 ^{ 2 }
25 = 5 ^{ 2 }
HCF = 1
LCM = 2 ^{ 3 } x 3 ^{ 2 } x 5 ^{ 2 } = 1800
Q4 Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:
As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have
HCF (306, 657) x LCM (306, 657) = 306 x 657
Q5 Check whether can end with the digit 0 for any natural number n.
Answer:
By prime factorizing we have
6 ^{ n } = 2 ^{ n } x 3 ^{ n }
A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6 ^{ n } we can conclude that for no value of n 6 ^{ n } will end with the digit 0.
Q6 Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers.
Answer:
7 x 11 x 13 + 13
= (7 x 11 + 1) x 13
= 78 x 13
= 2 x 3 x 13 ^{ 2 }
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5
= (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5
= 5 x 1008
After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.
Answer:
The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.
Time taken by Sonia = 18 = 2 x 3 ^{ 2 }
Time taken by Ravi = 12 = 2 ^{ 2 } x 3
LCM(18,12) = 2 ^{ 2 } x 3 ^{ 2 } = 36
Therefore they would again meet at the starting point after 36 minutes.
NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.3
Answer:
Let us assume is rational.
It means can be written in the form where p and q are co-primes and
Squaring both sides we obtain
From the above equation, we can see that p ^{ 2 } is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number.
Therefore p can be written as 5r
p = 5r
p ^{ 2 } = (5r) ^{ 2 }
5q ^{ 2 } = 25r ^{ 2 }
q ^{ 2 } = 5r ^{ 2 }
From the above equation, we can see that q ^{ 2 } is divisible by 5, Therefore q will also be divisible by 5 as 5 is a prime number.
From (i) and (ii) we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that is rational was wrong. Hence proved that is irrational.
Answer:
Let us assume is rational.
This means can be wriiten in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
Since p and q are co-prime integers will be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 (2) Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
Q3 (3) Prove that the following are irrationals :
Answer:
Let us assume is rational.
This means can be written in the form where p and q are co-prime integers.
As p and q are integers would be rational, this contradicts the fact that is irrational. This contradiction arises because our initial assumption that is rational was wrong. Therefore is irrational.
NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.4
Answer:
The denominator is of the form 2 ^{ a } x 5 ^{ b } where a = 0 and b = 5. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is of the form 2 ^{ a } x 5 ^{ b } where a = 3 and b = 0. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2 ^{ a } x 5 ^{ b } . Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2 ^{ a } x 5 ^{ b } where a = 6 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2 ^{ a } x 5 ^{ b } . Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2 ^{ a } x 5 ^{ b } where a = 3 and b = 2. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2 ^{ a } x 5 ^{ b } . Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
The denominator is of the form 2 ^{ a } x 5 ^{ b } where a = 0 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is of the form 2 ^{ a } x 5 ^{ b } where a = 1 and b = 1. Therefore the given rational number will have a terminating decimal expansion.
Answer:
The denominator is not of the form 2 ^{ a } x 5 ^{ b } . Therefore the given rational number will have a non-terminating repeating decimal expansion.
Answer:
decimal expansions of rational numbers are
(i)
(ii)&nbsnbsp;
(iv)
(vI)
(viii)
(ix)
Answer:
The denominator is of the form 2 ^{ a } x 5 ^{ b } where a = 9 and b = 9. Therefore the given number is rational and has a terminating decimal expansion.
Answer:
Since the decimal part of the given number is non-terminating and non-repeating we can conclude that the given number is irrational and cannot be written in the form where p and q are integers.
Answer:
As the decimal part of the given number is non-terminating and repeating, the number is rational but its denominator will have factors other than 2 and 5.
NCERT solutions for class 10 maths chapter wise
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables |
Chapter 4 | NCERT solutions for class 10 maths chapter 4 Quadratic Equations |
Chapter 5 | NCERT solutions for class 10 chapter 5 Arithmetic Progressions |
Chapter 6 | |
Chapter 7 | NCERT solutions for class 10 maths chapter 7 Coordinate Geometry |
Chapter 8 | NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry |
Chapter 9 | NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles |
Chapter 13 | NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes |
Chapter 14 | |
Chapter 15 |
NCERT solutions of class 10 subject wise
How to use NCERT Solutions for class 10 maths chapter 1 Real Numbers?
- Now you have gone through the solutions of NCERT class 10 maths chapter 1 real numbers and have a good knowledge of answer writing in a structural manner. Its time to practice various kinds of problems based on real numbers.
After the completion of the NCERT syllabus, you can check past 5-year papers of board exams. Previous year papers will increase your dealing ability with a variety of questions.
NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERTs and the previous year papers, you can jump to the next chapters.
Keep working hard & happy learning!
Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Question: What is the weightage of the chapter real numbers for CBSE board exam
Answer:
As per the CBSE the total weightage of real number is 4 marks in the final board exam.
Question: Where can I find the complete solutions of NCERT class 10 maths
Answer:
Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.
Question: What are the important topics of class 10 maths chapter 1 real numbers.
Answer:
Euclid's division algorithm, Euclid's division algorithm to find highest common factor, prime & composite numbers, rational & irrational numbers are the important topics from the chapter real numbers.
Question: How many chapters are there in the class 10 maths ?
Answer:
There are 15 chapters in the class 10 maths NCERT. Chapter 1- Real Numbers, Chapter 2- Polynomials, Chapter 3- Pair of Linear Equations in Two Variables, Chapter 4- Quadratic Equations, Chapter 5- Arithmetic Progressions, Chapter 6- Triangles, Chapter 7- Coordinate Geometry, Chapter 8- Introduction to Trigonometry, Chapter 9- Some Applications of Trigonometry, Chapter 10- Circles, Chapter 11- Constructions, Chapter 12- Areas Related to Circles, Chapter 13- Surface Areas and Volumes, Chapter 14- Statistics, Chapter 15- Probability are the chapters in the NCERT class 10 maths.
Question: Which is the official website of NCERT ?
Answer:
http://ncert.nic.in/ is the official website of the NCERT where you can get NCERT textbooks from class 1 to 12 and syllabus from class 1 to 12 for all the subjects.
Question: How the NCERT solutions are helpful in the board exam ?
Answer:
Students are advised to solve every problems on their own but they may find difficulties while solving the problems. You can take help from the these NCERT solutions where you will get following benefits-
- It will help you to get the conceptual clarity.
- These solutions are provided by the experts who knows how best to write in board exam in order to get good marks.
- These solutions are provided in very simple language, so it will very easy for you to understand the concepts.
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