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NCERT solutions for Class 10 Maths chapter 1 Real Numbers are discussed here. In Mathematics, there are two kinds of numbers - Real number and Imaginary number. Real numbers are those numbers that can be represented on the number line. NCERT solutions for Class 10 Maths Chapter 1 Real Numbers will help the students to prepare for the exams in a better way. The problems and solutions are prepared by expert teachers keeping in mind the Latest updated CBSE syllabus 2023. Students can practice these NCERT solutions for class 10 to get good hold on the concepts.
In this Class 10 Maths Chapter 1, we are going to talk about real numbers. The real numbers class 10 carry detailed explanations for each and every question in simple, comprehensive, and step by step explanation. NCERT solutions for other subjects and classes can be downloaded by clicking on the above link.
Also, see-
Types of Numbers:
Rational and Irrational Numbers:
Real Numbers: Numbers located on the number line, used in real-world applications.
Euclid's Division Algorithm (Lemma):
Fundamental Theorem of Arithmetic:
HCF and LCM by Prime Factorization:
HCF (a, b) × LCM (a, b) = a × b
Free download NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers PDF for CBSE Exam.
Real Numbers class 10 Exercise: 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.
Real Numbers class 10 Exercise: 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.
Class 10 Maths Chapter 1 Solutions Exercise: 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.
Class 10 Maths Chapter 1 Solutions Exercise: 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.
The following points have been examined by us:
Chapter No. | Chapter Name |
Chapter 1 | Real Numbers |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
If interested students can also check exercises here:
Now you have gone through the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers and have a good knowledge of answer writing in a structural manner. It's 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 NCERT and the previous year papers, you can jump to the next chapters.
Using the NCERT Class 10 Maths solutions chapter 1, students will be able to confirm the right answers once they are done solving the questions themselves.
Class 10 Maths Chapter 1 NCERT solutions are solved by the subject matter experts. Therefore, the answers to all the questions are reliable.
NCERT Class 10 solutions for Chapter 1 Maths gives the step by step explanations to all the questions which makes it easy for the students to understand.
NCERT Solutions for Class 10 Maths Real Numbers includes topics like Euclid's Division Algorithm, Euclid's Division Algorithm to Find the Highest Common Factor, Prime & Composite Numbers, and Rational & Irrational Numbers. NCERT Solutions for ncert class 10 maths chapter 1 is designed by our experts to provide students with a solid foundation in the fundamental concepts and techniques of mathematics. it will help students to score well on tests as well as in board exams.
Yes, NCERT Solutions for maths class 10 chapter 1 are important for several reasons. Firstly, these solutions provide detailed explanations of the various concepts and principles covered in the chapter.
Secondly, the NCERT Solutions for chapter 1 maths class 10 often include a large number of examples and practice problems, which can help students apply their knowledge to solve real-world problems.
Thirdly, the NCERT Solutions for class 10 chapter 1 maths are typically written by experienced teachers or subject matter experts.
Overall, real numbers class 10 solutions are essential resources for students who want excel in maths.
As per the CBSE, the total weightage of class 10 maths ch 1 Real Number is 4 marks in the final board exam. To get a good score follow the NCERT syllabus and problems from the NCERT book and the previous year's papers. Interested students can use class 10 maths chapter 1 solutions to command the concepts.
The NCERT solutions for all 15 chapters of Class 10 Mathematics can be downloaded from the careers360 website. these solutions are listed chapter-wise in tabulate format above in this article. students can visit maths ch 1 class 10 solutions if they are facing any issues while solving the maths ch 1 class 10 problems.
There are 15 chapters in Class 10 Maths NCERT syllabus. it includes Real Numbers, Polynomials, Pair of Linear Equations in Two Variables, Quadratic Equations, Arithmetic Progressions, Triangles, Coordinate Geometry, Introduction to Trigonometry, Some Applications of Trigonometry, Circles, Constructions, Area Related to Circles, Surface Areas and Volumes, Statistics, and Probability.
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