NCERT solutions for class 10 maths chapter 1 Real Numbers In mathematics, there are two kinds of numbers one is the real number and another one is the imaginary number. In this particular chapter, we are going to talk about real numbers. Solutions of NCERT class 10 maths chapter 1 real numbers carry detailed explanations for each and every question. CBSE NCERT solutions for class 10 maths chapter 1 real numbers can assist you while doing homework as well as for the board exam preparation. Real numbers are those numbers that can be represented in the number line. If anyone wants to go deeper into the number system then studying real numbers is the first step to master number system. Real numbers include the indepth classification of numbers, its applications, and properties related to various kinds of numbers. CBSE NCERT solutions for class 10 maths chapter 1 real numbers concepts are very important when you will be appearing in the competitive examinations. Apart from this, NCERT solutions for other subjects and classes can be downloaded by clicking on the link.
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 coprimes 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 coprimes. 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 coprime 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 coprime integers.
Since p and q are coprime 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 wriiten in the form where p and q are coprime 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 coprime 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 nonterminating 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 nonterminating 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 nonterminating 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 nonterminating 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 nonterminating and nonrepeating 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 nonterminating and repeating, the number is rational but its denominator will have factors other than 2 and 5.
Chapter No. 
Chapter Name 
Chapter 1 
CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers 
Chapter 2 
NCERT solutions for class 10 maths chapter 2 Polynomials 
Chapter 3 
Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4 
CBSE 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 
CBSE 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 
Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10 

Chapter 11 

Chapter 12 
Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13 
CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14 

Chapter 15 
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Q1 (10) Without actually performing the long division, state whether the following rational
numbers will have a terminating decimal expansion or a nonterminating repeating decimal
expansion: 77 / 210
Q 1 (2) Express each number as a product of its prime factors: 156
Q1 (3) Without actually performing the long division, state whether the following rational
numbers will have a terminating decimal expansion or a nonterminating repeating decimal
expansion: 64 / 455