# 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- 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 in-depth 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.

## 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

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.

38220 > 196. Applying Euclid's Division algorithm we get

since remainder = 0 we conclude the HCF of  38220 and 196 is 196.

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.

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.

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.

[Hint : Let x be any positive integer then 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]

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

x2 = (3q)2

x2 = 9q2

x2 = 3(3q2)

x2 = 3m

Case 2:

For r = 1 we have

x2 = (3q+1)2

x2 = 9q2 + 6q +1

x2 = 3(3q2 + 2q) + 1

x2 = 3m + 1

Case 3:

For r = 2 we have

x2 = (3q+2)2

x2 = 9q2 + 12q +4

x2 = 3(3q2 + 4q + 1) + 1

x2 = 3m + 1

Hence proved.

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

x3 = (3q)3

x3 = 27q3

x3 = 9(3q3)

x3 = 9m

Case 2:

For r = 1 we have

x3 = (3q+1)3

x3 = 27q3 + 27q2 + 9q + 1

x3 = 9(3q3 + 3q2 +q) + 1

x3 = 3m + 1

Case 3:

For r = 2 we have

x3 = (3q+2)3

x3 = 27q3 + 54q2 + 36q + 8

x3 = 9(3q3 + 6q2 +4q) + 8

x3 = 3m + 8

Hence proved.

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.2

The number can be as a product of its prime factors as follows

The given number can be expressed as follows

The number is expressed as the product of the prime factors as follows

The given number can be expressed as the product of its prime factors as follows.

The given number can be expressed as the product of their prime factors as follows

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

The number can be expressed as the product of prime factors as

510 = 2 x 3 x 5 x 17

92 = 2x 23

HCF(510,92) = 2

LCM(510,92) = 22 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

336 is expressed as the product of its prime factor as

336 = 24 x 3 x 7

54 is expressed as the product of its prime factor as

54 = 2 x 33

HCF(336,54) = 2 x 3 = 6

LCM(336,54) = 24 x 3x 7 = 3024

HCF x LCM = 6 x 3024 = 18144

336 x 54 = 18144

336 x 54 = HCF x LCM

Hence Verified

The numbers can be written as the product of their prime factors as follows

12 = 22 x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 22 x 3 x 5 x 7 = 420

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

The given numbers are written as the product of their prime factors as follows

8 = 23

9 = 32

25 = 52

HCF = 1

LCM = 23 x 32 x 52 = 1800

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

By prime factorizing we have

6n = 2n x 3n

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 6n we can conclude that for no value of n 6n will end with the digit 0.

7 x 11 x 13 + 13

= (7 x 11 + 1) x 13

= 78 x 13

= 2 x 3 x 132

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.

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 32

Time taken by Ravi = 12 = 22 x 3

LCM(18,12) = 22 x 32 = 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

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 p2 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

p2 = (5r)2

5q2 = 25r2

q2 = 5r2

From the above equation, we can see that q2 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.

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.

(i)

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.

(ii)

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.

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

The denominator is of the form 2a x 5b where a = 0 and b = 5. Therefore the given rational number will have a terminating decimal expansion.

The denominator is of the form 2a x 5b where a = 3 and b = 0. Therefore the given rational number will have a terminating decimal expansion.

The denominator is not of the form 2a x 5b. Therefore the given rational number will have a non-terminating repeating decimal expansion.

The denominator is of the form 2a x 5b where a = 6 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

The denominator is not of the form 2a x 5b. Therefore the given rational number will have a non-terminating repeating decimal expansion.

The denominator is of the form 2a x 5b where a = 3 and b = 2. Therefore the given rational number will have a terminating decimal expansion.

The denominator is not of the form 2a x 5b. Therefore the given rational number will have a non-terminating repeating decimal expansion.

The denominator is of the form 2a x 5b where a = 0 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

The denominator is of the form 2a x 5b where a = 1 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

The denominator is not of the form 2a x 5b. Therefore the given rational number will have a non-terminating repeating decimal expansion.

decimal expansions of  rational numbers are

(i)

(ii)&nbsnbsp;

(iv)

(vI)

(viii)

(ix)

The denominator is of the form 2a x 5b where a = 9 and b = 9. Therefore the given number is rational and has a terminating decimal expansion.

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.

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 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 Solutions of NCERT class 10 maths chapter 6 Triangles 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 CBSE NCERT solutions class 10 maths chapter 10 Circles Chapter 11 NCERT solutions  for class 10 maths chapter 11 Constructions 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 NCERT solutions for class 10 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 10 maths chapter 15 Probability

## 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!