# NCERT Solutions for Class 10 Maths Chapter 1

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers all exercises in Hindi and English medium PDF format updated for the new academic year 2020-2021. NCERT Textbook solutions are useful for both UP Board (High School) Schools, and CBSE Board Schools as Uttar Pradesh Board is now following NCERT Books for 2020-21. So, students can download UP Board Solutions for Class 10 Maths Chapter 1 from this page. Download Class 10 Maths App or Class 10 Ganit App or Offline Apps for other subjects for offline use in UP Board or CBSE or other boards based on NCERT Book’s Syllabus.

Visit Discussion Forum to share your knowledge and ask your doubts. In Chapter 1 of 10th Maths, we observe that Euclid’s division algorithm has to do with divisibility of integers. The result obtained from Euclid’s Lemma is quite easy to state and understand. In Class 10 Maths, there are many applications related to the divisibility properties of integers. Here, we will try a few of them to compute the HCF of two positive integers.## NCERT Solutions for Class 10 Maths Chapter 1

Class: 10 | Maths (English and Hindi Medium) |

Chapter 1: | Real Numbers |

### Class 10 Maths Chapter 1 Solution All Pages

### 10th Maths Chapter 1 Solutions

Class 10 Maths Chapter 1 NCERT Solutions 2020-21 and study material related to 10th Maths chapter 1 is modified according to latest CBSE Syllabus 2020-2021. UP Board students also use this solution for their exams. The fundamental theorem of arithmetic and Euclid’s division lemma are the main topics of this chapter (Real Numbers). The Fundamental Theorem of Arithmetic and Euclid’s division lemma has many applications, both within mathematics and in other fields. NCERT Solutions of all other subjects are also available in PDF form.

### 10th Maths Chapter 1 Exercise 1.1

### 10th Maths Chapter 1 Exercise 1.2

### 10th Maths Chapter 1 Exercise 1.3

### 10th Maths Chapter 1 Exercise 1.4

#### Changes in CBSE Syllabus 2020-2021 Class 10 Maths Chapter 1

CBSE syllabus change due to corona-virus are given below for class 10 Maths Chapter 1. CBSE has issued a new syllabus by reducing about 30 percent of previous syllabus. See all the changes made for Class 10 Maths Syllabus for 2020-2021.

##### Revised CBSE Syllabus issued on July 7, 2020

The updated CBSE Syllabus for Class 10 Maths chapter 1 is as follows:

Fundamental Theorem of Arithmetic – statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality of √2, √3, √5. Decimal representation of rational numbers interms of terminating/non-terminating recurring decimals.

###### Deleted Section

Euclid’s division lemma.

#### Class 10 Maths Exercise 1.1 & 1.2 Solutions in Video

#### Class 10 Maths Exercise 1.3 & 1.4 Solutions in Video

#### Class 10 Maths Chaper 1 Exercise 1.1 and 1.2 Solution in Hindi

#### Class 10 Maths Chaper 1 Exercise 1.1 and 1.2 Solution in Hindi

## Difference between algorithm and lemma.

### Algorithm

An algorithm is a series of well defined steps which gives a procedure for solving a type of problem.

### Lemma

A lemma is a proven statement used for proving another statement.

### Class 10 Maths Chapter 1: Real Numbers

#### Class 10 Maths Exercise 1.1

In Exercise 1.1 of Class 10 Maths Chapter 1, there are only five questions. All the questions are based on Euclid’s Division Lemma and its applications. In Questions number 1 we have to find HCF applying Euclid’s Division Lemma. In Questions number 2, 4 and 5 also apply step by step Euclid’s Division Algorithms to prove the questions. In Questions 3, we have to find out the maximum number of columns in which they can march, that is, HCF of the two numbers.

#### Class 10 Maths Exercise 1.2

In Exercise 1.2 of Class 10 Maths, the Fundamental Theorem of Arithmetic is main to solve the questions. Question number 1, 2, 3 are based on the prime factorisation method of LCM and HCF. In Question 4, we can use direct formula a × b = LCM × HCF. Question number 6 is an important question as per exams point of view and based on Example Number 5. One can solve Question 6 by taking common or directly solving each number and showing that it has more than two factors. At last in Question 7, we have to find out a number which is divisible by both 12 and 18 that is LCM of the two numbers.

#### Class 10 Maths Exercise 1.3

In Exercise 1.2 of Class 10 Maths, only three questions are there. We know that the square root of a non-perfect square number is an irrational number. Here, we have to prove the same fact with a various set of numbers. In each of these questions, first of all, we assume that the given number is a rational number with numerator and denominator as coprime integers. Later on, we receive a false result due to the wrong assumption, so we conclude that the given number is irrational.

#### Class 10 Maths Exercise 1.4

Class 10 Maths Exercise 1.4 deals with that decimal representation of rational numbers. Questions of Exercise 1.4 are based on Theorem 1.5, Theorem 1.6 and Theorem 1.7 given in NCERT Book of Class 10 Maths Chapter 1. Using these theorems, we can find whether the given rational number is either terminating or non-terminating repeating or non-terminating non-repeating. Most often, one mark questions are asked in CBSE Board exams or school tests also from Exercise 1.4 of Class 10 Maths.

#### Euclid’s division algorithm

The step by step way to find the HCF of two positive integers, say c and d, with c > d.

- Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 £ r < d.
- Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
- Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

##### What is algorithm?

An algorithm is a series of well-defined steps which gives a procedure for solving a type of problem.

##### What is a Lemma?

A lemma is a proven statement used for proving another statement.

##### Define Euclid’s Division Lemma?

Euclid’s Division Lemma:

For the given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, r = 0 or 0 < r < b.

##### What do you understand by Fundamental Theorem of Arithmetic?

Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. In other word, the prime factorisation of a natural number is unique, except for the order of its factors.

#### Test & Assignments

Download the practice test series with answers. Level 1 Test 1 contains basic questions for practicing the chapter Real Numbers. Most of the questions of Level 1 Test 2 are easy to understand and provides good practice. Answers of these test series will be available on website free to use.

##### Historical Facts on Real Numbers

The word algorithm come from the name of the name of the 9th century Persian mathematician al-Khwarizmi. The word ‘algebra’ is derived from a book, he wrote, called Hisab al-jabr w’al-muqabala.

An equivalent version of Fundamental theorem of Arithmetic was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic.

However, the first correct proof was given by Carl Friedrich Gauss in his al-Khwarizmi.

Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton. He has made fundamental contributions to both mathematics and science.

###### Prove that the square of an integer is of the form 9k or 3k + 1.

### Important Questions on Class 10 Maths Chapter 1

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 × 45 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45.

हम जानते हैं कि किसी परिमेय संख्या p/q

(जहाँ p और q सहअभाज्य हैं) में, यदि q का अभाज्य गुणनखंडन 2^m 5^n के रूप में है,

जहाँ m और n ऋणेतर पूर्णांक हैं, तो उसका दशमलव प्रसार सांत होता है।

3125 का अभाज्य गुणनखंडन = 55

क्योंकि यह गुणनखंडन 5^n के रूप में है,

इसलिए इसका दशमलव प्रसार सांत होगा ।

Then, by Euclid’s algorithm,

a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2p + 1, where p is a positive integer 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2m + 1, where m is an integer 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2n + 1, where n is an integer Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers. Therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5.

We can use Euclid’s algorithm to find the HCF.

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.

= 3×3×5×5×17

510 = 2×3×5×17

92 = 2×2×23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

दो संख्याओं का गुणनफल = 510×92 = 46920

HCF×LCM = 2×23460 = 46920

इस प्रकार, दो संख्याओं का गुणनफल = HCF×LCM

12 = 2×2×3 = 2^2×3

15 = 3×5

21 = 3×7

HCF = 3

LCM = 2^2 × 3 × 5 × 7 = 420

या दूसरे शब्दों यह संख्या 2 और 5 से विभाजित होगी।

क्योंकि 10 = 2 × 5

6^n का अभाज्य गुणनखंडन = (2 ×3)^n = 2^n×3^n।

6^n के अभाज्य गुणनखंडन में 5 नहीं है। इसलिए 6^n, 5 से विभाजित नहीं होगा।

अंकगणित की आधारभूत प्रमेय की अद्वितीयता हमें यह निश्चित कराती है कि 6^n के गुणनखंड में 2 और 3 के अतिरिक्त और कोई अभाज्य गुणनखंड नहीं है।

अतः, किसी भी प्राकृत संख्या n के लिए, संख्या 6^n अंक 0 पर समाप्त नहीं हो सकती है।

It can be observed that:

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)

= 13 × 78

= 13 ×13 × 6

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 ×1009

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

18 = 2 × 3 × 3

And, 12 = 2 × 2 × 3

LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.