# NCERT Solutions for Class 10 Maths Chapter 1

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers all exercises in Hindi & English medium PDF form updated for new academic year 2020-21. These solutions are for UP Board (High School) Schools as well as CBSE Board Schools whoever is following NCERT Books 2020-21.

Download Class 10 Maths App or कक्षा 10 गणित App or Apps for other subjects for offline use. Visit to Discussion Forum to share your knowledge and ask your doubts.## NCERT Solutions for Class 10 Maths Chapter 1

Class: | 10 |

Subject: | Maths – गणित |

Chapter 1: | Real Numbers |

### 10th Maths Chapter 1 Solutions

These NCERT Solutions 2020-21 and study material related to this chapter is modified according to latest CBSE Syllabus 2020-2021. 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

##### 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.