# NCERT Solutions for Class 10 Maths Chapter 1

Download NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers exercise 1.4, 1.3, 1.2 & 1.1 in Hindi (English & हिंदी medium) in PDF form for UP Board (High School) Schools as well as CBSE Board Schools. Download Class 10 Maths App or कक्षा 10 गणित App or Apps for other subjects for offline use.

 Class 10: Maths – गणित Chapter 1: Real Numbers Go Back to Class 10 Maths Main Page

## NCERT Solutions for Class 10 Maths Chapter 1

These solutions and study material related to this chapter is modified according to latest CBSE Syllabus 2019-20. 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.

### NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

#### English Medium/ Hindi Medium Solutions

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 uploaded in July-August 2018.

#### What is 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.

#### 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.

Example of Euclid’s division lemma
Prove that the square of an integer is of the form 9k or 3k + 1. 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.

#### Historical Facts

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.

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#### Use Euclid’s division algorithm to find the HCF of 135 and 225.

135 and 225
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.

#### बिना लम्बी विभाजन प्रक्रिया किये बताइए कि निम्नलिखित परिमेय संख्या के दशमलव प्रसार सांत हैं या असांत आवर्ती हैं: 13/3125.

3125 = 5×5×5×5×5 = 55
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 ।

#### Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Let a be any positive integer and b = 6.
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.

#### An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

HCF (616, 32) will give the maximum number of columns in which they can march.
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.

3825
= 3×3×5×5×17

#### 510 और 92 के HCF और LCM ज्ञात कीजिए तथा इसकी जाँच कीजिए कि दो संख्याओं का गुणनफल = HCF×LCM है।

510 और 92
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

#### Find the LCM and HCF of 12, 15 and 21 by applying the prime factorisation method.

12, 15 and 21
12 = 2×2×3 = 2^2×3
15 = 3×5
21 = 3×7
HCF = 3
LCM = 2^2 × 3 × 5 × 7 = 420

#### जाँच कीजिए कि क्या किसी प्राकृत संख्या n के लिए, संख्या 6^n अंक 0 पर समाप्त हो सकती है।

यदि कोई संख्या अंक 0 पर समाप्त हो सकती है, तो वह 10 से विभाजित होती है
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।

#### Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Numbers are of two types – prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
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.

#### There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
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.

## 1 thought on “NCERT Solutions for Class 10 Maths Chapter 1”

1. Gagan deep says:

kuch aise question jiske hal nahi hua hai ham uska unsoled ki pdf lena chahte hai