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RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers

In this article we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers prepared by our experts to help students to prepare better for their examinations.

Ananya Gupta19 Aug, 2024

RS Aggarwal Solutions for Class 10 Maths Chapter 1: RS Aggarwal Solutions for Class 10 Maths Chapter 1, "Real Numbers," helps students understand important math concepts in a clear and simple way.

This chapter covers topics like the Euclidean Division Algorithm, the Fundamental Theorem of Arithmetic, and how to find the Highest Common Factor (HCF) and Least Common Multiple (LCM). The solutions provide step-by-step explanations and examples, making it easier for students to learn and solve problems. These solutions are great for building a strong math foundation and preparing for exams.

RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers Overview

These notes, created by the experts at Physics Wallah, give an overview of RS Aggarwal Solutions for Class 10 Maths Chapter 1, "Real Numbers." This chapter covers key topics like the Euclidean Division Algorithm, the Fundamental Theorem of Arithmetic, and how to find the Highest Common Factor (HCF) and Least Common Multiple (LCM). The solutions are clearly explained with step-by-step instructions, making it easy for students to understand and solve problems. These notes are a valuable resource for exam preparation and help students build a strong foundation in mathematics.

RS Aggarwal Solutions for Class 10 Maths Chapter 1 PDF

The PDF link below has RS Aggarwal Solutions for Class 10 Maths Chapter 1, "Real Numbers." The solutions are easy to follow with step-by-step explanations, helping students understand and solve problems. This PDF is a great resource for studying and building strong math skills.

RS Aggarwal Solutions for Class 10 Maths Chapter 1 PDF

What is Real Numbers?

Real numbers include all the numbers that can be found on the number line. This includes:

Natural Numbers : Counting numbers like 1, 2, 3, and so on.
Whole Numbers : Natural numbers plus zero (0, 1, 2, 3, ...).
Integers : Whole numbers and their negatives (..., -3, -2, -1, 0, 1, 2, 3, ...).
Rational Numbers : Numbers that can be expressed as fractions (like 1/2, 3/4) or as repeating or terminating decimals (like 0.5 or 0.333...).
Irrational Numbers : Numbers that cannot be expressed as simple fractions and have non-repeating, non-terminating decimals (like √2 or π).

Real numbers include all rational and irrational numbers, covering everything from simple counting numbers to complex decimals.

RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.1

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.2

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.3

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.4

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.5

RS Aggarwal Solutions for Class 10 Maths Chapter 1 Exercise Real Numbers

Here we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers for the ease of students so that they can prepare better for their exams.

Question 1.

Solution:

For any two given positive integers a and b, there exist unique whole numbers q and r such that a = bq + r, when 0 ≤ r < b. Here, a is called the dividend, b is a divisor, q is the quotient, and r is the remainder. Dividend = (Divisor x Quotient) + Remainder.

Question 2.

Solution:

Using Euclid’s division Lemma Dividend = (Divisor x Quotient) + Remainder = (61 x 27) + 32 = 1647 + 32 = 1679 Required number = 1679

Question 3.

Solution:

Let the required divisor = x Then by Euclid’s division Lemma, Dividend = (Divisor x Quotient) + remainder 1365 = x x 31 + 32 => 1365 = 31x + 32 => 31x= 1365 – 32 = 1333 x = 1331 31 = 43 Divisor = 43

Question 4.

Solution:

Let n be a given positive odd integer. On dividing n by 6, let m be the quotient and r be the remainder, then by Euclid’s division Lemma. n = 6m + r, where 0 ≤ r < 6 => n = 6m + r, where r = 0, 1, 2, 3, 4, 5 => n = 6m or (6m + 1) or (6m + 2) or (6m + 3) or (6m + 4) or (6m + 5) But n = 6m, (6m + 2) and (6m + 4) are even. Thus when n is odd, it will be in the form of (6m + 1) (6m + 3), or (6m + 5) for some integer m.

Question 5.

Solution:

Let n be an arbitrary odd positive integer. On dividing by 4, let m be the quotient and r be the remainder. So to Euclid’s division lemma, n = 4m + r, where 0 ≤ r < 4 n = 4m or (4m + 1) or (4m + 2) or (4m + 3) But 4m and (4m + 2) are even integers. Since n is odd, so n ≠ 4m or n ≠ (4m + 2) n = (4m + 1) or (4m + 3) for some integer m. Hence any positive odd integer is of the form (4m + 1) or (4m + 3) for some integer m.

Question 6.

Solution:

Let a = n ³ – n => a = n (n ² – 1) => a = n (n – 1) (n + 1) [(a ² – b ² ) = (a – b) (a + b)] => a = (n – 1 ) n (n + 1) We know that, (i) If a number is completely divisible by 2 and 3, then it is also divisible by 6. (ii) If the sum of digits of any number is divisible by 3, then it is also divisible by 3. (iii) If one of the factors of any number is an even number, then it is also divisible by 2. a = (n – 1) n (n + 1) [From Eq. (i)] Now, the sum of the digits = n – 1 + n + n + 1 = 3n = Multiple of 3, where n is any positive integer. and (n – 1) n (n +1) will always be even, as one out of (n – 1) or n or (n + 1) must be even. Since, conditions (ii) and (iii) completely satisfy Eq. (i). Hence, by condition (i) the number n3 – n is always divisible by 6, where n is any positive integer. Hence proved.

Question 7.

Solution:

Let x = 2m + 1 and y = 2m + 3 are odd positive integers, for every positive integer m. Then, x ² + y ² = (2m + 1) ² + (2m + 3) ² = 4m ² + 1 + 4 m + 4m ² + 9 + 12m [(a + b) ² = a ² + 2ab + b ² ] = 8m ² + 16m + 10 = even = 2(4m ² + 8m + 5) or 4(2m ² + 4m + 2) + 1 Hence, x ² + y ² is even for every positive integer m but not divisible by 4.

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 1

Clear Explanations : Each solution is provided with step-by-step explanations, making it easier for students to understand complex concepts and solve problems effectively.
Comprehensive Coverage : The solutions cover all topics in the chapter, including the Euclidean Division Algorithm, the Fundamental Theorem of Arithmetic, and methods to find HCF and LCM, ensuring thorough preparation.
Problem-Solving Skills : By working through these solutions, students can improve their problem-solving skills and learn how to approach different types of questions confidently.
Exam Preparation : The solutions help students prepare for exams by providing practice with a variety of problems similar to those they might encounter in their tests.
Foundation Building : These solutions help build a strong foundation in mathematics, which is essential for understanding more advanced topics in higher classes.
Self-Learning : The detailed explanations and step-by-step approach enable students to study and learn on their own, promoting independent learning.
Error Reduction : By following these solutions, students can identify and correct their mistakes, leading to a better understanding of the subject and improved accuracy.

RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers FAQs

What are Real Numbers?

Real numbers include all rational and irrational numbers. They can be represented on the number line and include integers, fractions, decimals, and more.

What is the Euclidean Division Algorithm?

The Euclidean Division Algorithm is a method used to find the quotient and remainder when dividing one integer (dividend) by another (divisor).

What is the Fundamental Theorem of Arithmetic?

The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be uniquely expressed as a product of prime numbers.

How do you find the Highest Common Factor (HCF) and Least Common Multiple (LCM)?

The HCF of two or more numbers is the largest number that divides each of them without leaving a remainder. The LCM is the smallest number that is a multiple of each of the numbers.

What are Rational and Irrational Numbers?

Rational numbers can be expressed as fractions or as terminating or repeating decimals. Irrational numbers cannot be expressed as fractions and have non-repeating, non-terminating decimals.

Why are Real Numbers important?

Real numbers are fundamental in mathematics as they form the basis for solving equations, representing quantities, and understanding relationships between numbers.

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RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers Overview

RS Aggarwal Solutions for Class 10 Maths Chapter 1 PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 1 PDF

What is Real Numbers?

Real numbers include all the numbers that can be found on the number line. This includes:

Natural Numbers : Counting numbers like 1, 2, 3, and so on.
Whole Numbers : Natural numbers plus zero (0, 1, 2, 3, ...).
Integers : Whole numbers and their negatives (..., -3, -2, -1, 0, 1, 2, 3, ...).
Rational Numbers : Numbers that can be expressed as fractions (like 1/2, 3/4) or as repeating or terminating decimals (like 0.5 or 0.333...).
Irrational Numbers : Numbers that cannot be expressed as simple fractions and have non-repeating, non-terminating decimals (like √2 or π).

Real numbers include all rational and irrational numbers, covering everything from simple counting numbers to complex decimals.

RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.1

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.2

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.3

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.4

RS Aggarwal Solutions for Class 10 Maths Chapter-1 Exercise 1.5

RS Aggarwal Solutions for Class 10 Maths Chapter 1 Exercise Real Numbers

Here we have provided RS Aggarwal Solutions for Class 10 Maths Chapter 1 Real Numbers for the ease of students so that they can prepare better for their exams.

Question 1.

Solution:

Question 2.

Solution:

Using Euclid’s division Lemma Dividend = (Divisor x Quotient) + Remainder = (61 x 27) + 32 = 1647 + 32 = 1679 Required number = 1679

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Benefits of RS Aggarwal Solutions for Class 10 Maths Chapter 1

Clear Explanations : Each solution is provided with step-by-step explanations, making it easier for students to understand complex concepts and solve problems effectively.
Comprehensive Coverage : The solutions cover all topics in the chapter, including the Euclidean Division Algorithm, the Fundamental Theorem of Arithmetic, and methods to find HCF and LCM, ensuring thorough preparation.
Problem-Solving Skills : By working through these solutions, students can improve their problem-solving skills and learn how to approach different types of questions confidently.
Exam Preparation : The solutions help students prepare for exams by providing practice with a variety of problems similar to those they might encounter in their tests.
Foundation Building : These solutions help build a strong foundation in mathematics, which is essential for understanding more advanced topics in higher classes.
Self-Learning : The detailed explanations and step-by-step approach enable students to study and learn on their own, promoting independent learning.
Error Reduction : By following these solutions, students can identify and correct their mistakes, leading to a better understanding of the subject and improved accuracy.