RD Sharma Class 6 Maths Chapter 2 Exercise 2.6

RD Sharma Solutions for Class 6 Maths Chapter 2 Exercise 2.6: Exercise 2.6 of RD Sharma Class 6 Maths focuses on the concept of prime factorization using the division method. This topic is important as it forms the base for HCF and LCM, which are frequently tested in school exams. 

Understanding this exercise helps students align with the Class 6 Maths syllabus and prepares them well for the exam pattern followed in most CBSE schools.

Practicing these problems also aids in solving similar questions asked in previous year papers, ensuring a strong foundation in number theory and arithmetic reasoning.

What is Covered in  Class 6 Chapter 2 Playing with Numbers Exercise 2.6?

Below, we have provided the topics that students will learn in this exercise -

•  Prime Factorization using Division Method

•  Step-by-step breakdown of numbers into prime factors

•  Identification of prime and composite numbers

•  Practice on expressing numbers as a product of prime numbers

•  Building understanding to find HCF and LCM later

•  Reinforcement of divisibility rules indirectly

•  Foundation for solving advanced number system problems in higher classes

RD Sharma Solutions for Class 6 Chapter 2 Playing with Numbers Exercise 2.6

RD Sharma Solutions for Class 6 Chapter 2 Playing with Numbers Exercise 2.6 provide step-by-step explanations for prime factorization using the division method.

These solutions follow the latest CBSE syllabus and exam pattern, helping students build a strong foundation and prepare effectively for tests and previous year exam questions.

1. Find the HCF of the following numbers using prime factorization method:

(i) 144, 198

(ii) 81, 117

(iii) 84, 98

(iv) 225, 450

(v) 170, 238

(vi) 504, 980

(vii) 150, 140, 210

(viii) 84, 120, 138

(ix) 106, 159, 265

Solution:

(i) 144, 198

We know that the prime factorization of 144 = 2 × 2 × 2 × 2 × 3 × 3

The same way prime factorization of 198 = 2 × 3 × 3 × 11

Hence, HCF of 144, 198 is 2 × 3 × 3 = 18

(ii) 81, 117

We know that prime factorization of 81 = 3 × 3 × 3 × 3

The same way prime factorization of 117 = 3 × 3 × 13

Hence, HCF of 81, 117 = 3 × 3 = 9

(iii) 84, 98

We know that prime factorization of 84 = 2 × 2 × 3 × 7

The same way prime factorization of 98 = 2 × 7 × 7

Hence, HCF of 84, 98 = 2 × 7 = 14

(iv) 225, 450

We know that prime factorization of 225 = 3 × 3 × 5 × 5

The same way prime factorization of 450 = 2 × 3 × 3 × 5 × 5

Hence, HCF of 225, 450 = 3 × 3 × 5 × 5 = 225

(v) 170, 238

We know that prime factorization of 170 = 2 × 5 × 17

The same way prime factorization of 238 = 2 × 7 × 17

Hence, HCF of 170, 238 = 2 × 17 = 34

(vi) 504, 980

We know that the prime factorization of 504 = 2 × 2 × 2 × 3 × 3 × 7

The same way prime factorization of 980 = 2 × 2 × 5 × 7 × 7

Hence, HCF of 504, 980 = 2 × 2 × 7 = 28

(vii) 150, 140, 210

We know that prime factorization of 150 = 2 × 3 × 5 × 5

The same way prime factorization of 140 = 2 × 2 × 5 × 7

Prime factorization of 210 = 2 × 3 × 5 × 7

Hence, HCF of 150, 140, 210 = 2 × 5 = 10

(viii) 84, 120, 138

We know that prime factorization of 84 = 2 × 2 × 3 × 7

The same way prime factorization of 120 = 2 × 2 × 2 × 3 × 5

Prime factorization of 138 = 2 × 3 × 23

Hence, HCF of 84, 120, 138 = 2 × 3 = 6

(ix) 106, 159, 265

We know that prime factorization of 106 = 2 × 53

The same way prime factorization of 159 = 3 × 53

Prime factorization of 265 = 5 × 53

Hence, HCF of 106, 159, 265 = 53

2. What is the HCF of two consecutive

(i) Numbers

(ii) even numbers

(iii) odd numbers

Solution:

(i) We know that the common factor of two consecutive numbers is 1.

Hence, HCF of two consecutive numbers is 1.

(ii) We know that the common factors of two consecutive even numbers are 1 and 2.

Hence, HCF of two consecutive even numbers is 2.

(iii) We know that the common factors of two consecutive odd numbers is 1.

Hence, HCF of two consecutive odd numbers is 1.

3. HCF of co-prime numbers 4 and 15 was found as follows:

4 = 2 × 2 and 15 = 3 × 5

3. Since there is no common prime factor. So, HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Solution:

No. It is not correct.

The HCF of two co-prime numbers is 1.

We know that 4 and 15 are co-prime numbers having common factor 1.

Therefore, HCF of 4 and 15 is 1.

RD Sharma Solutions for Class 6 Maths Chapter 2 Exercise 2.6 PDF Download

If you're looking for clear and accurate solutions to Exercise 2.6 of Chapter 2 Playing with Numbers from the RD Sharma Class 6 Maths book, you're in the right place. These solutions follow the latest  CBSE Class 6 syllabus and match the exam pattern followed in schools.

Practicing with these solutions helps in better understanding of prime factorization and prepares students for questions asked in previous year papers. Download the PDF below to access the complete solutions.

Key Features of Using RD Sharma Class 6 Maths Chapter 2 Exercise 2.6

Below we have provided key features of using these solutions - 

Step-by-step Solutions: Clear, detailed steps help students understand the logic behind prime factorization.

Concept Clarity: Strengthens understanding of prime and composite numbers through practical application.

Exam-oriented Approach: Aligned with the latest  CBSE syllabus and exam pattern, ideal for test preparation.

Helpful for Previous Year Questions: Prepares students for similar questions seen in previous year papers.

Foundation for Higher Concepts: Builds a strong base for topics like HCF, LCM, and number theory in higher classes.

Practice and Revision Friendly: Excellent for self-study, revision, and daily practice.