Class 6 maths Exercise 3.6 solutions

NCERT mathematics class 6 chapter 3 Playing with Numbers exercise 3.6 solutions are given. You should study the textbook lesson Playing with Numbers very well. You should also observe and practice all example problems and solutions given in the textbook. You can observe the solutions given below and try them in your own method. NCERT maths class 6 solutions Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 Exercise 3.6 Exercise 3.7 CA foundation maths solutions 

Chapter 3 exercise 3.6 solutions Playing with Numbers class 6 Maths NCERT

CBSE class 6 maths solutions  Chapter 3 Playing with Numbers Exercise 3.6

NCERT chapter 3 class 6 maths exercise 3.6 Playing with Numbers

6th class maths solutions  Problem 1 1. Find the H C F of the following numbers: a. 18, 48,      b. 30, 42,     c. 18, 60 d. 27, 63,      e. 36, 84,      f. 34, 102 g. 70, 105, 175,                  h. 91, 112, 49 i. 18, 54, 81,                       j. 12, 45, 75 Solutions: a. 18, 48 The HCF of 18, 48 can be found by prime factorization. The factors of 18 18 = 2 × 9 18 = 2 × 3 ×3 The factors of 48 48 = 2 × 24 48 = 2 × 2 × 12 48 = 2 × 2 × 2 × 6 48 = 2 × 2 × 2 × 2 × 3 The common factor of 18 and 48 is 2 x 3. Thus, HCF of 18 and 48 is 6. b. 30, 42 The HCF of 30, 42 can be found by prime factorization. The factors of 30 30 = 2 × 15 30 = 2 × 3 ×5 The factors of 42 42 = 2 × 21 42 = 2 × 3 × 7 The common factor of 30 and 42 is 2 x 3. Thus, HCF of 30 and 42 is 6. c. 18, 6o The HCF of 18, 60 can be found by prime factorization. The factors of 18 18 = 2 × 9 18 = 2 × 3 ×3 The factors of 60 60 = 2 × 30 48 = 2 × 2 × 15 48 = 2 × 2 × 3 × 5 The common factor of 18 and 60 is 2 x 3. Thus, HCF of 18 and 60 is 6. d. 27, 63 The HCF of 27, 63 can be found by prime factorization. The factors of 27 27 = 3 × 9 18 = 3 × 3 ×3 The factors of 63 63 = 3 × 21 63 = 3 × 3 × 7 The common factor of 27 and 63 is 3 x 3. Thus, HCF of 27 and 63 is 9. e. 36, 84 The HCF of 36, 84 can be found by prime factorization. The factors of 36 36 = 2 × 18 36 = 2 × 2 × 9 36 = 2 × 2 × 3 × 3 The factors of 84 84 = 2 × 42 84 = 2 × 2 × 21 84 = 2 × 2 × 3 × 7 The common factor of 36 and 84 is 2 x 3 × 3. Thus, HCF of 36 and 84 is 12. f. 34, 102 The HCF of 34, 102 can be found by prime factorization. The factors of 34 34 = 2 × 17 The factors of 102 102 = 2 × 51 102 = 2 × 3 × 17 The common factor of 34 and 102 is 2 x 17. Thus, HCF of 34 and 102 is 34. g. 70, 105, 175 The HCF of 70, 105 and 175 can be found by prime factorization. The factors of 70 70 = 2 × 35 70 = 2 × 5 ×7 The factors of 105 105 = 3 × 35 105 = 3 × 5 × 7 The factors of 175 175 = 5 × 35 175 = 5 × 5 × 7 The common factor of 70, 105 and 175 is 5 × 7. Thus, HCF of 70, 105 and 175 is 35. h. 91, 112, 49 The HCF of 91, 112 and 49 can be found by prime factorization. The factors of 91 91 = 7 × 13 The factors of 112 112 = 2 × 56 112 = 2 × 2 × 28 112 = 2 × 2 × 2 × 14 112 = 2 × 2 × 2 × 2 × 7 The factors of 49 49 = 7 × 7 The common factor of 91, 112 and 49 is 7. Thus, HCF of 91, 112 and 49 is 7. i. 18, 54, 81 The HCF of 18, 54 and 81 can be found by prime factorization. The factors of 18 18 = 2 × 9 18 = 2 × 3 ×3 The factors of 54 54 = 2 × 27 54 = 2 × 3 × 9 54 = 2 × 3 × 3 × 3 The common factor of 18, 54 and 81 is 3 x 3. Thus, HCF of 18, 54 and 81 is 9. j. 12, 45, 75 The HCF of 12, 45 and 75 can be found by prime factorization. The factors of 12 12 = 2 × 6 12 = 2 × 2 ×3 The factors of 45 45 = 3 × 15 45 = 3 × 3 × 5 The factor of 75 75 = 3 × 25 75 = 3 × 5 × 5 The common factor of 12, 45 and 75 is 3. Thus, HCF of 12, 45 and 75 is 3. Problem 2 2. What is the HCF of two consecutives a. numbers?        b. even numbers?        c. odd numbers.? Solutions: a. Numbers The two consecutive numbers be 2 and 3 The factors of 2 = 1, 2 The factors of 3 = 1, 3 The common factor = 1 Therefore, the H C F of two consecutive numbers be 1. b. Even numbers The two consecutive even numbers be 4 and 6 The factors of 4 =1, 2 ,4 The factors of 6 = 1, 2, 3, 6 The common factor = 1 × 2 =1 Therefore, the H C F of two consecutive even numbers be 2. c. odd numbers The two consecutive odd numbers be 5 and 7. The factors of 5 = 1, 5 The factors of 7 = 1, 7 The common factor = 1 Therefore, the HCF of two consecutive odd numbers be 1. Problem 3 3. HCF of co-prime numbers 4 and 15 was found as follows by factorization: 4 = 2 × 2 and 15 = 3 × 5 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: It is not the correct answer. 4 = 1 × 2 × 2 15 = 1 × 3 × 5 The common factor is 1. The HCF of the co -prime numbers be 1. Note: Observe the solutions and try them in your own method. Inter maths 1A solutions SSC maths class 10 solutions NCERT maths class 7 solutions NIOS maths 311 book 2 solutions for some chapters