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3. Playing with Numbers Mathematics class 6 in English Medium ncert book solutions Exercise 3.6
3. Playing with Numbers Exercise 3.6 – Complete NCERT Book Solutions for Class 6 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 3. Playing with Numbers Exercise 3.6 to help you master concepts and score higher.
3. Playing with Numbers Mathematics class 6 in English Medium ncert book solutions Exercise 3.6
NCERT Solutions for Class 6 Mathematics play an important role in helping students understand the concepts of the chapter 3. Playing with Numbers clearly. This chapter includes the topic Exercise 3.6, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 6 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 3. Playing with Numbers is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 3.6. By studying these updated NCERT Solutions for Class 6 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
3. Playing with Numbers
Exercise 3.6
Exercise -3.6
Q1. Find the H.C.F. of the following numbers:
(f) 34, 102
Solutions:
Prime factorization of 34
= 2 × 17
Prime factorization of 102
= 2 × 3 × 17
HCF of 34 and 102 is 2 × 17
= 34
(g) 70, 105, 175
Solution:
Prime factorization of 70
= 2 × 5 × 7
Prime factorization of 105
= 3 × 5 × 7
Prime factorization of 175
= 5 × 5 × 7
H.C.F. (70, 105, 175)
= 5 × 7
= 35
(h) 91, 112, 49
Solution:
Prime factorization of 91
= 7 × 13
Prime factorization of 112
= 2 × 2 × 2 × 2 × 7
Prime factorization of 49
= 7 × 7
H.C.F. (91, 112, 49) = 7
(i) 18, 54, 81
Solution:
Prime factorization of 18
= 2 × 3 × 3
Prime factorization of 54
= 2 × 3 × 3 × 3
Prime factorization of 81
= 3 × 3 × 3 × 3
H.C.F. = 3 × 3 = 9
(j) 12, 45, 75
Solution:
Prime factorization of 12
= 2 × 2 × 3
Prime factorization of 45
= 3 × 3 × 5
Prime factorization of 75
= 3 x 5 x 5
H.C.F. = 1 × 3
= 3
Q2. What is the H.C.F. of two consecutive:
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution:
(a) H.C.F. of two consecutive numbers be 1.
(b) H.C.F. of two consecutive even numbers be 2.
(c) H.C.F. of two consecutive odd numbers be 1.
Q3. H.C.F. 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 H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F.?
Solution: No. The correct H.C.F. is 1.
See other sub-topics of this chapter:
1. Exercise 3.1 2. Exercise 3.2 3. Exercise 3.3 4. Exercise 3.4 5. Exercise 3.5 6. Exercise 3.6 7. Exercise 3.7