RD Sharma Class 6 ex 2.6 Solutions Chapter 2 Playing With Numbers

In this chapter, we provide RD Sharma Class 6 ex 2.6 Solutions Chapter 2 Playing With Numbers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 2.6 Solutions Chapter 2 Playing With Numbers Maths pdf, Now you will get step by step solution to each question.

Chapter 2: Playing With Numbers – Exercise 2.6

Question: 1

Find the H.C. F of the following numbers using prime factors using prime factorization method:

Solution:

(i) 144 and 198

Prime factorization of 144 = 2 × 2 × 2 × 3 × 3

Prime factorization of 198 = 2 × 3 × 3 ×11

Therefore, HCF = 2 × 2 × 3 = 18

(ii) 81 and 117

Prime factorization of 81 = 3 × 3 × 3 × 3

Prime factorization of 117 = 3 × 3 ×13

Therefore, HCF = 3 × 3 = 9

(iii) 84 and 98

Prime factorization of 84 = 2 × 2 × 3 × 7

Prime factorization of 98 = 2 × 7 × 7

Therefore, HCF = 2 × 7 = 14

(iv) 225 and 450

Prime factorization of 225 = 3 × 3 × 5 × 5

Prime factorization of 198 = 2 × 3 × 3 × 5 × 5

Therefore, HCF = 3 × 3 × 5 × 5 = 225

(v) 170 and 238

Prime factorization of 170 = 2 × 5 × 17

Prime factorization of 238 = 2 × 7 × 17

Therefore, HCF = 2 × 17 = 34

(vi) 504 and 980

Prime factorization of 504 = 2 × 2 × 2 × 3 × 3 × 7

Prime factorization of 980 = 2 × 2 × 5 × 7 × 7

Therefore, HCF = 2 × 2 × 7 = 28

(vii) 150, 140 and 210

Prime factorization of 150 = 2 × 3 × 5 × 5

Prime factorization of 140 = 2 × 2 × 5 × 7

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

Therefore, HCF = 2 × 5 = 10

(viii) 84, 120 and 138

Prime factorization of 84 = 2 × 2 × 3 × 7

Prime factorization of 120 = 2 × 2 × 2 × 3 × 5

Prime factorization of 138 = 2 × 3 × 23

Therefore, HCF = 2 × 3 = 6

(i×) 106, 159 and 265

Prime factorization of 106 = 2 × 53

Prime factorization of 159 = 2 × 53

Prime factorization of 265 = 5 × 53

Therefore, HCF = 53

Question: 2

What is the H.C.F of two consecutive?

Solution:

(i) The common factor of two consecutive numbers is always 1.

Therefore, HCF of two consecutive numbers = 1

(ii) The common factors of two consecutive even numbers are 1 and 2.

Therefore, HCF of two consecutive even numbers = 2

(iii) The common factor of two consecutive odd numbers is 1.

Therefore, HCF of two consecutive odd numbers = 1

Question: 3

H.C.F of co-primes numbers 4 and 15 was found as follows:

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, it is not correct.

We know that HCF of two co-prime number is 1.

4 and 15 are co-prime numbers because the only factor common to them is 1.

Thus, HCF of 4 and 15 is 1.

All Chapter RD Sharma Solutions For Class 6 Maths

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good