# Mathematics (RD Sharma Solution) - Class 6 / Grade 6

- Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.7)
### Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.7)

Tags: RD Sharma Class 6 Math Solution Chapter 2 Playing with Numbers Exercise 2.7, Solution of RD Sharma for class 6, RD Sharma solutions Grade VI, Solution of RD Sharma exercises for sixth standard, R.D. Sharma Mathematics solutions and practice pages, Playing With Numbers Exercise 2.7 (Continued Division Method - Euclid's Algorithm)Playing With Numbers

Exercise 2.7 (Continued Division Method – Euclid’s Algorithm)

Question 1 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

Clearly, the last divisor is 150.

Hence the H.C.F of the given number is 150.

Clearly, the last divisor is 19.

Hence the H.C.F of the given number is 19.

Clearly, the last divisor is 95.

Hence the H.C.F of the given number is 95.

Question 2 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 59, 97

Factors of 59 - 1 and 59

Factors of 97 - 1 and 97

Common Factor = 1

Therefore, H.C.F is 1. So, 59 and 97 are co-prime.

ii. 875, 1859

Factors of 875 – 1, 5, 7, 25, 35, 125, 175, 875

Factors of 1859 – 1, 1859

Common Factor = 1

Therefore, H.C.F is 1. So, 875 and 1859 are co-prime.

iii. 288, 1375

Factors of 288 – 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 72, 96, 144, 288.

Factors of 1375 – 1,5,11,25,55,125,275,1375

Common Factor = 1

Therefore, H.C.F is 1. So, 288 and 1375 are co-prime.

Question 3 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

H.C.F of two consecutive numbers is 1.

__Example:__3,4Factors of 3 – 1, 3

Factors of 4 – 1, 2, 4

Common Factor – 1

Therefore, H.C.F = 1

Question 4 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. True

ii. False

iii. True

iv. True

v. False