NCERT Class 6 Math Exercise 3.6 Solution of Chapter 3 Playing with Numbers with explanation. Here we provide Class 6 Maths all Chapters in Hindi for cbse, HBSE, Mp Board, UP Board and some other boards.

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NCERT Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.6 Solution in english Medium.

**Class 6 Maths Chapter 3 Exercise 3.6 Solution**

**Question 1.**

**Find the HCF of te 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, 75**

**I. 18, 54, 81**

**J. 12, 45, 75**

**Answer.**

A.

Factors of 18 = 2x3x3

Factors of 48 = 2x2x2x2x3

HCF ( 18, 48) = 2×3 = 6

B.

Factors of 30 = 2x3x5

Factors of 42 = 2x3x7

HCF (30, 42) = 2×3 = 6

C.

Factors of 18 = 2x3x3

Factors of 60 = 2x2x3x5

HCF (18, 60) = 2×3 = 6

D.

Factors of 27 = 3x3x3

Factors of 63 = 3x3x7

HCF (27, 63) = 3×3 = 9

E.

Factors of 36 = 2x2x3x3

Factors of 84 = 2x2x3x7

HCF (36, 84) = 2x2x3 = 12

F.

Factors of 34 = 2×17

Factors of 102 = 2x3x17

HCF (34, 102) = 2X17 = 34

G.

Factors of 70 = 2x5x7

Factors of 105 = 3x5x7

Factors of 175 = 5x5x7

HCF (70, 105, 175) = 5X7 = 35

H.

Factors of 91 = 7×13

Factors of 112 = 2x2x2x2x7

Factors of 49 = 7×7

HCF (91, 112, 49) = 7

I.

Factors of 18 = 2x3x3

Factors of 54 = 2x3x3x3

Factors of 81 = 3x3x3x3

HCF (18, 54, 81) = 3×3 = 9

J.

Factors of 12 = 2x2x3

Factors of 45 = 3x3x5

Factors of 75 = 3x5x5

HCF (12, 45, 75) = 3

**Question 2.**

**What is the HCF of two consecutive :**

**A. numbers?**

**B. even numbers?**

**C. odd numbers?**

**Answer.**

A. HCF of two consecutive numbers is 1.

B. HCF of two consecutive even numbers is 2.

C. HCF of two consecutive numbers is 1.

**Question 3.**

**HCF of co-prime numbers 4 and 15 was found as follows by factorisation:**

**4 = 2×2 and 15 = 3×5 since there is no common prime factor m so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?**

**Answer.**

No.

The correct HCF is 1.