# NCERT Solutions for Class 6 Maths Chapter 3

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.1, Exercise 3.2, Exercise 3.3, Exercise 3.4, Exercise 3.5, Exercise 3.6 and Exercise 3.7 in English and Hindi Medium for session 2020-2021.

Solutions of Prashnavali 3.1, Prashnavali 3.2, Prashnavali 3.3, Prashnavali 3.4, Prashnavali 3.5, Prashnavali 3.6 and Prashnavali 3.7 in Hindi Medium in PDF file format as well as study online without downloading. Video Format Solution with explanation of 6 Maths Chapter 3 Playing With Numbers is also available in English Medium free to download. Download Class 6 Maths App for Offline use.

## NCERT Solutions for Class 6 Maths Chapter 3

Class: 6 | Maths (English and Hindi Medium) |

Chapter 3: | Playing With Numbers |

### Class 6 Maths Chapter 3 All Exercises Solution

- NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.1
- NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.2
- NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.3
- NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.4
- NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.5
- NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.6
- NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7

### Download Class 6 Maths Chapter 3 Solutions in PDF Format

NCERT Solutions for Class 6 Maths Chapter 3 in Hindi Medium and English Medium are given below, free to use. All the solutions are updated for new academic session 2020-2021 based on latest NCERT Books 2020-21.

### Class 6 Maths Chapter 3 Solutions in English Medium

### Class 6 Maths Chapter 3 Solutions in Hindi Medium

#### Class 6 Maths Chapter 3 Exercise 3.1 Solution and Explanation

#### Class 6 Maths Chapter 3 Exercise 3.2 Solution and Explanation

#### Class 6 Maths Chapter 3 Exercise 3.3 Solution and Explanation

#### Class 6 Maths Chapter 3 Exercise 3.4 Solution and Explanation

#### Class 6 Maths Chapter 3 Exercise 3.5 Solution and Explanation

#### Class 6 Maths Chapter 3 Exercise 3.6 Solution and Explanation

#### Class 6 Maths Chapter 3 Exercise 3.7 Solution and Explanation

##### About 6 Maths Chapter 3 Solutions

In 6 Maths Chapter 3 Playing With Numbers, we will study about the factors and multiples of numbers. A factor of a number is an exact divisor of that number. The following the some facts about a factor:

1 is a factor of every number.

Every number is a factor of itself.

Every factor of a number is an exact divisor of that number.

Every factor is less than or equal to the given number.

Number of factors of a given number are finite.

Following facts are related to a multiple of a number:

Every multiple of a number is greater than or equal to that number.

The number of multiples of a given number is infinite.

Every number is a multiple of itself.

Perfect Number: A number for which sum of all its factors is equal to twice the number is called a perfect number. The numbers 6 are perfect numbers as 6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14.

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### Important Questions on 6 Maths Chapter 3

= 1 x 24

= 2 x 12

= 3 x 8

= 4 x 6

= 6 x 4

Factors of 24 = 1, 2, 3, 4, 6, 12, 24

28 के गुणनखंड = 1, 2, 4, 7, 14, 28

सार्व गुणनखंड = 1, 2, 4

5 x 2 = 10,

5 x 3 = 15,

5 x 4 = 20,

5 x 5 = 25

First five multiples of 5 are 5, 10, 15, 20, 25.

9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

उदाहरण:

1 + 3 = 4,

3 + 5 = 8

37 और 73;

79 और 97

Sum of the digits at odd places = 4 + 5 = 9

Sum of the digits at even places = 4 + 5 = 9

Difference of both sums = 9 – 9 = 0

Since the difference is 0, therefore, the number is divisible by 11.

2, 3, 5, 7, 11, 13, 17, 19

भाज्य संख्याएँ:

4, 6, 8, 9, 10, 12, 14, 15, 16, 18

The prime factors of 9999 are 3 × 3 × 11 × 101.

The prime factors of 10000 are 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.

Example:

(i) 2 x 3 x 4 = 24

(ii) 4 x 5 x 6 = 120

5 और 7;

11 और 13

7 + 13 = 20;

3 + 17 = 20;

2 + 13 = 15;

5 + 5 = 10

Factors of 75 = 3 x 5 x 5

Factors of 69 = 3 x 69

H.C.F. = 3

Therefore the required weight is 3 kg.

5 + 7 = 12 and 12 is divisible by 4.

7 + 9 = 16 and 16 is divisible by 4.

9 + 11 = 20 and 20 is divisible by 4.

Hence, the required number is 2 x 3 x 5 x 7 = 210

Factors of 825 = 3 x 5 x 5 x 11

Factors of 675 = 3 x 5 x 5 x 3 x 3

Factors of 450 = 2 x 3 x 3 x 5 x 5

H.C.F. = 3 x 5 x 5 = 75 cm

Therefore, the longest tape is 75 cm.