# NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.6

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.6 (Ex. 3.6) Playing with Numbers updated for CBSE as well as other state board like UP Board, MP Board. All the solutions are free to use and updated for academic year 2021-22.

All the steps are mentioned during the solution of questions. Questions are explained well and easy to understand.## Class 6 Maths Chapter 3 Exercise 3.6 Solution

### CBSE NCERT Class 6 Maths Chapter 3 Exercise 3.6 Solution in Hindi and English

Class: 6 | Mathematics |

Chapter: 3 | Playing with Numbers |

Exercise: 3.6 | NCERT Solution in Hindi and English |

### Class 6 Maths Chapter 3 Exercise 3.6 Solution in Videos

### Class 6 Maths Exercise 3.6 Important Questions

##### For 48 and 60, verify the property: Product of numbers = Product of their HCF and LCM.

We compute the HCF and LCM of 48 and 60 as under:

HCF of 48, 60

We can factorise 48 as 2 x 2 x 2 x 2 x 3

An d factors of 60 are 2 x 2 x 3 x 5

Here highest common factor is 2 x 2 x 3 =12

LCM of

2 | 48, 60 |

2 | 24, 30 |

2 | 12, 15 |

2 | 6, 15 |

3 | 3, 15 |

1, 5 |

LCM = 2 x 2 x 2 x 2 x 3 x 5 = 240

Now product of numbers = LCM x HCF

48 x 60 = 12 x 240

2880 = 2880 proved

##### The HCF and LCM of two numbers are 23 and 1449 respectively. If one of the numbers is 161, find the other.

Here, HCF = 23, LCM = 1449 and one number = 161.

We know that: One number × The other number = HCF × LCM

Other number = (23 x 1449)/ 161

So required number = 207.

#### Properties of HCF and LCM of Given Numbers

(i) The HCF of given numbers is not greater than any of the given numbers.

(ii) The LCM of given numbers is not less than any of the given numbers.

(iii) The HCF of two co-prime numbers is 1.

(iv) The LCM of two co-prime numbers is equal to their product.

(v) If a number, say a, is a factor of another number, say b, then HCF of a and b is a and their LCM is b.

(vi) The HCF of given numbers is always a factor of their LCM.

For example, the HCF of 8 and 12 is 4. the LCM of 8 and 12 is 24.

Since, 4 is a factor of 24, HCF of 8 and 12 is a factor of their LCM.

(vii) The product of the HCF and the LCM of two numbers is equal to the product of the given numbers, i.e., if a and b are two numbers, then

a × b = HCF × LCM or,

LCM = (a x b)/HCF or, HCF = (a x b)/LCM

### Class 6 Maths Exercise 3.6 Extra Questions with Answer

##### What is the basic property of HCF?

(i) The HCF of given numbers is not greater than any of the given numbers.

(ii) The HCF of two co-prime numbers is 1.

##### What is the basic property of LCM?

(i) The LCM of given numbers is not less than any of the given numbers.

(ii) The LCM of two co-prime numbers is equal to their product.

##### Find the HCF and LCM of the following pairs of numbers: (i) 57, 6720

HCF of 57, 6720

Factors of 57 are 3 x 19

And factors of 6720 are 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3 x 7

Hence, HCF of 57, 6720 is 3

Now LCM of 57, 6720 is

3 | 57, 6720 |

2 | 19, 2240 |

2 | 19, 1120 |

2 | 19, 524 |

2 | 19, 262 |

2 | 19, 132 |

2 | 19, 66 |

3 | 19, 33 |

19, 11 |

LCM 3 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 11 x 19 = 120384