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

Class: 6 Mathematics
Chapter: 3Playing with Numbers
Exercise: 3.6NCERT Solution in Hindi and English

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Class 6 Maths Chapter 3 Exercise 3.6 Solution in Videos

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




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

On which topic questions of exercise 3.6 of class 6th Maths are based?

Questions of exercise 3.6 of class 6th Maths are based on the topic named Highest Common Factor (HCF). The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors. It is also known as Greatest Common Divisor (GCD).

How do students score full marks in exercise 3.6 of class 6 Maths?

Exercise 3.6 of class 6th Maths is very important for the exams. In exercise 3.6, there are 3 questions. The first question has 10 parts and the second question has 3 parts. All questions of this exercise are important. Students should practice all questions of this exercise to score full marks in exercise 3.6 of grade 6th Maths.

Is exercise 3.6 of class 6th Maths hard to solve and understand?

Exercise 3.6 of class 6th Maths is not at all hard to solve and understand. Exercise 3.6 is the easiest exercise. Only three questions are there in this exercise and all questions of this exercise are very easy. Questions of this exercise are straightforward. Students enjoy doing this exercise.

How much time students needed to complete exercise 3.6 of class 6th Maths if students give 1 hour per day to this exercise?

If students give 1 hour per day to exercise 3.6 of grade 6th Maths, then they need only 2 days to complete exercise 3.6 of class 6th Maths. The number of days can vary because no students have the same working speed, capability, etc.

NCERT Solutions for Class 6 Maths Chapter 3 exercise 3.6 in english
Class 6 Maths Chapter 3 Exercise 3.6 in Hindi