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Interactive video lesson plan for: NCERT Solutions for Class 10 Maths Real Numbers Ex 1.2 | CBSE Class 10 Maths Solutions Real Numbers

Activity overview:

NCERT Solutions for Class 10 Maths Real Numbers Ex 1.2 | CBSE Class 10 Maths Solutions Real Numbers

http://www.learncbse.in/ncert-class-10-math-solutions/
http://www.learncbse.in/ncert-solutions-class-10th-maths-chapter-1-real-numbers/
http://www.learncbse.in/rd-sharma-class-10-solutions-chapter-1-real-numbers/

In this video, how to find the prime factors of given numbers using factor tree is explained.LCM and HCF of two numbers is obtained by prime factorization of numbers.Basic Algebra is included.That is Find the LCM and HCF of two numbers by the prime factorisation method.

HCF is the Product of the smallest power of each common prime factor in the numbers.
LCM is the Product of the greatest power of each prime factor, involved in the numbers.

Fundamental Theorem of Arithmetic: Every composite number can be expressed or factorised as a product of primes, and this factorisation is prime factorisation.
The prime factorisation of a natural number is unique, except for the order of its factors.

00:02 CBSE class 10 maths solutions Chapter 1 Real Numbers Ex 1.2 Q1. Express each number as a product of its prime factors:
00:10 (iii) 3825 (v) 7429
"Prime factorization " is finding which prime numbers multiply together to make the original number.
00:48 Factor tree.
01:29 (iv) 5005
02:00 Factor tree, Product of prime factors
02:25 (v) 7429

03:10 CBSE class 10 maths solutions Chapter 1 Real Numbers Ex 1.2 Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF =
product of the two numbers.
03:25 (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
03:34 Product of prime factors using factor tree.
04:05 Obtain HCF
04:40 Take LCM
05:46 Verification:HCF (a , b ) × LCM ( a , b ) = a × b

06:44 (ii) 510 and 92
10:15 (iii) 336 and 54

14:33 CBSE class 10 maths solutions Chapter 1 Real Numbers Ex 1.2 Q3. Find the LCM and HCF of the following integers by applying the prime factorisation
method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
14:59 Write prime factor of 12, 15 and 21.
15:53 HCF of 12, 15 and 21 = Product of the smallest power of each common prime factor in the numbers
LCM of 12, 15 and 21 = Product of the greatest power of each prime factor, involved in the numbers .
17:27 Find the LCM and HCF of 17, 23 and 29
19:38 Find the LCM and HCF of 8, 9 and 25

21:51 CBSE class 10 maths solutions Chapter 1 Real Numbers Ex 1.2 Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
22:31 HCF × LCM = Product of two numbers

23:55 CBSE class 10 maths solutions Chapter 1 Real Numbers Ex 1.2 Q5. Check whether 6^n can end with the digit 0 for any natural number n .
23:59 take two which ends with zero
24:14 Factorization of two numbers
25:26 observe the results and make a conclusion that given statement is false

CBSE class 10 maths solutions Chapter 1 Real Numbers Ex 1.2 Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
26:32 defination of composite number
composite number has atleast one factor other than 1 and the number itself.


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learncbse.in
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CBSE class 10 maths NCERT Solutions Chapter 1 Real Numbers
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NCERT solutions for class 10 maths Real Numbers
NCERT solutions for CBSE class 10 maths Real Numbers Ex 1.1
CBSE class 10 maths Solutions Real Numbers Exercise 1.4
Common core Math Real Numbers
ICSE Class 10 Maths

Tagged under: ncert solutions,learncbse.,gyanpub,NCERT solutions,Math Real Numbers,Real Numbers Exercise 1.2,composite number,HCF × LCM,Product numbers,Product prime factors,prime factorisation,prime factors,Verification:HCF,Fundamental Theorem,LCM,HCF

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