We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 1
Chapter 1 - Real Numbers Exercise Ex. 1.2
Solution 1
Concept Insight: Since the given number needs to be expressed as the product of prime factors so in order to solve this problem knowing prime numbers is required. Do not forget to put the exponent in case a prime number is repeating.
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 2
Solution 2
Concept Insight: HCF is the product of common prime factors raised to least power, while LCM is product of prime factors raised to highest power. HCF is always a factor of the LCM.
Do not skip verification product of two numbers = HCF x LCM as it can help in cross checking the answer.
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 3
Solution 3
Concept Insight: HCF is the product of common prime factors of all three numbers raised to least power, while LCM is product of prime factors of all here raised to highest power. Use the fact that HCF is always a factor of the LCM to verify the answer. Note HCF of (a,b,c) can also be calculated by taking two numbers at a time i.e HCF (a,b) and then HCF (b,c) .
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 4
Solution 4
Concept Insight: This problem must be solved using product of two numbers = HCF x LCM rather then prime factorisation
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 5
Solution 5
If any number ends with the digit 0, it
should be divisible by 10 or in other words its prime factorisation must
include primes 2 and 5 both
Prime factorisation of 6n = (2 x 3)n
By Fundamental Theorem of Arithmetic Prime factorisation of a number is unique. So 5 is not a prime factor of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
Concept Insight: In order solve such problems the concept used is if a number is to end with zero then it must be divisible by 10 and the prime factorisation of a number is unique.
Prime factorisation of 6n = (2 x 3)n
By Fundamental Theorem of Arithmetic Prime factorisation of a number is unique. So 5 is not a prime factor of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
Concept Insight: In order solve such problems the concept used is if a number is to end with zero then it must be divisible by 10 and the prime factorisation of a number is unique.
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 6
Solution 6
Numbers are of two types - prime and
composite. Prime numbers has only two factors namely 1 and the number
itself whereas composite numbers have factors other than 1 and itself.
It can be observed that
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x (1008 + 1)
= 5 x 1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.
Concept Insight: Definition of prime numbers and composite numbers is used. Do not miss the reasoning.
It can be observed that
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x (1008 + 1)
= 5 x 1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.
Concept Insight: Definition of prime numbers and composite numbers is used. Do not miss the reasoning.
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 7
Solution 7
It can be observed that Ravi and Sonia does
not take same amount of time Ravi takes lesser time than Sonia for
completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia.
i.e When Sonia completes one round then ravi completes 1.5 rounds. So they will meet first time at the time which is a common multiple of the time taken by them to complete 1 round
i.e LCM of 18 minutes and 12 minutes.
Now
18 = 2 x 3 x 3 = 2 x 32
And, 12 = 2 x 2 x 3 = 22 x 3
LCM of 12 and 18 = product of factors raised to highest exponent =
22 x 32 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.
Concept Insight: In
order to solve the word problems first step is to interpret the problem
and identify what is to be determined. The problem asks for
simultaneous reoccurrence of events so we need to find LCM. The key word
for simultaneous reoccurrence of events is LCM. Do not forget to write
the final answer.
Thank you for reading this blog thank you very much
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 1
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 2
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 3
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 4
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 5
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 6
We the team of subject wise have provided detaiiled solution of class 10 ncert exercise 1.2 so please read this post completly if you have any doubt please comment it down also follow our youtube channel.
Chapter 1 - Real Numbers Exercise Ex. 1.2
here it is solution of question 7
if you not read the previous blog please read it. it is the solution to exercise 1.1.
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