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Edited By Ramraj Saini | Updated on Nov 01, 2023 02:54 PM IST | #CBSE Class 10th

**NCERT solutions for class 10 maths chapter 1 exercise 1.2** are discussed here that particularly deals with Fundamental Theorem of Arithmetic. The NCERT solutions for class 10 maths ex 1.2 include seven questions which are easy-to-solve questions that also cover the basics of HCF and LCM, Real Numbers, factorization, the prime numbers, real numbers, and many more.

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- NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers (Ex-1.2)
- Download PDF of NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers
- Access Exercise 1.2 Class 10 Maths Answers
- More About NCERT 10th Class Maths Exercise 1.2 Answers
- Benefits Of 10th Class Maths Exercise 1.2 Answers
- NCERT Solutions of Class 10 Subject Wise
- NCERT Solutions for Class 10 Maths
- Subject Wise NCERT Exemplar Solutions
- NCERT Exemplar Class 10 Maths

NCERT solutions for exercises are created by expert team at Careers360 considering the latest syllabus and pattern of CBSE 2023-24, covering step by step, comprehensive solutions of every problem. Students can find list of exercised below and practice problems to command the concepts.

**Real Numbers Class 10**** Chapter 1 - ****Exercise 1.2**

**Q1 (1) ** Express each number as a product of its prime factors: 140

** Answer: **

The number can be as a product of its prime factors as follows

** Q 1 (2) ** Express each number as a product of its prime factors: 156

** Answer: **

The given number can be expressed as follows

** Q1 (3) ** Express each number as a product of its prime factors: 3825

** Answer: **

The number is expressed as the product of the prime factors as follows

** Q1 (4) ** Express each number as a product of its prime factors: 5005

** Answer: **

The given number can be expressed as the product of its prime factors as follows.

** Q1 (5) ** Express each number as a product of its prime factors: 7429

** Answer: **

The given number can be expressed as the product of their prime factors as follows

** Answer: **

26 = 2 x 13

91 = 7 x 13

HCF(26,91) = 13

LCM(26,91) = 2 x 7 x 13 = 182

HCF x LCM = 13 x 182 = 2366

26 x 91 = 2366

26 x 91 = HCF x LCM

Hence Verified

** Answer: **

The number can be expressed as the product of prime factors as

510 = 2 x 3 x 5 x 17

92 = 2 ^{ 2 } x 23

HCF(510,92) = 2

LCM(510,92) = 2 ^{ 2 } x 3 x 5 x 17 x 23 = 23460

HCF x LCM = 2 x 23460 = 46920

510 x 92 = 46920

510 x 92 = HCF x LCM

Hence Verified

** Answer: **

336 is expressed as the product of its prime factor as

336 = 2 ^{ 4 } x 3 x 7

54 is expressed as the product of its prime factor as

54 = 2 x 3 ^{ 3 }

HCF(336,54) = 2 x 3 = 6

LCM(336,54) = 2 ^{ 4 } x 3 ^{ 3 } x 7 = 3024

HCF x LCM = 6 x 3024 = 18144

336 x 54 = 18144

336 x 54 = HCF x LCM

Hence Verified

** Answer: **

The numbers can be written as the product of their prime factors as follows

12 = 2 ^{ 2 } x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 2 ^{ 2 } x 3 x 5 x 7 = 420

** Answer: **

The given numbers are written as the product of their prime factors as follows

17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 17 x 23 x 29 = 11339

** Answer: **

The given numbers are written as the product of their prime factors as follows

8 = 2 ^{ 3 }

9 = 3 ^{ 2 }

25 = 5 ^{ 2 }

HCF = 1

LCM = 2 ^{ 3 } x 3 ^{ 2 } x 5 ^{ 2 } = 1800

** Q4 ** Given that HCF (306, 657) = 9, find LCM (306, 657).

** Answer: **

As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have

HCF (306, 657) x LCM (306, 657) = 306 x 657

** Q5 ** Check whether can end with the digit 0 for any natural number n.

** Answer: **

By prime factorizing we have

6 ^{ n } = 2 ^{ n } x 3 ^{ n }

A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6 ^{ n } we can conclude that for no value of n 6 ^{ n } will end with the digit 0.

** Q6 ** Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite numbers.

Answer:

7 x 11 x 13 + 13

= (7 x 11 + 1) x 13

= 78 x 13

= 2 x 3 x 13 ^{ 2 }

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

= (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5

= 5 x 1008

After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.

** Answer: **

The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.

Time taken by Sonia = 18 = 2 x 3 ^{ 2 }

Time taken by Ravi = 12 = 2 ^{ 2 } x 3

LCM(18,12) = 2 ^{ 2 } x 3 ^{ 2 } = 36

Therefore they would again meet at the starting point after 36 minutes.

This Exercise 1.2 Class 10 Maths gives questions on the factorization of composite numbers. Composite numbers are numbers that have more than two factors. Mathematically, the natural numbers which are not prime numbers come under composite numbers as they can be divided by more than two numbers. This class 10 maths ex 1.2 mainly focused on the concepts of HCF and LCM. Except this ex 1.2 class 10 solutions students can also refer Real Numbers Class 10 Notes to command the concepts in detail.

- Students can use NCERT solutions for Class 10 Maths exercise 1.2 to help them solve and revise all of the exercise problems.
- You will be able to achieve better scores if you thoroughly practise the NCERT solution for Class 10th Maths chapter 1 exercise 1.2.
- Exercise 1.1 Class 10 Maths, is based on the Fundamental Theorem of Arithmetic, as well as HCF and LCM, which are important concepts in the chapter.

**Also see-**

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

1. Is 0 a Composite Number?

Zero is neither prime nor a composite number because it does not have any prime factors.

2. Write the prime factors of 144

The prime factors of 144,

144 = 2 × 2 × 2 × 2 × 3 × 3

Students can practice problems discussed in class 10 ex 1.2 to command these concepts.

3. What is the full form of HCF of LCM in Maths?

The full form of HCF is “Highest Common Factor” and LCM is “Least common multiple”.

4. What are the methods used in finding LCM and HCF?

The methods used in finding LCM and HCF are the prime factorization method and long division method.

5. Find the highest common factor of 104 and 160

104 = 2 × 2 × 2 × 13

160 = 2 × 2 × 2 × 2 × 2 × 5

The common factors of 104 and 160 are 2 × 2 × 2 = 8

Therefore, HCF (104, 160) = 8

6. Find the least common multiple of 21 and 28 .

21=3×7

28=4×7

Now, LCM (21, 28)=7×4×3=84

7. Write the formula which involves both HCF and LCM.

The formula that involves both HCF and LCM is

(HCF of the two numbers) x (LCM of the two numbers)= Product of Two numbers

8. The HCF of two numbers is 7 and their LCM is 84 . If one of the numbers is 21 then find the other number.

Let the other number be a.

We know that,

(HCF of the two numbers) x (LCM of the two numbers)= Product of Two numbers

Now, 7×84=21×a

588=21a

Which on dividing both sides by 21 ,

We get, a=28

The other number is 28 .

9. According to NCERT solutions for Class 10 Maths chapter 1 exercise 1.2 , What is the Fundamental Theorem of Arithmetic ?

Theorem: Every composite number may be factored as a product of primes, which means that any composite number can be expressed as a product of primes

10. What kinds of questions do the NCERT solutions for Class 10 Maths chapter 1 exercise 1.2 cover ?

The questions are based on the concept of the Fundamental Theorem of Arithmetic, in finding HCF and LCM, questions related to the relationship of HCF and LCM.

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6 minHave a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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