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NCERT Solutions for Exercise 1.2 Class 10 Maths Chapter 1 - Real Numbers

NCERT Solutions for Exercise 1.2 Class 10 Maths Chapter 1 - Real Numbers

Edited By Ramraj Saini | Updated on Nov 01, 2023 02:54 PM IST | #CBSE Class 10th
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NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers (Ex-1.2)

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.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers (Ex-1.2)
  2. Download PDF of NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers
  3. Access Exercise 1.2 Class 10 Maths Answers
  4. More About NCERT 10th Class Maths Exercise 1.2 Answers
  5. Benefits Of 10th Class Maths Exercise 1.2 Answers
  6. NCERT Solutions of Class 10 Subject Wise
  7. NCERT Solutions for Class 10 Maths
  8. Subject Wise NCERT Exemplar Solutions
  9. NCERT Exemplar Class 10 Maths
NCERT Solutions for Exercise 1.2 Class 10 Maths Chapter 1 - Real Numbers
NCERT Solutions for Exercise 1.2 Class 10 Maths Chapter 1 - Real Numbers

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.

Download PDF of NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers

Download PDF

Access Exercise 1.2 Class 10 Maths Answers

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

\\140=2\times 2\times 5\times 7\\ 140=2^{2}\times 5\times 7

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

Answer:

The given number can be expressed as follows

\\156=2\times 2\times 3\times 13\\ 156=2^{2}\times 3\times 13

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

\\3825=3\times 3\times 5\times 5\times 17\\ 3825=3^{2}\times 5^{2}\times 17

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.

5005=5\times 7\times 11\times 13

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

7429=17\times 19\times 23

Q2 (1) Find the LCM and HCF of the following pairs of integers and verify that LCM \times HCF = product of the two numbers: 26 and 91

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

Q2 (2) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 510 and 92

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

Q2 (3) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 336 and 54

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

Q3 (1) Find the LCM and HCF of the following integers by applying the prime factorization method. 12, 15 and 21

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

Q3 (2) Find the LCM and HCF of the following integers by applying the prime factorisation method. 17, 23 and 29

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

Q3 (3) Find the LCM and HCF of the following integers by applying the prime factorization method. 8, 9 and 25

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

\\ LCM (306, 657) = \frac{306\times 657}{HCF (306, 657) }\\ LCM (306, 657) = \frac{306\times 657}{9}\\ LCM (306, 657) =22338

Q5 Check whether 6 ^n 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 \times 11 \times 13 + 13 and 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 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.

Q7 There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

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.

More About NCERT 10th Class Maths Exercise 1.2 Answers

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.

Benefits Of 10th Class Maths Exercise 1.2 Answers

  • 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.
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Frequently Asked Questions (FAQs)

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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