NCERT Solutions Class 10 Maths Chapter 1 Real Numbers: Exercise 1.2

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2 is important to learn the prime factorisation concept and the Fundamental Theorem of Arithmetic. Class 10 Maths chapter 1 exercise 1.2 includes important concepts like the factorisation of composite numbers and their uniqueness. It is important to master this exercise to get a solid base in number theory, which is essential for learning complex topics of maths. Let us discuss these concepts in detail as per the new CBSE class 10 syllabus and examination protocols.

1.0Download NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2

2.0Introduction to the Fundamental Theorem of Arithmetic

The fundamental theorem of Arithmetic states that every composite number can be expressed uniquely as a product of prime numbers. In simple words, every natural number can be uniquely expressed as a product of prime factors. However, the order of these prime factors can be written in different ways. For example, 2 × 3 × 5 × 7 is the same as 3 × 5 × 7 × 2 or any other order. 

3.0Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 - Overview

Students can find the below list of important concepts covered under NCERT Solutions Class 10 Chapter 1 Real Numbers - Exercise 1.2.

Prime Factorisation

Prime factorisation is simply the process of breaking down a given composite number into its prime factors. All these prime factors are then raised to a certain power, which basically shows the number of times a prime is multiplied in the factorisation. Mathematically, prime factorisation can be expressed as: 

Given a composite number, say x, by factorising the number x = p₁, p₂, p₃, …., p_n. Here, p₁, p₂, p₃, …., p_n is written in ascending order or the prime numbers. By combining the same prime factors, we determine the powers of those primes. For example:  

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2³ × 3² × 5 × 7 × 13

Here, we can observe that the LCM and HCF are related to prime factorisation through the fact that we employ the prime factors to find the smallest common multiple (LCM) or the greatest common divisor (HCF) of two or more numbers.

Relation of Numbers and their HCF and LCM

The product of any two numbers, say a and b, is equal to the product of HCF and LCM of these numbers. Mathematically, the equation can be written as: 

HCF x LCM = a x b

Note that the above-mentioned equation holds true for a pair of any two positive natural numbers.

4.0NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2: Detailed Solutions

1. Express each number as product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solution:

(i) 140 = 2² × 5 × 7

(ii) 156 = 2² × 3 × 13

(iii) 3825 = 3² × 5² × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Solution:

(i) 26 and 91

26 = 2 × 13

91 = 7 × 13

Therefore, LCM(26, 91) = 2 × 7 × 13 = 182

HCF(26, 91) = 13

Verification:

LCM × HCF = 182 × 13 = 2366

26 × 91 = 2366

i.e., LCM × HCF = product of two numbers.

(ii) 510 and 92

510 = 2 × 3 × 5 × 17

92 = 2² × 23

LCM(510, 92) = 2² × 3 × 5 × 17 × 23 = 23460

HCF(510, 92) = 2

Verification:

LCM × HCF = 23460 × 2 = 46920

510 × 92 = 46920

i.e., LCM × HCF = product of two numbers.

(iii) 336 and 54

336 = 2⁴ × 3 × 7

54 = 2 × 3³

LCM = 2⁴ × 3³ × 7 = 3024

HCF = 2 × 3 = 6

Verification:

LCM × HCF = 2⁴ × 3³ × 7 × 2 × 3 = 18144

Product of two numbers = 336 × 54 = 18144

i.e., LCM × HCF = product of two numbers.

3. 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

Solution:

(i) 12, 15 and 21

12 = 2² × 3

15 = 3 × 5

21 = 3 × 7

Therefore, HCF(12, 15, 21) = 3

LCM(12, 15, 21) = 2² × 3 × 5 × 7 = 420

(ii) 17, 23, 29

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

LCM = 1 × 17 × 23 × 29 = 11339

HCF = 1

(iii) 8, 9, 25

8 = 2 × 2 × 2

9 = 3 × 3

25 = 5 × 5

LCM = 2³ × 3² × 5² = 1800

HCF = 1

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

Solution:

LCM × HCF = product of two numbers.

LCM(306, 657) = (306 × 657) / HCF(306, 657)

LCM(306, 657) = (306 × 657) / 9 = 22338

5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution:

If the number 6ⁿ, for any natural number n, ends with digit 0, then it would be divisible by 5. That is, the prime factorisation of 6ⁿ would contain the prime number 5. This is not possible because 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ; so the only primes in the factorisation of 6ⁿ are 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6ⁿ. So, there is no natural number n for which 6ⁿ ends with the digit zero.

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

Solution:

(i) Let 7 × 11 × 13 + 13 = (7 × 11 + 1) × 13 = (77 + 1) × 13 = 78 × 13 = 2 × 3 × 13 × 13 = 2 × 3 × 13² is a composite number as powers of prime occur and also having more than two distinct factors.

(ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 4 × 3 × 2 × 1 + 1) × 5 = 1009 × 5

Since, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 can be expressed as a product of primes, other than 1 therefore it is a composite number.

7. 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?

Solution:

LCM of 18 and 12.

18 = 2 × 3²

12 = 2² × 3

LCM(18, 12) = 2² × 3² = 36

Thus, after 36 minutes they will meet again at the starting point.

5.0Revisiting Important Topics in Details

• Prime Factorisation: The method of writing a composite number in terms of primes. This is one of the basics of number theory and an essential part of determining numbers' properties.
• Uniqueness of Factorisation: Uniqueness of prime factorisation refers to the fact that regardless of how you factorise a number, its set of prime factors will remain unchanged as long as order is not taken into consideration.
• Applications of Theorem: The theorem is useful in various practical fields such as the determination of the Highest Common Factor (HCF), least common multiple (LCM), and problem-solving in algebra and cryptography. 

6.0Benefits of Class 10 Maths NCERT Solutions Ch 1 - Real Number Exercise 1.2

• NCERT Solutions for Class 10 Maths Real Numbers (Ex 1.2) serve as a valuable resource for students aiming for conceptual clarity, exam success, and strong mathematical skills.
• Exercise 1.2 focuses on the Fundamental Theorem of Arithmetic, which states that every composite number can be expressed as a product of prime numbers in a unique way.
• The exercise includes problems that require breaking down numbers into prime factors, which enhances analytical thinking. By practicing NCERT Solutions, students develop a strong base for higher-level problems in algebra and number theory.
• The solutions are framed in a simple and precise manner, making learning easy.
• Students can refer to these solutions while practicing on their own to verify their answers.