NCERT Solutions for Class 10 Maths Chapters 1 Real Number Exercise 1.2

Class 10 Maths Chapters 1 Real Numbers Exercise 1.2:- Chapter 1 of Class 10 Maths, "Real Numbers," explains basic concepts about numbers that are important for understanding maths. One of the key parts of this chapter is Exercise 1.2 Class 10, which focuses on the Euclidean algorithm and how to find the Highest Common Factor (HCF). In this exercise, you will learn how to use the Euclidean division method, which is a simple way to calculate the HCF of two positive numbers. These concepts are very useful for solving tricky maths problems and for understanding real numbers better.

If you're looking for Class 10 Maths Chapter 1 Exercise 1.2 Solutions, these are the detailed and easy-to-follow answers that are in line with CBSE rules. These  class 10 maths chapter 1 exercise 1.2 solutions are perfect for students who want to understand the topic clearly and practise more effectively. With the help of these step-by-step explanations, working through Class 10 Maths Chapter 1 Exercise 1.2 becomes much easier.

Check out the full NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2 below. Whether you're preparing for exams or just practising, these detailed answers for Exercise 1.2 Class 10 will make your learning simple and easy.

NCERT Solutions for Class 10 Maths Chapters 1 Real Number Exercise 1.2

Check out the class 10 maths chapter 1 solutions below:-

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

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer:

(i) 140 = 2 x 2 x 5 x 7 = 22 x 5 x 7

(ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13

(iii) 3825 = 3 x 3 x 5 x 17 = 32 x 52 x x 17

(iv) 5005 = 5 x 7 x 11 x 13

(v) 7429 = 17 x 19 x 23 

Check out: NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER-1 REAL NUMBERS EXERCISE 1.1

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

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer:

(i) 26 and 91

a = 26, b = 91

∴ H.C.F = 13

L.C.M = 2 × 7 × 13

= 14 × 13 = 182

∴ H.C.F × L.C.M = a × b

13 × 182 = 26 × 96

2366 = 2366

(ii) 510 and 92

a = 510, b = 92

(iii) 336 and 54

a = 336, b = 54Read more on Sarthaks.com - https://www.sarthaks.com/661852/find-the-lcm-and-the-following-pairs-integers-and-verify-that-lcm-hcf-product-the-two-numbers

336= 2 x 2 x 2 x 2 x 3 x 7 = 24 x 3 x 7

54 = 2 x 3 x 3 x 3 = 2 x 33

HCF = 2 x 3 = 6

LCM = 24 x 33 x 7= 3024

Product of two numbers 336 and 54 = 336 x 54= 18144

3024 x 6= 18144

Hence, product of two numbers = 18144

Check out: CBSE Class 10th Books

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

Answer:

(i) 12, 15 and 21

12 = 2 × 2 × 3 = 22× 3

15 = 3 × 5

and 21 = 3 × 7

For HCF, we find minimum power of prime factor

H.C.F. = (3)1= 3

For LCM, taking maximum power of prime factors

L.C.M. = 22 × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420

So,H.C.F. = (3)1= 3

and L.C.M. =  420

(ii) 17, 23 and 29

17 = 1 × 17

23 = 1 x 23

29 = 1 x 29

For HCF, common factor is 1

HCF = 1

For LCM taking maximum power of prime factor.

L.C.M. = 1 × 17 × 23 × 29 = 11339

So H.C.F. = 1

L.C.M. = 11339

(iii) 8, 9 and 25

8 = 2 × 2 × 2 × 1 = 23× 1

9 = 3 × 3 = 32

and 25 = 5 × 5 = 52

For HCF common factor is 1

H.C.F. = 1

For LCM, taking maximum power of prime factors

L.C.M. = 23× 32 × 52

= 8 × 9 × 25 = 1800

So H.C.F. = 1

L.C.M. = 1800

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

Answer:

HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

L.C.M x H.C.F = first Number x Second Number

L.C.M x 9 = 306 x 657

LCM = 22338

Check out: CBSE Class 10th Previous Year Papers

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

Answer:

TO CHECK: Whether 62can end with the digit 0 for any natural number n.

We know that

62 = (2 × 3)n

62 = (2)n ×(3)n

Therefore, prime factorization of 6ndoes not contain 5 and 2 as a factor together.

Hence 6ncan never end with the digit 0 for any natural number n

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

Answer:

So, the given expression has 6 and 13 as its factors. Therefore, we can conclude that it is a composite number.

Similarly,

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) [taking 5 out- common]

= 5 x (1008 + 1)

= 5 x 1009

Since, 1009 is a prime number the given expression has 5 and 1009 as its factors other than 1 and the number itself.

Hence, it is also 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?

Answer:

It can be observed that 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. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

LCM of 12 and 18= 2 x 2 x 3 x 3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Check out: CBSE Class 10th Question Banks

Class 10 Maths Chapters 1 Real Numbers Exercise 1.2 Summary

• Exercise 1.2 is centred on understanding the Euclidean algorithm and the concept of the Highest Common Factor (HCF).

• The Euclidean division algorithm is a systematic method used to find the HCF of two positive integers. This exercise explains the step-by-step algorithm application procedure to determine the HCF.

• By completing this exercise, students will develop the ability to use the Euclidean algorithm to calculate the HCF of any two given positive integers. This involves performing successive divisions and interpreting the remainders until the remainder is zero.

• The ability to find the HCF is crucial for simplifying fractions, solving problems related to divisibility, and addressing various real-world mathematical scenarios. Exercise 1.2 provides practice problems that enhance students' proficiency in these areas.

• Mastering the Euclidean algorithm and HCF is essential for a strong foundation in mathematics. These concepts are not only fundamental for Class 10 but also serve as building blocks for more advanced topics in higher classes. Understanding these principles aids in developing logical thinking and problem-solving skills that are vital for academic success.

• The exercise includes a variety of problems, from straightforward calculations to more complex applications, ensuring that students get comprehensive practice and can apply these techniques in different contexts.

Check out: CBSE Class 10th Sample Papers

Class 10 Maths Chapters 1 Real Numbers Exercise 1.2 FAQs

Q1. What is the Euclidean algorithm?

Ans. The Euclidean algorithm is a method used to find the Highest Common Factor (HCF) of two positive integers by repeatedly dividing the larger number by the smaller number and taking remainders until a remainder of zero is obtained.

Q2. How do I apply the Euclidean algorithm to find the HCF?

Ans. To apply the Euclidean algorithm, divide the larger number by the smaller number, take the remainder, and then repeat the process using the smaller number and the remainder. Continue until the remainder is zero. The last non-zero remainder is the HCF.

Q3. Why is finding the HCF important?

Ans. Finding the HCF is important because it helps in simplifying fractions, solving problems related to divisibility, and understanding the properties of numbers, which are fundamental in various mathematical applications.

Q4. Can the Euclidean algorithm be used for any pair of positive integers?

Ans. Yes, the Euclidean algorithm can be applied to any pair of positive integers to determine their HCF.

Q5. What is the difference between HCF and LCM?

Ans. The HCF (Highest Common Factor) is the largest number that divides two or more numbers without leaving a remainder, while the LCM (Lowest Common Multiple) is the smallest number that is a multiple of two or more numbers.