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

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NCERT Solutions for Maths Class 10 Exercise 1.2 of Chapter 1 Real Numbers - Free PDF Download

NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers is focused on providing the students with a reliable study tool. The Vedantu solutions for CBSE Maths Chapter 1 Exercise 1.2 Class 10 allow students to understand the basics that are included in the chapter and help understand and identify different questions in Maths Chapter 1 Exercise 1.2 Class 10. Our team of experts have developed the NCERT Solution Class 10 Maths Exercise 1.2 to understand more around the basics and complex problems of Real Numbers. Moreover, these Class 10 Maths Chapter 1 Exercise 1.2 helps students understand the schematics around marking system as per the latest syllabus as well. 

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Table of Content
1. NCERT Solutions for Maths Class 10 Exercise 1.2 of Chapter 1 Real Numbers - Free PDF Download
2. Glance of NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.2 Real Numbers
3. Topics Covered in Class 10 Maths Chapter 1 Exercise 1.2
4. Access NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers
5. NCERT Solutions Class 10 Maths Chapter 1 All the Other Exercises
6. CBSE Class 10 Maths Chapter 1 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs


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

  • This exercise revisits the idea of real numbers, which encompass both rational (like 1/2 or -3/7), and irrational numbers (like pi or the square root of 2).

  • You will practice depicting real numbers on the number line. Every real number has a unique corresponding position on the line.

  • The exercise likely involves questions on performing basic operations like addition, subtraction, multiplication, and division on real numbers.

  • Ordering Real Numbers deals with comparing real numbers using the greater than (<), less than (>), equal to (=) signs symbols.

  • Use concepts of HCF (Highest Common Factor) and LCM (Least Common Multiple).

  • Exercise Focus

  • Distinguishing between rational and irrational numbers.

  • Plotting real numbers on the number line.

  • Performing basic calculations like addition, subtraction, multiplication, and division on real numbers.

  • Comparing and ordering real numbers using inequalities.

  • There are links to video tutorials explaining class 10 chapter 1  Exercise 1.2 Real Numbers for better understanding.

  • There is one exercise (7 fully solved questions) in Class 10th Maths, Chapter 1, Exercise 1.2 Real Numbers.


Topics Covered in Class 10 Maths Chapter 1 Exercise 1.2

  1. Review of Real Numbers

  2. Representing Real Numbers on the Number Line

  3. Ordering of Real Numbers 

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Access NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers

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

(i) $140$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 140=2\times 2\times 5\times 7$ 

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

Therefore, the prime factors of $140$ are $2,5,7$.

(ii) $156$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 156=2\times 2\times 3\times 13$ 

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

Therefore, the prime factors of $156$ are $2,3,13$.

(iii) $3825$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 3825=3\times 3\times 5\times 5\times 17$ 

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

Therefore, the prime factors of $3825$ are $3,5,17$.

(iv) $5005$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

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

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

Therefore, the prime factors of $5005$ are $5,7,11,13$.

(v) $7429$ 

Ans: We know that the procedure of writing a number as the product of prime numbers is known as the prime factorization. Prime numbers that can be multiplied to obtain the original number are known as prime factors.

$\Rightarrow 7429=17\times 19\times 23$ 

$\therefore 7429=17\times 19\times 23$ 

Therefore, the prime factors of $7429$ are $17,19,23$.


2. Find the LCM and HCF of the following pairs of integers and verify that $LCM\times HCF=\text{Product of two numbers}$.

(i) $26$ and $91$ 

Ans: First we write the prime factors of $26$ and $91$. We get

$ 26=2\times 13$ and 

$91=7\times 13$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $26$ and $91$ is $13$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $26$ and $91$ will be

$ 2\times 7\times 13=182$ 

Therefore, the LCM of $26$ and $91$ is $182$.

Now, the product of two numbers is 

$ 26\times 91=2366$ 

Product of LCM and HCF is

$ 13\times 182=2366$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

(ii) $510$ and $92$

Ans: First we write the prime factors of $510$ and $92$. We get

$510=2\times 3\times 5\times 17$ and 

$92=2\times 2\times 23$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $510$ and $92$ is $2$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $510$ and $92$ will be

$2\times 2\times 3\times 5\times 17\times 23=23460$ 

Therefore, the LCM of $510$ and $92$ is $23460$.

Now, the product of two numbers is 

$510\times 92=46920$ 

Product of LCM and HCF is

$2\times 23460=46920$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.

(iii) $336$ and $54$

Ans: First we write the prime factors of $336$ and $54$. We get

\[336=2\times 2\times 2\times 2\times 3\times 7\] and 

$54=2\times 3\times 3\times 3$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $336$ and $54$ is $2\times 3=6$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $336$ and $54$ will be

$ 2\times 2\times 2\times 2\times 3\times 3\times 3\times 7=3024$ 

Therefore, the LCM of $336$ and $54$ is $3024$.

Now, the product of two numbers is 

$336\times 54=18144$ 

Product of LCM and HCF is

$ 6\times 3024=18144$

We get $LCM\times HCF=\text{Product of two numbers}$.

The desired result has been verified.


3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) $12,15$ and $21$

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $12,15$ and $21$ are as follows:

\[12=2\times 2\times 3\] 

\[15=3\times 5\] and 

$21=3\times 7$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $12,15$ and $21$ is $3$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $12,15$ and $21$ will be

$2\times 2\times 3\times 5\times 7=420$

Therefore, the LCM of $12,15$ and $21$ is $420$.

(ii) $17,23$ and $29$

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $17,23$ and $29$ are as follows:

\[17=17\times 1\] 

\[23=23\times 1\] and 

$29=29\times 1$

Now, we know that HCF is the highest factor, among the common factors of two numbers.

Therefore, the HCF of $17,23$ and $29$ is $1$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $17,23$ and $29$ will be

$\times 23\times 29=11339$ 

Therefore, the LCM of $17,23$ and $29$ is $11339$.

(iii) $8,9$ and $25$ 

Ans: The procedure of writing a number as the product of prime numbers is known as the prime factorization.

The prime factors of $8,9$ and $25$ are as follows:

\[8=2\times 2\times 2\] 

\[9=3\times 3\] and 

$25=5\times 5$

Now, we know that HCF is the highest factor, among the common factors of two numbers. as there is no common factor.

Therefore, the HCF of $8,9$ and $25$ is $1$.

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $8,9$ and $25$ will be

$2\times 2\times 2\times 3\times 3\times 5\times 5=1800$ 

Therefore, the LCM of $8,9$ and $25$ is $1800$.


4. Given that HCF $\left( 306,657 \right)=9$, find LCM $\left( 306,657 \right)$.

Ans: We have been given the HCF of two numbers $\left( 306,657 \right)=9$.

We have to find the LCM of $\left( 306,657 \right)$.

Now, we know that $LCM\times HCF=\text{Product of two numbers}$

Substitute the values, we get

$LCM\times 9=306\times 657$

$\Rightarrow LCM=\dfrac{306\times 657}{9}$ 

$\therefore LCM=22338$ 

Therefore, the LCM of $\left( 306,657 \right)=22338$.


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

Ans: We have to check whether ${{6}^{n}}$ can end with the digit $0$ for any natural number $n$.

By divisibility rule we know that if any number ends with the digit $0$, it is divisible by $2$ and $5$.

Thus, the prime factors of ${{6}^{n}}$ is

$ {{6}^{n}}={{\left( 2\times 3 \right)}^{n}}$

Now, we will observe that for any value of $n$, ${{6}^{n}}$ is not divisible by $5$.

Therefore, ${{6}^{n}}$ cannot end with the digit $0$ for any natural number $n$.


6. 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.

Ans: The given numbers are $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$.

We can rewrite the given numbers as

$7\times 11\times 13+13=13\times \left( 7\times 11+1 \right)$

$\Rightarrow 7\times 11\times 13+13=13\times \left( 77+1 \right)$ 

$\Rightarrow 7\times 11\times 13+13=13\times 78$

$\Rightarrow 7\times 11\times 13+13=13\times 13\times 6$

And, 

$7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times \left( 7\times 6\times 4\times 3\times 2\times 1+1 \right)$

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times \left( 1008+1 \right)$

$\Rightarrow 7\times 6\times 5\times 4\times 3\times 2\times 1+5=5\times 1009$

Here, we can observe that the given expression has its factors other than $1$ and the number itself.

A composite number has factors other than $1$ and the number itself.

Therefore, $7\times 11\times 13+13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1+5$ are composite numbers.


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?

Ans: It can be observed that Ravi takes less time than Sonia for completing the $1$ round of the circular path. Both are going in the same direction, they will meet again when Ravi will have completed $1$ round of that circular path with respect to Sonia. 

The total time taken for completing this $1$ round of circular path will be the LCM of time taken by Sonia and Ravi for ending the $1$ round of circular path respectively, i.e., LCM of $18$ minutes and $12$ minutes.

The prime factors of $12$ and $18$ are as follows:

\[12=2\times 2\times 3\]  and 

$18=2\times 3\times 3$

Now, we know that LCM is the least common multiple. To find the LCM multiplies each factor to the number of times it occurs in any number.

Then the LCM of $12$ and $18$ will be

$2\times 2\times 3\times 3=36$

Therefore, Ravi and Sonia meet again at the starting point after $36$ minutes.


Conclusion

Exercise 1.2 in Chapter 1 of the Class 10 NCERT textbook focuses on the practical application of Euclid’s Division Lemma to find the highest common factor (HCF) of two numbers. This exercise contains seven questions, providing students with the best practice to understand and apply this fundamental concept. It is important for students to focus on the method of using Euclid’s Division Lemma iteratively, as this strengthens their problem-solving skills and lays the groundwork for more advanced topics in number theory. In previous years, questions from this exercise frequently appeared in exams, highlighting its significance. By mastering this exercise, students will build a strong foundation in arithmetic that is crucial for their further studies in mathematics.


NCERT Solutions Class 10 Maths Chapter 1 All the Other Exercises

Chapter 1 - Real Numbers All Exercises in PDF Format

Exercise 1.1

5 Questions & Solutions

Exercise 1.3

3 Questions & Solutions

Exercise 1.4

3 Questions & Solutions


CBSE Class 10 Maths Chapter 1 Other Study Materials


Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for class 10th maths chapter 1 exercise 1.2 Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


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

1. Why is it important to learn Euclid’s Division Lemma in class 10 math exercise 1.2?

Learning Euclid’s Division Lemma is important because it forms the basis for finding the HCF of two numbers and is fundamental for understanding more advanced mathematical concepts in number theory and algebra.

2. What are the different aspects covered in Exercise 1.2 Class 10 Maths Solution?

The solutions of Class 10th Maths Chapter 1 Exercise 1.2 revolves around several questions that help students prepare the best for their exams.

The suggested topics for practice include: 

  • Proving number as a product of its prime factors (Important Question For Class 10 Chapter 1 Maths Exercise 1.2)

  • Evaluating the HCF and LCM of two given numbers (a repeated question of Exercise 1.2 Class 10th Maths, over the years)

  • Finding LCM of two numbers when HCF is given. (Important Question From Class 10 Maths Chapter 1 Ex 1.2)

  • Evaluate HCF and LCM of integers using a method of prime factorization (an important question of Exercise 1.2 Class 10th Maths to be practised)

3. How many questions are there in Class 10 Maths Exercise 1.2?

The First Chapter given in the syllabus of Class 10 Mathematics is Real Numbers. Exercise 1.2 of this chapter includes seven questions for students to practice. These seven questions are based on the Fundamental Theorem of Arithmetic. Through this exercise, students can also understand the concept of the Highest Common Factor (HCF) and Least Common Multiple (LCM).

4. Which examples are important for solving Exercise 1.2 of Class 10 Mathematics?

All the examples provided in the NCERT Book for Class 10 Mathematics are equally important. Therefore, students must practice solving the examples as well. Four examples are based on exercise 1.2. These include examples five, six, seven, and eight. All of these examples hold equal importance. All of them should be studied to score well in the exam.

5. Is  Class 10 Maths chapter 1 difficult?

No, it is not. Regular practice of this chapter helps students achieve a proper understanding of all concepts. Referring to NCERT solutions for Class 10 Maths chapter 1  on Vedantu can further assist them to get a better hold of this chapter. You can also refer to revision notes of this chapter to revise before the exam. These notes are prepared by experts with decades of experience and hence are 100% reliable. 

6. Where can I find NCERT Solutions for Class 10 Maths Exercise 1.2?

Students can find the NCERT Solutions for Class 10 Maths Exercise 1.2 on Vedantu. NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.2 helps the students to solve the exercise easily. These solutions have been prepared in a step-by-step manner by the subject-matter experts. With the help of these NCERT Solutions, students will be able to enhance their understanding, and they can solve any kind of difficulties that they might face while practising Class 10 Maths.

7. Do I need to practice all the questions provided in Class 10 Maths Chapter 1 NCERT Solutions?

Question papers in your Class 10 exams are designed on the basis of the NCERT books. Questions can be asked from any exercise of the chapter, and there is no particular way to find which of them can be asked. Ignoring any questions that are provided in the NCERT can lead to a loss of marks in the exam. Hence, it is best to practice all the questions provided in the NCERT Solutions for Class 10 Maths Chapter 1.

8. How does Exercise 1.2 class 10 maths help in understanding number theory?

Exercise 1.2 Class 10 Maths NCERT textbook introduces students to Euclid’s Division Lemma, a key concept in number theory. By solving problems that require the application of this lemma to find the Highest Common Factor (HCF), students gain a foundational understanding of divisibility, crucial for more complex topics such as prime factorization and the fundamental theorem of arithmetic. This exercise not only deepens students’ knowledge of mathematical theory but also enhances their problem-solving skills, which are essential for both academic and real-world situations.

9. What skills are developed by practicing class 10 math exercise 1.2?

Practicing Exercise 1.2 from Chapter 1 of Class 10's NCERT textbook helps students enhance their problem-solving and logical thinking skills by applying Euclid's Division Lemma to find the Highest Common Factor (HCF). This exercise teaches them to systematically analyze and solve numerical problems, fostering a deeper understanding of mathematical principles and improving their analytical abilities.

10. Are questions from class 10 math exercise 1.2 frequently asked in exams?

Yes, questions based on Euclid’s Division Lemma and finding the HCF are commonly asked in exams, making this exercise highly relevant for exam preparation.

11. What is the main focus of Exercise 1.2 in Chapter 1 of Class 10 NCERT?

The main focus of Exercise 1.2 is to apply Euclid’s Division Lemma to find the Highest Common Factor (HCF) of two given numbers.