NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers

Class 10 students struggle to balance many subjects and assignments as they advance in their education. For this reason, ALLEN has developed comprehensive NCERT Solutions for Maths Class 10 Chapter 1 Real Numbers. The answers adhere to the most recent CBSE curriculum. ALLEN's subject matter experts have meticulously researched NCERT Class 10 Maths - Chapter 1 to ensure ease of comprehension. You can master complex problem-solving skills by understanding real numbers at the fundamental level by going through these NCERT class 10 Maths chapter 1 pdf solutions.

1.0NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers : Download PDF

Mathematics introduces us to a wide range of numbers, including real numbers. To comprehend higher mathematics, it makes sense to start by thoroughly examining the number system, and this is exactly what this book has accomplished. The NCERT book has an intriguing feature that breaks up the monotony of study material by providing some historical notes. 

Additionally, the solutions include some pointers and advice to help students make learning easier. The following are links to the NCERT Solutions Class 10 Maths PDF downloads, which are available for free:

2.0Class 10 Maths Chapter 1 Real Numbers: Breakdown of Exercises

3.0Class 10 Maths Chapter 1 Real Numbers: Important Topics

Real Numbers is one of the important chapters covered in class 10. It has four exercises. Below is a summary of the significant topics addressed in the chapter. Students are advised to carefully review these topics to understand, apply, and become proficient with this chapter and the problems based on real numbers.

• Introduction to Real Numbers
• Fundamental Theorem of Arithmetic (H.C.F. and L.C.M.) 
• Revisiting Irrational Numbers

Brief Overview Of Class 10 Chapter 1 Real Numbers

1. Introduction to Real Numbers

• Definition: Real Numbers are the union of rational and irrational numbers, which can be represented on the number line. Real numbers include both rational and irrational numbers.
• Rational numbers are numbers that can be expressed as the ratio of two integers (a/b, where b ≠ 0).
• Irrational numbers are numbers that cannot be expressed as a fraction of two integers (such as √2, π).
• Examples:
• • Rational: 2/3, -7, 1.5
  • Irrational: √2, π, e

2. The Fundamental Theorem of Arithmetic

• Statement: Every integer greater than 1 is either a prime number or can be uniquely factorized as a product of prime numbers (except for the order of the factors).
• Application: This theorem is the foundation of prime factorization, which helps in solving problems involving the HCF and LCM.
• Example: Factorization of 30 = 2 × 3 × 5 (prime factorization).

3. HCF and LCM (Highest Common Factor and Least Common Multiple)

• Definition of HCF: The largest number that divides two or more numbers without leaving a remainder.
• Definition of LCM: The smallest number that is divisible by two or more numbers.
• Methods to find HCF and LCM:
• • Prime Factorization Method: Express both numbers as a product of primes and then find the HCF and LCM.
  • Division Method: Use division to calculate the HCF and LCM.
• Relation between HCF and LCM: HCF × LCM = Product of the two numbers $\text{HCF}\times \text{LCM}=\text{Product of the two numbers}$

4. Revisiting Irrational Numbers

• Definition: Irrational numbers are numbers that cannot be expressed as fractions (e.g., √2, π).
• Properties of Irrational Numbers:
• • They cannot be expressed as a ratio of two integers.
  • Their decimal representation is non-terminating and non-repeating.
• Examples of Irrational Numbers: √2, √3, π, e
• Proving Irrationality: A common method involves proof by contradiction. For example, proving that √2 is irrational by assuming the opposite and deriving a contradiction.

4.0Sample NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

1. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational, or not. If they are rational, and of the form p/q, what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120120012000120000…...

(iii) 43.123456789 (with a bar over 123456789 indicating repetition)

Solution:

(i) 43.123456789

Since the decimal expansion terminates, the given real number is rational.

It can be expressed in the form p/q as:

43.123456789 = 43123456789 / 1000000000

43.123456789 = 43123456789 / 10⁹

43.123456789 = 43123456789 / (2 × 5)⁹

43.123456789 = 43123456789 / (2⁹ × 5⁹)

Therefore, q = 2⁹ × 5⁹.

The prime factorization of q is of the form 2ⁿ ⋅ 5ᵐ, where n = 9 and m = 9.

(ii) 0.120120012000120000....

Since the decimal expansion is neither terminating nor non-terminating, the given real number is irrational.

(iii) 43.123456789 (with a bar over 123456789)

Since the decimal expansion is non-terminating and repeating, the given real number is rational.

However, since the decimal is non-terminating repeating, q is not of the form 2ᵐ × 5ⁿ.

The prime factors of q will contain factors other than 2 and 5.

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Solution:

(i) 13/3125 = 13/5⁵ = (13 × 2⁵)/(5⁵ × 2⁵) = 416/10⁵ = 0.00416

(ii) 17/8 = 17/2³ = (17 × 5³)/(2³ × 5³) = 2125/10³ = 2.125

(iv) 15/1600 = 3/(2⁶ × 5) = (3 × 5⁵)/(2⁶ × 5⁶) = 9375/10⁶ = 0.009375

(vi) 23/(2³ × 5²) = (23 × 5)/(2³ × 5³) = 115/10³ = 0.115

(viii) 6/15 = 2/5 = 4/10 = 0.4

(ix) 35/50 = 7/10 = 0.7

3. Prove that 3 + 2√5 is irrational.

Solution:

Let us assume, to the contrary, that 3 + 2√5 is rational. So, we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = a/b, b ≠ 0, a, b ∈ I.

Therefore, a/b - 3 = 2√5 ⇒ (a - 3b)/b = 2√5 ⇒ (a - 3b)/(2b) = √5.

Since a and b are integers, we get (a - 3b)/(2b) is rational, and so √5 is rational.

But this contradicts the fact that √5 is irrational. This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is rational.

So, we conclude that 3 + 2√5 is irrational.

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.

5.0Benefits Of NCERT Solutions For Class 10 Chapter 1 Real Numbers

1. Foundation for Advanced Mathematics: Real Numbers lay the groundwork for higher mathematics, including algebra, calculus, and number theory.
2. Improved Problem-Solving Skills: It enhances logical thinking and problem-solving abilities, especially in topics like HCF, LCM, and prime factorization.
3. Understanding the Number System: The chapter helps in understanding the entire number system, including rational and irrational numbers.
4. Practical Applications: Real numbers are used in everyday life for calculations in finance, measurements, science, and more.
5. Competitive Exam Preparation: Mastering this chapter strengthens skills required for exams like JEE, NEET, and other competitive tests.
6. Strengthens Analytical Thinking: It promotes logical reasoning and analytical thinking, especially through proof-based questions.
7. Real-World Relevance: Understanding real numbers is crucial in fields like engineering, technology, and physics, where accurate calculations are essential.
8. Conceptual Clarity: It helps in understanding the relationships between different types of numbers, improving overall mathematical clarity.