CBSE Class 10 Mathematics: NCERT Solutions Class 10 Maths Chapter 1 Real Numbers (2025–26)

Mastering CBSE Class 10 Mathematics begins with Chapter 1: Real Numbers. For students preparing for the 2025–26 board exam 10, this chapter carries 4 to 6 guaranteed marks in the question paper. Under the rationalized NCERT syllabus, Chapter 1 focuses on two core pillars: the Fundamental Theorem of Arithmetic (with applications in HCF and LCM) and Revisiting Irrational Numbers through formal contradiction proofs. These comprehensive NCERT Solutions Class 10 Maths Chapter 1 Real Numbers break down each theorem, formula, and step-by-step solution to ensure top scores in your board examinations.

Key Concepts in Class 10 Real Numbers

The revised CBSE syllabus streamlines the chapter by focusing on essential number theory concepts. Below are the core principles every student must master before solving exercise problems.

1. The Fundamental Theorem of Arithmetic (FTA)

The Fundamental Theorem of Arithmetic is the foundation of prime factorisation. It states:

"Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur."

Mathematically, any composite integer n can be written uniquely as:

n = p₁^a₁ × p₂^a₂ × p₃^a₃ × ... × p_k^a_k

where p₁, p₂, ..., p_k are distinct prime numbers in ascending order and a₁, a₂, ..., a_k are positive integers.

2. Computing HCF and LCM Using Prime Factorisation

Using the prime power representation of two or more numbers:

• Highest Common Factor (HCF): The product of the smallest power of each common prime factor involved in the numbers.
• Lowest Common Multiple (LCM): The product of the greatest power of each prime factor involved in the numbers.

Crucial Relationship Formula: For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

Important Note for CBSE Exams: This product property holds true ONLY for two integers. For three numbers a, b, c, HCF(a, b, c) × LCM(a, b, c) ≠ a × b × c.

3. Real-World Applications of HCF and LCM

Understanding when to calculate HCF versus LCM is key to solving CBSE competency-based word problems:

• Apply HCF when: The problem asks to divide or distribute different quantities into the maximum possible equal groups, stacks, or container sizes without any remainder.
• Apply LCM when: The problem involves periodic events happening together in the future (e.g., runners completing laps on a circular track, toll bells ringing at different intervals, or traffic lights flashing simultaneously).

4. Revisiting Irrational Numbers (Proof by Contradiction)

A real number is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.

Fundamental Lemma: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

All irrationality proofs in CBSE Mathematics rely on the method of contradiction: assuming the given number is rational, expressing it in simplest co-prime form, and arriving at a logical impossibility.

Important CBSE Questions with Answers (NCERT Solutions)

The following solved questions represent the exact types and standard marking patterns seen in official CBSE board papers.

Question 1 (2 Marks): Expressing Primes and Verifying HCF-LCM Product

Problem: Express 140 and 504 as products of their prime factors. Find their HCF and LCM, and verify that HCF(140, 504) × LCM(140, 504) = 140 × 504.

Solution:

Step 1: Prime factorisation of the given numbers:

• 140 = 2 × 2 × 5 × 7 = 2² × 5¹ × 7¹
• 504 = 2 × 2 × 2 × 3 × 3 × 7 = 2³ × 3² × 7¹

Step 2: Calculate HCF and LCM:

• HCF(140, 504) = Smallest powers of common prime factors (2 and 7) = 2² × 7¹ = 4 × 7 = 28
• LCM(140, 504) = Highest powers of all prime factors = 2³ × 3² × 5¹ × 7¹ = 8 × 9 × 5 × 7 = 2520

Step 3: Verification:

• HCF × LCM = 28 × 2520 = 70,560
• Product of numbers = 140 × 504 = 70,560

Since HCF × LCM = Product of the two numbers (70,560 = 70,560), the relationship is verified.

Question 2 (3 Marks): Standard Irrationality Proof

Problem: Prove that √5 is an irrational number.

Solution:

1. Assumption: Let us assume, to the contrary, that √5 is a rational number.
2. Therefore, we can find two co-prime integers a and b (where b ≠ 0 and HCF(a, b) = 1) such that:
   √5 = a / b ⇒ a = √5 b
3. Squaring both sides:
   a² = 5b²   --- (Equation 1)
4. Since 5 divides 5b², 5 divides a².
   By theorem, if a prime p divides a², then p divides a.
   Therefore, 5 divides a.
5. We can write a = 5c for some integer c.
   Substituting a = 5c into Equation 1:
   (5c)² = 5b²
   25c² = 5b²
   b² = 5c²   --- (Equation 2)
6. This means 5 divides b², and hence 5 divides b.
7. From steps (4) and (6), 5 is a common factor of both a and b.
8. This contradicts our initial assumption that a and b are co-prime (having no common factor other than 1).
9. This contradiction arises because of our incorrect assumption that √5 is rational.

Conclusion: Hence, √5 is an irrational number. (Proved)

Question 3 (2 Marks): Linear Combination Irrationality

Problem: Given that √3 is irrational, prove that 5 - 2√3 is irrational.

Solution:

1. Let us assume, to the contrary, that 5 - 2√3 is rational.
2. Then, there exist co-prime integers a and b (b ≠ 0) such that:
   5 - 2√3 = a / b
3. Rearranging terms to isolate √3:
   2√3 = 5 - (a / b)
   2√3 = (5b - a) / b
   √3 = (5b - a) / 2b
4. Since a and b are integers, (5b - a) / 2b is a rational number.
5. This implies that √3 must also be a rational number.
6. However, this contradicts the given fact that √3 is irrational.

Conclusion: Our assumption was false. Therefore, 5 - 2√3 is irrational.

Question 4 (3 Marks): Real-World Word Problem on Stacking and Area

Problem: A sweets merchant has 420 Kaju barfis and 130 Badam barfis. She wants to stack them such that each stack has the same number of barfis and they occupy the least area on the tray. What is the maximum number of barfis that can be placed in each stack?

Solution:

To occupy the least area of the tray, the number of barfis in each stack must be the maximum possible, and the number of barfis in each stack must divide both 420 and 130 exactly. Therefore, the required number of barfis is the HCF(420, 130).

• 420 = 2 × 2 × 3 × 5 × 7 = 2² × 3¹ × 5¹ × 7¹
• 130 = 2 × 5 × 13 = 2¹ × 5¹ × 13¹

Common prime factors are 2 and 5 with their lowest powers:

HCF(420, 130) = 2¹ × 5¹ = 10

Answer: The merchant can make stacks of 10 barfis each to take up the minimum tray area.

Question 5 (1 Mark): Checking Ending Digits (Competency Question)

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

Solution:

If any number ends with the digit 0, it must be divisible by 10, meaning its prime factorisation must contain both 2 and 5.

Prime factorisation of 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ

The only prime factors in 6ⁿ are 2 and 3. By the uniqueness of the Fundamental Theorem of Arithmetic, no other prime factor can exist in the expansion of 6ⁿ. Since 5 is not a factor, 6ⁿ can never end with the digit 0 for any natural number n.

How to Prepare for This Topic in CBSE Board Exam 2025–26

To secure full marks in Chapter 1 during your board examinations, follow these teacher-recommended revision tactics:

• Write Formal Proof Statements: CBSE evaluators deduct half a mark if you do not explicitly declare that a and b are co-prime integers where b ≠ 0 in contradiction proofs.
• Always Verify with the HCF-LCM Product Formula: In 2-mark questions, use HCF × LCM = a × b as a quick self-check to ensure your arithmetic is 100% error-free.
• Master Factor Tree Method: Maintain clear branching while prime factorising larger 3-digit numbers to avoid missing repeated factors like 2 or 3.
• Categorise Word Problems Instantly: Look for keywords like "meet again" or "simultaneously" for LCM, and "maximum capacity" or "largest divisor" for HCF.

Where to Practice More

Consistent practice with official CBSE pattern questions is the most effective way to eliminate exam anxiety and build speed. For chapter-wise question banks, competency-based questions, and previous years' solved papers aligned with the 2025–26 syllabus, generate customised practice sets on Theorify QPTool (qptool.theorify.in).