Real Numbers — Class 10 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Real Numbers" — 9 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Real Numbers" — 9 important questions with detailed answers for CBSE board exam pr…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

State the Fundamental Theorem of Arithmetic.
Prove that √2 is irrational.
Find the HCF of 96 and 404 using Euclid's division algorithm.
Find the LCM and HCF of 12 and 18 using prime factorisation method.
Express 0.3̄5̄ (0.353535...) as a fraction in simplest form.
Find the HCF of 84, 120, and 138.
+ 3 more questions in the full chapter

Solutions Summary:

Question	Status
State the Fundamental Theorem of Arithmetic.	✓ Solved
Prove that √2 is irrational.	✓ Solved
Find the HCF of 96 and 404 using Euclid's division algori…	✓ Solved
Find the LCM and HCF of 12 and 18 using prime factorisati…	✓ Solved
Express 0.3̄5̄ (0.353535...) as a fraction in simplest form.	✓ Solved
Find the HCF of 84, 120, and 138.	✓ Solved

Showing 6 of 9 questions

Q1: State the Fundamental Theorem of Arithmetic.

The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of prime numbers, and this factorisation is unique, apart from the order in which the prime factors occur. For example: - 60 = 2² × 3 × 5 - 84 = 2² × 3 × 7 This means that if we write a composite number as a product of primes in two different ways, the prime factors (with their powers) must be exactly the same.

Q2: Prove that √2 is irrational.

Proof by contradiction: Assume √2 is rational. Then √2 = p/q, where p and q are coprime integers (HCF(p,q) = 1) and q ≠ 0. Squaring both sides: 2 = p²/q² Therefore: 2q² = p² ... (1) From equation (1), p² is even, which means p must be even. Let p = 2m for some integer m. Substituting in equation (1): 2q² = (2m)² 2q² = 4m² q² = 2m² This shows q² is even, so q must be even. But if both p and q are even, then HCF(p,q) ≥ 2, which contradicts our assumption that p and q are coprime. Therefore,...

Q3: Find the HCF of 96 and 404 using Euclid's division algorithm.

Using Euclid's division algorithm: Step 1: Divide 404 by 96 404 = 96 × 4 + 20 Step 2: Divide 96 by 20 96 = 20 × 4 + 16 Step 3: Divide 20 by 16 20 = 16 × 1 + 4 Step 4: Divide 16 by 4 16 = 4 × 4 + 0 Since the remainder is 0, the HCF is 4. Therefore, HCF(96, 404) = 4

Q4: Find the LCM and HCF of 12 and 18 using prime factorisation method.

Step 1: Find prime factorisation of both numbers 12 = 2² × 3 18 = 2 × 3² Step 2: Find HCF (take lowest powers of common prime factors) Common prime factors: 2 and 3 HCF = 2¹ × 3¹ = 6 Step 3: Find LCM (take highest powers of all prime factors) LCM = 2² × 3² = 4 × 9 = 36 Verification: HCF × LCM = 6 × 36 = 216 and 12 × 18 = 216 ✓ Therefore, HCF(12, 18) = 6 and LCM(12, 18) = 36

Q5: Express 0.3̄5̄ (0.353535...) as a fraction in simplest form.

Let x = 0.353535... Since two digits repeat, multiply both sides by 100: 100x = 35.353535... Subtract the original equation: 100x - x = 35.353535... - 0.353535... 99x = 35 x = 35/99 Check if this can be simplified: HCF(35, 99) = HCF(5 × 7, 9 × 11) = 1 Therefore, 0.3̄5̄ = 35/99

Q6: Find the HCF of 84, 120, and 138.

Using prime factorisation: 84 = 2² × 3 × 7 120 = 2³ × 3 × 5 138 = 2 × 3 × 23 HCF = Product of lowest powers of common prime factors Common prime factors: 2 and 3 HCF = 2¹ × 3¹ = 6 Therefore, HCF(84, 120, 138) = 6

Showing 6 of 9 questions. Visit the full page for complete solutions.