Class 10 Maths Chapter 1 – Real Numbers | NCERT Solutions, Key Concepts & Complete Guide (2025–26)

 Class 10 CBSE Maths - Chapter 1: Real Numbers


🔹 1.1 Introduction

The chapter builds on your understanding of natural numbers, whole numbers, integers, rational and irrational numbers, and introduces:

  • Euclid's Division Lemma

  • Fundamental Theorem of Arithmetic

  • Properties of rational and irrational numbers

  • Decimal expansions of real numbers


🔹 1.2 Euclid’s Division Lemma

Concept:

For two positive integers a and b, there exist unique integers q and r such that:
a=bq+r,0r<ba = bq + r,\quad 0 \leq r < b

Used to find the HCF (Highest Common Factor) of two numbers.


📄 Exercise 1.1 Solutions

Q1. Use Euclid's division algorithm to find the HCF of:

(i) 135 and 225:

  • 225 = 135 × 1 + 90

  • 135 = 90 × 1 + 45

  • 90 = 45 × 2 + 0
    HCF = 45

(ii) 196 and 38220:

  • 38220 = 196 × 195 + 0
    HCF = 196

(iii) 867 and 255:

  • 867 = 255 × 3 + 102

  • 255 = 102 × 2 + 51

  • 102 = 51 × 2 + 0
    HCF = 51

Q2. Show that any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5:

Let a be an integer. According to Euclid's lemma:
a=6q+ra = 6q + r where 0 ≤ r < 6. Thus, r = 0, 1, 2, 3, 4, 5
Odd values of a occur when r = 1, 3, 5
Hence, a can be written as 6q + 1, 6q + 3 or 6q + 5.

Exercise 1.2 Solutions

Q1. Express the following numbers as a product of prime factors:

(i) 140 = 2 × 2 × 5 × 7 (ii) 156 = 2 × 2 × 3 × 13 (iii) 3825 = 3 × 3 × 5 × 5 × 17 (iv) 5005 = 5 × 7 × 11 × 13 (v) 7429 = 17 × 19 × 23

Q2. Find the LCM and HCF of:

(i) 12 and 18:

  • 12 = 2^2 × 3

  • 18 = 2 × 3^2

  • HCF = 2 × 3 = 6

  • LCM = 2^2 × 3^2 = 36

(ii) 8 and 9:

  • 8 = 2^3

  • 9 = 3^2

  • HCF = 1

  • LCM = 2^3 × 3^2 = 72

Q3. Verify HCF × LCM = Product of numbers:

(i) 6 and 20:

  • HCF = 2

  • LCM = 60

  • Product = 6 × 20 = 120

  • HCF × LCM = 2 × 60 = 120

Q4. Check whether the numbers are terminating or non-terminating repeating:

(i) → Denominator is 5^5 → Terminating (ii) → Denominator has prime 7 → Non-terminating repeating (iii) → Denominator is 2^6 × 5^2 → Terminating (iv) → Denominator has prime 7 → Non-terminating repeating (v) Terminating

🔹 1.3 Fundamental Theorem of Arithmetic

Every composite number can be uniquely expressed (apart from order) as a product of primes.


📄 Exercise 1.2 Solutions

Q1. Express the following numbers as a product of prime factors:

(i) 140 = 2 × 2 × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

Q2. Find the LCM and HCF of:

(i) 12 and 18:

  • 12 = 2^2 × 3

  • 18 = 2 × 3^2

  • HCF = 2 × 3 = 6

  • LCM = 2^2 × 3^2 = 36

(ii) 8 and 9:

  • 8 = 2^3

  • 9 = 3^2

  • HCF = 1

  • LCM = 2^3 × 3^2 = 72

Q3. Verify HCF × LCM = Product of numbers:

(i) 6 and 20:

  • HCF = 2

  • LCM = 60

  • Product = 6 × 20 = 120

  • HCF × LCM = 2 × 60 = 120







1.4 Revisiting Irrational Numbers

Concept:

An irrational number cannot be expressed as a fraction and its decimal form is non-terminating and non-repeating.



🔹 1.5 Revisiting Rational Numbers and Their Decimal Expansions

Rule:

If the denominator (in lowest terms) has only 2 and/or 5 as prime factors, decimal is terminating; otherwise, non-terminating repeating.

📘 CBSE Past Year Questions – Chapter: Real Numbers


1-Mark Questions

  1. (2020) Write the decimal expansion of 1458724×53\frac{14587}{2^4 \times 5^3} and state whether it terminates or not.
    📌 Answer: Yes, it terminates because the denominator has only prime factors 2 and 5.

  2. (2018) Write the HCF of 6 and 20 using prime factorisation.
    📌 Answer: HCF = 2.

  3. (2016) State whether the decimal expansion of 27210\frac{27}{210} terminates or not. Justify.
    📌 Answer: No, because the denominator has a factor of 7 (other than 2 and 5).


2-Mark Questions

  1. (2022) Use Euclid’s Division Algorithm to find the HCF of 216 and 132.
    📌 Answer:

  • 216=132×1+84216 = 132 \times 1 + 84

  • 132=84×1+48132 = 84 \times 1 + 48

  • 84=48×1+3684 = 48 \times 1 + 36

  • 48=36×1+1248 = 36 \times 1 + 12

  • 36=12×3+036 = 12 \times 3 + 0
    ✅ HCF = 12

  1. (2019) Find the largest number which exactly divides 1251, 9377, and 15628.
    📌 Answer: Use Euclid’s algorithm to find HCF of the three numbers step by step. Final HCF = 103.

  2. (2017) State whether 12922×3×57\frac{129}{2^2 \times 3 \times 5^7} will have a terminating decimal or not. Justify.
    📌 Answer: No. Denominator contains 3, so it’s non-terminating repeating.


3-Mark Questions

  1. (2020) Find the HCF of 65 and 117 and express it as a linear combination of 65 and 117.
    📌 Answer:

  • 117=65×1+52117 = 65 \times 1 + 52

  • 65=52×1+1365 = 52 \times 1 + 13

  • 52=13×4+052 = 13 \times 4 + 0
    So HCF = 13
    Now express 13 = 65 – 52 × 1
    ⇒ 13 = 65 – (117 – 65 × 1) × 1
    ⇒ 13 = 65 × 2 – 117 × 1
    ✅ Final answer: 13 = 65 × 2 – 117 × 1

  1. (2018) Show that 3\sqrt{3} is an irrational number.
    📌 Answer: Use contradiction method by assuming 3=pq\sqrt{3} = \frac{p}{q} and arrive at contradiction.

  2. (2017) Using the Fundamental Theorem of Arithmetic, find LCM and HCF of 306 and 657. Verify that LCM × HCF = product of the numbers.
    📌 Answer:

  • 306 = 2 × 3 × 3 × 17

  • 657 = 3 × 3 × 73

  • HCF = 3 × 3 = 9

  • LCM = 2 × 3 × 3 × 17 × 73 = 22338
    ✅ LCM × HCF = 22338 × 9 = 201042 = 306 × 657


4-Mark Questions

  1. (2023 Sample) Find the HCF of 81 and 237 using Euclid’s algorithm. Also find integers x and y such that HCF = 81x + 237y.
    📌 Answer:
    Follow division steps to get HCF = 3
    Then back-substitute using extended Euclidean algorithm.


5-Mark Questions

  1. (2020 - Internal Assignment)
    Prove that 2+5\sqrt{2} + \sqrt{5} is irrational.
    📌 Answer:
    Assume the sum is rational ⇒ subtract √2 ⇒ irrational = rational ⇒ contradiction.

  2. (2016)
    Use prime factorisation to find HCF and LCM of 5005 and 30030. Verify that HCF × LCM = product of numbers.
    📌 Answer:

  • 5005 = 5 × 7 × 11 × 13

  • 30030 = 2 × 3 × 5 × 7 × 11 × 13

  • HCF = 5 × 7 × 11 × 13 = 5005

  • LCM = product of all primes with highest powers
    ✅ LCM = 2 × 3 × 5 × 7 × 11 × 13 = 30030
    ✅ HCF × LCM = 5005 × 30030 = Product of given numbers


📌 Tips for Board Exam from This Chapter:

  • Practice Euclid's Algorithm for pairs like (867, 255), (65, 117)

  • Memorize terminating vs non-terminating decimal conditions

  • Be thorough with irrationality proofs like 2,3,5\sqrt{2}, \sqrt{3}, \sqrt{5}

  • Know how to express HCF as linear combination (x and y form)

  • Practice verification: HCF × LCM = product of two numbers


Comments

Popular posts from this blog

Rise of Nationalism in Europe – Class 10 CBSE Detailed Notes | Chapter 1 History

Nationalism in India – Class 10 CBSE History Notes | Chapter 2 | Key Events, Movements & Leaders

Resources and Development – Class 10 Geography Notes, Concepts, and Map Work