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:
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Euclid's Division Lemma
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Fundamental Theorem of Arithmetic
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Properties of rational and irrational numbers
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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:
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:
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225 = 135 × 1 + 90
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135 = 90 × 1 + 45
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90 = 45 × 2 + 0
HCF = 45
(ii) 196 and 38220:
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38220 = 196 × 195 + 0
HCF = 196
(iii) 867 and 255:
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867 = 255 × 3 + 102
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255 = 102 × 2 + 51
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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:
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:
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12 = 2^2 × 3
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18 = 2 × 3^2
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HCF = 2 × 3 = 6
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LCM = 2^2 × 3^2 = 36
(ii) 8 and 9:
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8 = 2^3
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9 = 3^2
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HCF = 1
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LCM = 2^3 × 3^2 = 72
Q3. Verify HCF × LCM = Product of numbers:
(i) 6 and 20:
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HCF = 2
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LCM = 60
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Product = 6 × 20 = 120
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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
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(2020) Write the decimal expansion of and state whether it terminates or not.
📌 Answer: Yes, it terminates because the denominator has only prime factors 2 and 5. -
(2018) Write the HCF of 6 and 20 using prime factorisation.
📌 Answer: HCF = 2. -
(2016) State whether the decimal expansion of terminates or not. Justify.
📌 Answer: No, because the denominator has a factor of 7 (other than 2 and 5).
✅ 2-Mark Questions
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(2022) Use Euclid’s Division Algorithm to find the HCF of 216 and 132.
📌 Answer:
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✅ HCF = 12
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(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. -
(2017) State whether will have a terminating decimal or not. Justify.
📌 Answer: No. Denominator contains 3, so it’s non-terminating repeating.
✅ 3-Mark Questions
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(2020) Find the HCF of 65 and 117 and express it as a linear combination of 65 and 117.
📌 Answer:
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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
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(2018) Show that is an irrational number.
📌 Answer: Use contradiction method by assuming and arrive at contradiction. -
(2017) Using the Fundamental Theorem of Arithmetic, find LCM and HCF of 306 and 657. Verify that LCM × HCF = product of the numbers.
📌 Answer:
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306 = 2 × 3 × 3 × 17
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657 = 3 × 3 × 73
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HCF = 3 × 3 = 9
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LCM = 2 × 3 × 3 × 17 × 73 = 22338
✅ LCM × HCF = 22338 × 9 = 201042 = 306 × 657
✅ 4-Mark Questions
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(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
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(2020 - Internal Assignment)
Prove that is irrational.
📌 Answer:
Assume the sum is rational ⇒ subtract √2 ⇒ irrational = rational ⇒ contradiction. -
(2016)
Use prime factorisation to find HCF and LCM of 5005 and 30030. Verify that HCF × LCM = product of numbers.
📌 Answer:
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5005 = 5 × 7 × 11 × 13
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30030 = 2 × 3 × 5 × 7 × 11 × 13
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HCF = 5 × 7 × 11 × 13 = 5005
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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:
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Practice Euclid's Algorithm for pairs like (867, 255), (65, 117)
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Memorize terminating vs non-terminating decimal conditions
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Be thorough with irrationality proofs like
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Know how to express HCF as linear combination (x and y form)
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Practice verification: HCF × LCM = product of two numbers
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