Class 10 Maths Ch 1 Real Numbers Ex 1.2 is based on two important concepts in number theory:
- Euclid’s Division Lemma
- Fundamental Theorem of Arithmetic (Prime Factorization Method)
In this exercise, students learn how to:
- Apply Euclid’s Division Algorithm to find the HCF of two numbers.
- Use prime factorization to find the HCF (Highest Common Factor) and LCM (Least Common Multiple) of two or more numbers.
- Verify the relation:
HCF × LCM = Product of the two numbers
The questions in this exercise help build a strong foundation in understanding how numbers are broken down into their basic prime factors and how this can be used to solve real-life problems involving divisibility and multiples.
Let’s now look at the detailed solutions to each question from Exercise 1.2.

Class 10 Maths Ch 1 Real Numbers Ex 1.2-Textbook Solutions
Question 1.
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:

Ex 1.2 Class 10 Maths Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers:
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
Solution:
Prime factorization:
26 = 2 × 13
91 = 7 × 13
Common factor = 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Verification:
HCF × LCM = 13 × 182 = 2366
26 × 91 = 2366
(ii) 510 and 92
Solution:
Prime factorization:
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
Common factor = 2
HCF = 2
LCM = (product of all prime factors without repetition) = 2² × 3 × 5 × 17 × 23 = 23460
Verification:
HCF × LCM = 2 × 23460 = 46920
510 × 92 = 46920
(iii) 336 and 54
Solution:
Prime factorization:
336 = 2⁴ × 3 × 7
54 = 2 × 3³
Common factors: 2 × 3 = 6
HCF = 6
LCM = 2⁴ × 3³ × 7 = 3024
Verification:
HCF × LCM = 6 × 3024 = 18144
336 × 54 = 18144
All verified: LCM × HCF = Product of the two numbers
Class 10 Maths Ch 1 Real Numbers Ex 1.2 – prime factorization method:
Ex 1.2 Class 10 Maths Question 3.
Find the LCM and HCF of the following integers by applying the prime factorization method:
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) 12, 15, and 21
Prime Factorization:
- 12 = 2² × 3
- 15 = 3 × 5
- 21 = 3 × 7
HCF = 1
LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
(ii) 17, 23, and 29
These are all prime numbers.
So:
- 17 = 17
- 23 = 23
- 29 = 29
HCF: No common factor other than 1 → HCF = 1
LCM: Product of all numbers
= 17 × 23 × 29 = 11339
(iii) 8, 9, and 25
Prime Factorization:
- 8 = 2³
- 9 = 3²
- 25 = 5²
HCF: No common prime factor →HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
Ex 1.2 Class 10 Maths Question 4:
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Step 1: Use the formula
HCF × LCM = Product of the two numbers
9 × LCM = 306 × 657
LCM = 201042 ÷ 9 = 22338
Ex 1.2 Class 10 Maths Question 5.
Check whether 6n can end with the digit 0 for any natural number n.
Solution:
A number ends with the digit 0 only if it is divisible by 10.
That means the number must have both 2 and 5 as factors.
Now, 6 = 2 × 3.
So, 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
This means 6ⁿ will always have 2 and 3 as factors, but never 5.
Since 5 is not a factor, 6ⁿ can never be divisible by 10.
Ex 1.2 Class 10 Maths Question 6.
Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Solution:
7 × 11 × 13 + 13
Step 1: Factor the expression
= 7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
Since the number is the product of two numbers (13 and 78), both greater than 1, it has more than two factors.
Therefore, it is a composite number.
Now,
=7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5040 + 5
= 5045
Step 2: Check the factors of 5045
= 5 × 1009
Since 5045 can be written as a product of 5 and 1009, it has more than two factors.
Therefore, it is a composite number.
Ex 1.2 Class 10 Maths Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
They will meet again at the starting point after the Least Common Multiple (LCM) of their times.
Step 1: Find LCM of 18 and 12
Prime factorization:
- 18 = 2 × 3 × 3 = 2 × 3²
- 12 = 2 × 2 × 3 = 2² × 3
LCM = 2² × 3² = 4 × 9 = 36
They will meet again at the starting point after 36 minutes.
Class 10 Maths Ch 1 Real Numbers Ex 1.2-Concept of Real Numbers
Introduction to Exercise 1.2 – Real Numbers
Exercise 1.2 of Class 10 Maths Ch 1 Real Numbers Ex 1.2 focuses on the Fundamental Theorem of Arithmetic, which states that every composite number can be uniquely expressed as a product of prime numbers (except for the order of the factors). This concept is crucial for solving problems related to factorization and the highest common factor (HCF) and lowest common multiple (LCM) of numbers.
In this section, we will cover:
- Prime factorization method
- Finding HCF and LCM using prime factorization
- Relationship between HCF and LCM
New Syllabus- Class 10 Maths Ch 1 Real Numbers Ex 1.2
Exercise 1.2 Solutions
Q1. Prove that √5 is irrational.
Assume √5 is rational.
Then, √5 = a/b, where a and b are integers with no common factor, and b ≠ 0.
Squaring both sides:
5 = a² / b² ⇒ a² = 5b²
So, 5 divides a² ⇒ 5 divides a
Let a = 5k for some integer k
Substitute: a² = 25k² ⇒ 5b² = 25k² ⇒ b² = 5k² ⇒ 5 divides b
So both a and b are divisible by 5, contradicting that they are coprime.
Hence, √5 is irrational.
Q2. Prove that 3 + 2√5 is irrational.
Assume 3 + 2√5 is rational.
Then, 3 + 2√5 = a/b, where a and b are integers and b ≠ 0.
Then, 2√5 = a/b − 3 ⇒ √5 = (a − 3b) / (2b)
This implies √5 is rational, which is false.
Hence, 3 + 2√5 is irrational.
Q3. Prove that the following are irrationals:
(i) 1 / √2
Assume 1 / √2 is rational.
Then √2 = 1 / (rational), which implies √2 is rational.
But √2 is irrational.
So, 1 / √2 is irrational.
(ii) 7√5
Assume 7√5 is rational.
Then √5 = rational / 7 ⇒ rational.
But √5 is irrational.
So, 7√5 is irrational.
(iii) 6 + √2
Assume 6 + √2 is rational.
Then √2 = rational − 6 ⇒ rational.
But √2 is irrational.
So, 6 + √2 is irrational.
Class 10 Maths Ch 1 Real Numbers Ex 1.2-Extra Questions and Answers
Question 1:
Find the LCM and HCF of 26 and 91 and verify that HCF × LCM = Product of the numbers.
Solution:
Step 1: Find the Prime Factorization
- 26=2×13
- 91=7×13
Step 2: Calculate HCF
- Common factor = 13
- So, HCF (26, 91) = 13
Step 3: Calculate LCM
- LCM = 2×7×13=182
- So, LCM (26, 91) = 182
Step 4: Verify HCF × LCM = Product of the Numbers
- HCF × LCM = 13 × 182 = 2366
- Product of numbers = 26 × 91 = 2366
- Since both values are equal, the verification is correct
Question 2:
Find the HCF and LCM of 510 and 92 using the prime factorization method.
Solution:
Step 1: Prime Factorization
- 510=2×3×5×17
- 92=2×2×23
Step 2: HCF (Greatest Common Factor)
- Common factor = 2
- So, HCF (510, 92) = 2
Step 3: LCM (Least Common Multiple)
- LCM = 22×3×5×17=23460
- So, LCM (510, 92) = 23460
Question 3:
Prove that HCF × LCM = Product of the two numbers for 336 and 54.
Solution:
Step 1: Prime Factorization
- 336=24×3×7
- 54=2×33
Step 2: Calculate HCF
- Common factors: 212^121 and 313^131
- So, HCF (336, 54) = 2 × 3 = 6
Step 3: Calculate LCM
- LCM =3024
- So, LCM (336, 54) = 3024
Step 4: Verification
- HCF × LCM = 6 × 3024 = 18144
- Product of numbers = 336 × 54 = 18144
- Since both values are equal, the verification is correct
Class-wise Solutions – Class 10 Maths Ch 1 Real Numbers Ex 1.2
Class 12:
Class 12 Physics – NCERT Solutions
Class 12 Chemistry – NCERT Solutions
Class 11:
- Class 11 Physics – NCERT Solutions
- Class 11 Chemistry – NCERT Solutions
- Class 11 Biology – NCERT Solutions
- Class 11 Math – NCERT Solutions
Class 10:
Class 9:
Class 8:
Class 7:
Class 6:
Subject-wise Solutions – Class 10 Maths Ch 1 Real Numbers Ex 1.2
Physics:
Chemistry:
Biology:
Math:
- Class 11 Math – NCERT Solutions
- Class 10 Math – NCERT Solutions
- Class 9 Math – NCERT Solutions
- Class 8 Math – NCERT Solutions
Science:
- Class 10 Science – NCERT Solutions
- Class 9 Science – NCERT Solutions
- Class 8 Science – Oxford Solutions
- Class 7 Science – Oxford Solutions
- Class 6 Science – Oxford Solutions
NEET BIOLOGY
- Evolution
- Breathing and Exchange of Gases
- Anatomy of Flowering Plants
- Body Fluids and Circulation
- Human Health and Disease
- Microbes in Human Welfare
- Cell Cycle and Cell Division
- Biotechnology and Its Applications
- Biodiversity and Conservation
- Morphology of Flowering Plants
For the official Class 8 Mathematics Solutions, you can visit:
- NCERT Textbooks (for Class 8):