NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers Exercise 1.2: Detailed Guide for Assam State Board Syllabus 2025-2026 (English Medium)

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Free Solutions for Assam State Board Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 in English Medium

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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 – Digital Pipal Academy

Digital Pipal Academy presents clear and easy-to-understand NCERT Solutions for Class 10 Maths Chapter 1, Exercise 1.2, covered under the Assam State Board Syllabus 2025-2026. Our expert faculty provides step-by-step solutions, focusing on the properties and applications of real numbers to enhance student understanding. Aligned with NCERT guidelines, our solutions ensure complete syllabus coverage and help students perform well in their exams.

Responsive Table

Real Number	Solution Link
Exercise 1.1 (Real Number)	Click Here
Exercise 1.2 (Real Number)	Click Here
Exercise 1.3 (Real Number)	Click Here
Exercise 1.4 (Real Number)	Click Here

Exercise 1.2

Question No. :1. Express each number as a product of its prime factors:

(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

(i) 140
Solution:
140
→ 2 × 70
→ 2 × 35
→ 7 × 5
∴ 140 = 2² × 5 × 7

(ii) 156
Solution:
156
→ 2 × 78
→ 2 × 39
→ 3 × 13
∴ 156 = 2² × 3 × 13

(iii) 3825
Solution:
3825
→ 5 × 765
→ 5 × 153
→ 3 × 51
→ 3 × 17
∴ 3825 = 5² × 3² × 17

(iv) 5005
Solution:
5005
→ 5 × 1001
→ 7 × 143
→ 11 × 13
∴ 5005 = 5 × 7 × 11 × 13

(v) 7429
Solution:
7429
→ 17 × 437
→ 19 × 23
∴ 7429 = 17 × 19 × 23

Question No. 2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

Solution:
Prime factorization:
26 = 2 × 13
91 = 7 × 13

HCF = 13
LCM = 2 × 7 × 13 = 182

Verification:
HCF × LCM = 13 × 182 = 2366
Product of the numbers = 26 × 91 = 2366
∴ HCF × LCM = Product of the numbers ✅

(ii) 510 and 92

Solution:
Prime factorization:
510 = 2 × 5 × 3 × 17
92 = 2² × 23

HCF = 2
LCM = 2² × 5 × 3 × 17 × 23 = 23,460

Verification:
HCF × LCM = 2 × 23,460 = 46,920
Product of the numbers = 510 × 92 = 46,920
∴ HCF × LCM = Product of the numbers ✅

(iii) 336 and 54

Solution:
Prime factorization:
336 = 2⁴ × 3 × 7
54 = 2 × 3³

HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3,024

Verification:
HCF × LCM = 6 × 3,024 = 18,144
Product of the numbers = 336 × 54 = 18,144
∴ HCF × LCM = Product of the numbers ✅

Question No. 3: Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15, and 21

Solution:
Prime factorization:
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7

HCF = 3 (Common factor in all three numbers)
LCM = 2² × 3 × 5 × 7 = 420

(ii) 17, 23, and 29

Solution:
Prime factorization:
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29

HCF = 1 (No common factors)
LCM = 1 × 17 × 23 × 29 = 11,339

(iii) 8, 9, and 25

Solution:
Prime factorization:
8 = 2³
9 = 3²
25 = 5²

HCF = 1 (No common factors)
LCM = 2³ × 3² × 5² = 1,800

Question No. 4: Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:
We know that for two numbers $a$ and $b$ :

\text{HCF} (a, b) \times \text{LCM} (a, b) = a \times b

Given:

HCF (306, 657) = 9
Find LCM (306, 657) = ?

\text{LCM} (306, 657) = \frac{306 \times 657}{\text{HCF}}

= \frac{306 \times 657}{9}

= 22,338

Thus, LCM (306, 657) = 22,338 ✅

Question No. 5: Check whether $6^n$ can end with the digit 0 for any natural number $n$ .

Solution:

If $6^n$ (where $n \in \mathbb{N}$ ) ends with the digit 0, then it must be divisible by 10.

6^n = 5 \times m + 0, \quad \text{for some } m \in \mathbb{N}

For a number to be divisible by 10, its prime factorization must include 5 as a factor.

Now, the prime factorization of 6 is:

6 = 2 \times 3

Since 6 is made up of only the prime factors 2 and 3, and does not contain 5, any power of 6 will also never contain 5 as a factor.

By the Fundamental Theorem of Arithmetic, the prime factorization of a number is unique, and since 5 is not a factor of $6^n$ , $6^n$ can never end with the digit 0.

✅ $6^n$ cannot end with the digit 0 for any natural number $n$ .

Question No. 6: Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.

Solution:

A composite number is a number that has more than two factors (i.e., it is not a prime number).

(i) $7 \times 11 \times 13 + 13$

$= 13 (7 \times 11 + 1)$
$= 13 (77 + 1)$
$= 13 \times 78$

Since this number has more than two factors (1, 13, and 78), it is a composite number.

(ii) $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

$= 5 (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$

Since this number is divisible by 5 and has more than two factors (1, 5, and another factor), it is a composite number.

✅ Both numbers have more than two factors, so they are composite numbers.

Question No. 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:

To find after how many minutes they will meet again, we need to determine the Least Common Multiple (LCM) of 18 and 12.

Step 1: Prime factorization
$18 = 2 \times 3^2$
$12 = 2^2 \times 3$

Step 2: Find the LCM

\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36

Sonia and Ravi will meet again at the starting point after 36 minutes. ✅

Question No. 8:

(i) The soldiers in a regiment can stand in rows consisting of 15, 20, or 25 soldiers. Find the least number of soldiers in the regiment.

Solution:

To find the least number of soldiers in the regiment, we need to determine the LCM (15, 20, 25).

Step 1: Prime factorization
$15 = 3 \times 5$
$20 = 2^2 \times 5$
$25 = 5^2$

Step 2: Find the LCM

\text{LCM} = 3 \times 2^2 \times 5^2 = 3 \times 4 \times 25 = 300

The least number of soldiers in the regiment is 300. ✅

(ii) A bell rings every 18 seconds, another bell rings every 60 seconds. If these two bells ring simultaneously at an instant, find after how many seconds they will ring together again.

Solution:

To find the required time, we need to determine the LCM (18, 60).

Step 1: Prime factorization
$18 = 2 \times 3^2$
$60 = 2^2 \times 3 \times 5$

Step 2: Find the LCM

\text{LCM} = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180

The bells will ring together again after 180 seconds (or 3 minutes). ✅

(iii) A radio station plays ‘Assam Sangeet’ once every two days. Another radio station plays the same song once every three days. How many times in 30 days will both radio stations play the song on the same day?

Solution:

Given,
The first radio station plays the song once every 2 days.
The second radio station plays the song once every 3 days.

To determine when both stations play the song together, we need to find the LCM (2, 3).

Step 1: Prime factorization
$2 = 2^1$
$3 = 3^1$

Step 2: Find the LCM

\text{LCM} = 2^1 \times 3^1 = 6

Step 3: Count the occurrences in 30 days
Since the LCM is 6, both stations will play the song together every 6 days.
Total occurrences in 30 days:

\frac{30}{6} = 5

Both radio stations will play the song on the same day 5 times in 30 days. ✅

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NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers Exercise 1.2: Detailed Guide for Assam State Board Syllabus 2025-2026 (English Medium)