**NCERT/CBSE MATHS Class 10 Ex- 1.2 Q No. 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**

Hy Friends Welcome On **NCERT MATHS SOLUTIONS** !! Today we are going to solve the Question : **”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” **of** Exercise 1.2 Question No 3 Solutions (Real Numbers)** of Class 10th, which will prove to be very helpful for you.

# CBSE/NCERT MATHS Class 10 Ex- 1.2 Q No.3 Solutions

If you are a student of CBSE, today we are going to give you the **CBSE / NCERT** **Chapter : Real Numbers Exercise 1.2 Question No – 3** **Solutions** . Hope you like this post about Class 10th Maths Solution.

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**Real Numbers CHAPTER : 1**

Symbol of real numbers (ℝ) In mathematics, the real number is the value presented to any amount corresponding to the simple line. Actual numbers include all rational numbers such as -5 and fractional numbers such as 4/3 and all irrational numbers such as √2 (1.41421356 …, square root of 2, an unregulated algebraic number). By incorporating the ample numbers in the actual numbers, they can be presented from the eternal points that can be attributed on a line in the form of a real number line.

### Exercise- 1.2 Question No. 1 : **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**

**Class 10 Exercise 1.2 Question No. 3 Solutions**

**(i) 12, 15 and 21**

12 = 2^{2}×3

15 = 2×3

21 = 3×7

HCF = 3

LCM = 2^{2} ×3 ×5 ×7 = 420

**(ii) 17, 13 and 29**

17 = 1×17

23 = 1×23

29 = 1×29

HCF = 1

LCM = 17 × 23 × 29 = 11339

**(iii) 8, 9 and 25**

8 = 2×2×2

9 = 3×3

25 = 5×5

HCF = 1

LCM = 2×2×2×3×3×5×5 = 1800

**CLICK ON RED BOX FOR CBSE Maths Class 10 Exercise 1.2 Real Numbers Next Question Answer :-**

**NCERT MATHS Class 10 Ex- 1.2 Q No. 4 : Given that HCF (306, 657) = 9, find LCM (306, 657).**

**SEE NCERT Maths Class 10 Chapter 1 Ex-1.2 All Questions Solutions **

**NCERT MATHS Class 10 Ex- 1.2 Q No. 1 : Express each number as product of its prime factors : 140,156****NCERT MATHS Class 10 Ex- 1.2 Q No. 2 : Find the LCM and HCF of the following pairs of integers and verify that LCM****NCERT/CBSE MATHS Class 10 Ex- 1.2 Q No. 3 : Find the LCM and HCF of the following integers by applying the prime****NCERT MATHS Class 10 Ex- 1.2 Q No. 4 : Given that HCF (306, 657) = 9, find LCM (306, 657).****CBSE/NCERT MATHS Class 10 Ex- 1.2 Q No. 5 : Check whether 6n can end with the digit 0 for any natural number n.****CBSE/NCERT MATHS Class 10 Ex- 1.2 Q No.6 Solutions : Explain why 7×11× 13 + 13 and****NCERT/CBSE MATHS Class 10 Ex- 1.2 Q No.7 Solutions :There is a circular path around a sports field. Sonia takes 18 minutes to drive**

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