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# 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.

Here we would like to tell you that there are other websites on which the students demanded the Solution of NCERT MATHS Chapters, we had promised them that we will soon send you the questions of Class 9, And we completed our promise to them.We asked the good knowledgeable teachers of mathematics and asked them to solve the issues of CBSE’s mathematical questions and also said that the answers to our questions are such that the students have no difficulty in solving all those questions.

### 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.

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

(i)  12, 15 and 21

12 = 22×3

15 = 2×3

21 = 3×7

HCF = 3

LCM = 22 ×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