Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 6

Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 7

Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 8

Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 9

Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 10

 

 

  1. Chapter 1 Class 10 Real Numbers (Term 1)
  2. Serial order wise

Transcript

Ex 1.2 , 2 (Method 1) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (ii) 510 and 92 510 = 2 × 3 × 5 × 17 = 21 × 31 × 51 × 171 92 = 2 × 2 × 23 = 22 × 231 H.C.F = Product of smallest power of each common prime factor = 21 = 2 L.C.M = Product of greatest power of each prime factor = 22 × 3 × 5 × 17 × 23 = 23460 Now, we have to verify that H.C.F × L.C.M = Product of 2 numbers H.C.F × L.C.M = 2 × 23460 = 46920 Product of two numbers = 510 × 92 = 46920 Since L.H.S = R.H.S Hence verified Ex 1.2 , 2 (Method 2) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (ii) 510 and 92 510 = 2 × 3 × 5 × 17 92 = 2 × 2 × 23 H.C.F = 2 Finding L.C.M L.C.M = 2 × 2 × 3 × 5 × 17 × 23 = 23460 Now, we have to verify that H.C.F × L.C.M = Product of 2 numbers H.C.F × L.C.M = 2 × 23460 = 46920 Product of two numbers = 510 × 92 = 46920 Since L.H.S = R.H.S Hence verified

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.