Ex 1.2, 2 - Find LCM and HCF of the following pairs (i) 26 & 91 (ii)

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

Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 3

Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 4
Ex 1.2, 2 - Chapter 1 Class 10 Real Numbers - Part 5

 

  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. 26 and 91 H.C.F = Product of smallest power of each common prime factor = 13 L.C.M = Product of greatest power of each prime factor = 2 × 13 × 7 = 182 Now, we have to verify that H.C.F × L.C.M = Product of 2 numbers H.C.F × L.C.M = 13 × 182 = 2386 Product of two numbers = 26 × 91 = 2386 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. 26 and 91 26 = 2 × 13 91 = 7 × 13 H.C.F = 13 Finding L.C.M L.C.M = 2 × 13 × 7 = 182 Now, we have to verify that H.C.F × L.C.M = Product of 2 numbers H.C.F × L.C.M = 13 × 182 = 2386 Product of two numbers = 26 × 91 = 2386 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.