1. Chapter 1 Class 10 Real Numbers
  2. Serial order wise
  3. Ex 1.1
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Ex 1.1, 4 - Chapter 1 Class 10 Real Numbers

Last updated at Dec. 13, 2024 by Teachoo

Transcript

Ex 1.1, 4 Given that HCF (306, 657) = 9, find LCM (306, 657). We know that H.C.F × L.C.M = Product of numbers 9 × L.C.M = 306 × 657 L.C.M = (306 × 657)/9 L.C.M = 22338 ∴ L.C.M of 306 & 657 is 22338
  1. Chapter 1 Class 10 Real Numbers
  2. Serial order wise

    3. Ex 1.1

    Ex 1.2 →
Ex 1.2 →
Chapter 1 Class 10 Real Numbers
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo