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  1. Chapter 6 Class 8 Squares and Square Roots
  2. Serial order wise
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Ex 6.3, 10 Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.Smallest square number divisible by 8, 15, 20 = L.C.M of 8, 15, 20 Or Multiple of L.C.M Finding L.C.M of 8, 15, 20 L.C.M of 8, 15, 20 = 2 × 2 × 2 × 3 × 5 = 4 × 6 × 5 = 4 × 30 = 120 Checking if 120 is a perfect square We see that, 120 = 2 × 2 × 2 × 3 × 5 Here, 2, 3 & 5 do not occur in pairs ∴ 120 is not a perfect square So, we multiply by 2, 3 and 5 to make pairs So, our number becomes 120 × 2 × 3 × 5 = 2 × 2 × 2 × 3 × 5 × 2 × 3 × 5 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 Now, it becomes a perfect square. So, required number is 3600

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.