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If (H) is the HCF and (L) is the LCM of (2^6\times3^2\times5^2) and (2^4\times3^5\times5), what is (\frac{L}{H})?
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Answer and explanation
Correct answer: (2^2\times3^3\times5)
Step 1: (H=2^4\times3^2\times5) and (L=2^6\times3^5\times5^2). Step 2: (\frac{L}{H}=2^{6-4}\times3^{5-2}\times5^{2-1}=2^2\times3^3\times5). Step 3: In division, subtract exponents of the same base.
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What is the correct answer to this question?
(2^2\times3^3\times5)
Why is this the correct answer?
Step 1: (H=2^4\times3^2\times5) and (L=2^6\times3^5\times5^2). Step 2: (\frac{L}{H}=2^{6-4}\times3^{5-2}\times5^{2-1}=2^2\times3^3\times5). Step 3: In division, subtract exponents of the same base.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: HCF and LCM using prime factorisation.
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