The HCF of two numbers is (2^2\times 3^2) and their LCM is (2^5\times 3^2\times 5\times 7). If one number is (2^5\times 3^2\times 5), what is the other number?
Answer and explanation
Correct answer: (2^2\times 3^2\times 7)
Step 1: The other number (=\frac{\text{HCF}\times \text{LCM}}{\text{first number}}). Step 2: Using exponents, (\frac{(2^2\times 3^2)(2^5\times 3^2\times 5\times 7)}{2^5\times 3^2\times 5}=2^2\times 3^2\times 7). Step 3: In such problems, simplify by subtracting exponents.
Frequently asked questions
What is the correct answer to this question?
(2^2\times 3^2\times 7)
Why is this the correct answer?
Step 1: The other number (=\frac{\text{HCF}\times \text{LCM}}{\text{first number}}). Step 2: Using exponents, (\frac{(2^2\times 3^2)(2^5\times 3^2\times 5\times 7)}{2^5\times 3^2\times 5}=2^2\times 3^2\times 7). Step 3: In such problems, simplify by subtracting exponents.
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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