0 reads0 ratings0 helpful
The HCF of two numbers is (15) and their LCM is (420). How many unordered pairs are possible?
Advertisement
Answer and explanation
Correct answer: (2)
Step 1: Let the numbers be (15m) and (15n), where (m) and (n) are coprime. Step 2: (15mn=420), so (mn=28=2^2\times7); the unordered coprime pairs are ((1,28)) and ((4,7)), so the count is (2). Step 3: (m) and (n) must not share a prime factor.
Frequently asked questions
What is the correct answer to this question?
(2)
Why is this the correct answer?
Step 1: Let the numbers be (15m) and (15n), where (m) and (n) are coprime. Step 2: (15mn=420), so (mn=28=2^2\times7); the unordered coprime pairs are ((1,28)) and ((4,7)), so the count is (2). Step 3: (m) and (n) must not share a prime factor.
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.
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
Advertisement