The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the number is :

The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the number is :
[A]3
[B]1
[C]4
[D]2

2
Let the number be 15x and 15y, where x and y are co-prime.
Since 15x \times 15y = 6300
=> xy = \frac{6300}{15\times 15} = 28
So, two pairs are (7, 4) and (14, 2)
Hence the right option is [D].

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