Question

The LCM and HCF of two numbers are $144$ and $12$ respectively. How many such pairs of numbers are possible?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$2$

Let the numbers be $12x;12y$, where $x,y$ are co-prime to each other.
We know that Product of two numbers $=LCM\times HCF$
Then $12x\times 12y=144\times 12$
$=>xy=12$
Now, we need to find co-prime numbers whose product is $12$
$12=2^2\times 3$
Therefore co-prime pairs will be $(1,12);(4,3)$
Therefore, only two pairs of such numbers satisfy the given condition.