Question

The HCF and LCM of two numbers are \(12\) and \(360\). If the numbers differ by \(12\), their sum will be:

• A. \(132\)
• B. \(348\)
• C. \(86\)
• D. \(246\)

Hint:

For two numbers, \(\text{Product}=\text{HCF}\times\text{LCM}\). If HCF is given, write numbers as HCF multiples.

Answer 1

The Correct Option is A

Let the two numbers be: \[ 12x \] and \[ 12y. \] Since the HCF is \(12\), \(x\) and \(y\) must be co-prime. We know: \[ \text{Product of two numbers}=\text{HCF}\times\text{LCM}. \] So: \[ (12x)(12y)=12\times360. \] \[ 144xy=4320. \] \[ xy=30. \] Also, the numbers differ by \(12\). So: \[ 12x-12y=12. \] \[ 12(x-y)=12. \] \[ x-y=1. \] Now we need two co-prime numbers whose product is \(30\) and whose difference is \(1\). The numbers are: \[ 5 \] and: \[ 6. \] Thus, the original numbers are: \[ 12\times5=60 \] and: \[ 12\times6=72. \] Their sum is: \[ 60+72=132. \] Therefore, the required sum is: \[ 132. \]