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

Step 1: Understanding the Question:
We are given the HCF and LCM of two positive integers along with the difference between them. We must find the sum of these two integers.

Step 2: Key Formula or Approach:
Let the two numbers be $a$ and $b$ (with $a > b$).
• The product of two numbers is equal to the product of their HCF and LCM: \[ a \times b = \text{HCF} \times \text{LCM} \]
• We are given: \[ a - b = 12 \]
• We can relate the sum and difference of two numbers using the algebraic identity: \[ (a + b)^2 = (a - b)^2 + 4ab \]

Step 3: Detailed Explanation:

• Calculate the product $ab$: \[ ab = 12 \times 360 = 4320 \]
• Substitute the values into our identity: \[ (a + b)^2 = (12)^2 + 4(4320) \] \[ (a + b)^2 = 144 + 17280 \] \[ (a + b)^2 = 17424 \]
• Calculate the square root to find $(a + b)$: \[ a + b = \sqrt{17424} \] Let us verify $132$: \[ 132 \times 132 = 17424 \] Thus, $a + b = 132$.

• Alternative Co-prime Method:
Since HCF = 12, let the numbers be $12x$ and $12y$, where $x$ and $y$ are co-primes. \[ \text{LCM} = 12xy = 360 \implies xy = 30 \] The difference is: \[ 12x - 12y = 12(x-y) = 12 \implies x-y = 1 \] We find two numbers that multiply to 30 and differ by 1. These are 6 and 5. So, the original numbers are $12(6) = 72$ and $12(5) = 60$. Their sum is $72 + 60 = 132$.

Step 4: Final Answer:
The sum of the two numbers is 132, matching Option (A).