The ratio of the sum, difference and product of two numbers is `11: 1: 120.` What is the sum of squares ?

Options

A. 549
B. 808
C. 656
D. 976

Answer

To solve the problem, we need to find the sum of squares of two numbers given the ratio of their sum, difference, and product.

Let's denote the two numbers as \( A \) and \( B \).

### Step 1: Set up the equations based on the given ratio
We know from the problem that:
- The sum of the numbers \( A + B \) is in the ratio of 11.
- The difference of the numbers \( A - B \) is in the ratio of 1.
- The product of the numbers \( A \cdot B \) is in the ratio of 120.

We can express these relationships as:
\[ A + B = 11k \]
\[ A - B = k \]
\[ A \cdot B = 120m \]

Here, \( k \) and \( m \) are constants that we will determine.

### Step 2: Solve for \( A \) and \( B \)
We can solve for \( A \) and \( B \) using the equations for sum and difference.

Adding the equations for sum and difference:
\[ (A + B) + (A - B) = 11k + k \]
\[ 2A = 12k \implies A = 6k \]

Subtracting the difference from the sum:
\[ (A + B) - (A - B) = 11k - k \]
\[ 2B = 10k \implies B = 5k \]

### Step 3: Substitute \( A \) and \( B \) into the product equation
Now, substitute \( A \) and \( B \) into the product equation:
\[ A \cdot B = (6k)(5k) = 30k^2 \]
According to the ratio, this should equal \( 120m \):
\[ 30k^2 = 120m \implies k^2 = 4m \implies k = 2\sqrt{m} \]

### Step 4: Find the sum of squares
The sum of squares of \( A \) and \( B \) is given by:
\[ A^2 + B^2 = (6k)^2 + (5k)^2 = 36k^2 + 25k^2 = 61k^2 \]

### Step 5: Substitute \( k^2 \) in terms of \( m \)
Using \( k^2 = 4m \):
\[ A^2 + B^2 = 61(4m) = 244m \]

### Step 6: Determine \( m \)
Since we don't have a specific value for \( m \), we can express the final answer in terms of \( m \).

### Conclusion
The sum of squares of the two numbers is \( 244m \). If we assume \( m = 1 \) for simplicity, then the sum of squares is \( 244 \).