The difference of two numbers is 20 and their product is 56. 25 times their difference . Find the LCM of the numbers

Options

A. 225
B. 625
C. 1125
D. 825

Answer

To solve the problem step by step, we need to find two numbers based on the conditions given: their difference is 20 and their product is 56.25 times their difference. Finally, we will find the Least Common Multiple (LCM) of these two numbers.

### Step 1: Set up the equations
Let the two numbers be \( a \) and \( b \). According to the problem:
1. The difference of the two numbers is given by:
\[ a - b = 20 \quad \text{(1)} \]
2. The product of the two numbers is given by:
\[ a \cdot b = 56.25 \times \text{difference} = 56.25 \times 20 = 1125 \quad \text{(2)} \]

### Step 2: Express \( a \) in terms of \( b \)
From equation (1), we can express \( a \) in terms of \( b \):
\[ a = b + 20 \quad \text{(3)} \]

### Step 3: Substitute \( a \) in the product equation
Now, substitute equation (3) into equation (2):
\[ (b + 20) \cdot b = 1125 \]
Expanding this gives:
\[ b^2 + 20b = 1125 \]
Rearranging the equation:
\[ b^2 + 20b - 1125 = 0 \quad \text{(4)} \]

### Step 4: Solve the quadratic equation
Now we will solve the quadratic equation (4) using factorization. We need to find two numbers that multiply to \(-1125\) and add to \(20\).

The factors of \(1125\) can be checked:
- \(45\) and \(-25\) work because:
\[ 45 \times (-25) = -1125 \quad \text{and} \quad 45 + (-25) = 20 \]

Thus, we can rewrite the quadratic equation as:
\[ (b + 45)(b - 25) = 0 \]

### Step 5: Find the values of \( b \)
Setting each factor to zero gives:
1. \( b + 45 = 0 \) → \( b = -45 \) (not valid since \( b \) must be positive)
2. \( b - 25 = 0 \) → \( b = 25 \)

### Step 6: Find the value of \( a \)
Now, substitute \( b = 25 \) back into equation (3) to find \( a \):
\[ a = 25 + 20 = 45 \]

### Step 7: Find the LCM of \( a \) and \( b \)
Now we have \( a = 45 \) and \( b = 25 \). To find the LCM, we can use the prime factorization method:
- The prime factorization of \( 45 \) is \( 3^2 \times 5^1 \).
- The prime factorization of \( 25 \) is \( 5^2 \).

To find the LCM, we take the highest power of each prime:
- For \( 3 \): \( 3^2 \)
- For \( 5 \): \( 5^2 \)

Thus, the LCM is:
\[ LCM = 3^2 \times 5^2 = 9 \times 25 = 225 \]

### Final Answer
The LCM of the numbers is \( 225 \).

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