The difference of two numbers is 20 and their product is 56.25 times their difference. Find the LCM of the numbers.

Options

A. 70
B. 1125
C. 225
D. 5

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript.

### Step 1: Define the Variables
Let the two numbers be \( a \) and \( b \).

### Step 2: Set Up the Equations
We know from the problem statement:
1. The difference of the two numbers is 20:
\[ a - b = 20 \quad \text{(Equation 1)} \]
2. The product of the two numbers is 56.25 times their difference:
\[ a \times b = 56.25 \times 20 \]
Calculating \( 56.25 \times 20 \):
\[ 56.25 \times 20 = 1125 \quad \text{(Equation 2)} \]

### Step 3: Express \( b \) in Terms of \( a \)
From Equation 1, we can express \( b \) in terms of \( a \):
\[ b = a - 20 \]

### Step 4: Substitute \( b \) in Equation 2
Substituting \( b \) into Equation 2:
\[ a \times (a - 20) = 1125 \]
Expanding this gives:
\[ a^2 - 20a = 1125 \]

### Step 5: Rearrange the Equation
Rearranging the equation to standard quadratic form:
\[ a^2 - 20a - 1125 = 0 \]

### Step 6: Factor the Quadratic Equation
Now we need to factor the quadratic equation:
\[ a^2 - 45a + 25a - 1125 = 0 \]
Grouping the terms:
\[ a(a - 45) + 25(a - 45) = 0 \]
Factoring out \( (a - 45) \):
\[ (a - 45)(a + 25) = 0 \]

### Step 7: Solve for \( a \)
Setting each factor to zero gives:
1. \( a - 45 = 0 \) → \( a = 45 \)
2. \( a + 25 = 0 \) → \( a = -25 \) (we reject this as we need positive numbers)

Thus, \( a = 45 \).

### Step 8: Find \( b \)
Now substituting \( a \) back into the equation for \( b \):
\[ b = 45 - 20 = 25 \]

### Step 9: Find the LCM of \( a \) and \( b \)
Now we need to find the LCM of \( 45 \) and \( 25 \).

**Prime Factorization:**
- \( 45 = 3^2 \times 5^1 \)
- \( 25 = 5^2 \)

**Finding LCM:**
The LCM is found by taking the highest power of each prime factor:
\[ \text{LCM} = 3^2 \times 5^2 = 9 \times 25 = 225 \]

### Final Answer
The LCM of the two numbers is \( \boxed{225} \).