The A.M. of two positive numbers is 15 and their G.M. is 12 what is the larger number?

Options

A. 8
B. 12
C. 6
D. 6

Answer

To solve the problem, we need to find the larger of two positive numbers given that their Arithmetic Mean (A.M.) is 15 and their Geometric Mean (G.M.) is 12.

### Step-by-Step Solution:

1. **Define the Variables**:
Let the two positive numbers be \( x \) and \( y \).

2. **Set Up the Equation for A.M.**:
The formula for the Arithmetic Mean of two numbers is:
\[ A.M. = \frac{x + y}{2} \]
Given that the A.M. is 15, we can write:
\[ \frac{x + y}{2} = 15 \]
Multiplying both sides by 2 gives:
\[ x + y = 30 \quad \text{(Equation 1)} \]

3. **Set Up the Equation for G.M.**:
The formula for the Geometric Mean of two numbers is:
\[ G.M. = \sqrt{xy} \]
Given that the G.M. is 12, we can write:
\[ \sqrt{xy} = 12 \]
Squaring both sides gives:
\[ xy = 144 \quad \text{(Equation 2)} \]

4. **Substitute Equation 1 into Equation 2**:
From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 30 - x \]
Now substitute this expression for \( y \) into Equation 2:
\[ x(30 - x) = 144 \]
Expanding this gives:
\[ 30x - x^2 = 144 \]
Rearranging it leads to:
\[ x^2 - 30x + 144 = 0 \quad \text{(Equation 3)} \]

5. **Solve the Quadratic Equation**:
We can solve Equation 3 using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -30 \), and \( c = 144 \):
\[ x = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot 144}}{2 \cdot 1} \]
Simplifying further:
\[ x = \frac{30 \pm \sqrt{900 - 576}}{2} \]
\[ x = \frac{30 \pm \sqrt{324}}{2} \]
\[ x = \frac{30 \pm 18}{2} \]
This gives us two possible values for \( x \):
\[ x = \frac{48}{2} = 24 \quad \text{and} \quad x = \frac{12}{2} = 6 \]

6. **Find the Corresponding Values of \( y \)**:
If \( x = 24 \):
\[ y = 30 - 24 = 6 \]
If \( x = 6 \):
\[ y = 30 - 6 = 24 \]

7. **Determine the Larger Number**:
In both cases, the larger number is \( 24 \).

### Final Answer:
The larger number is **24**.