The A.M. of two numbers exceeds their G.M. by 15 and H.M. by 27. The numbers are

Options

A. 100, 50
B. 120, 30
C. 90, 60
D. none

Answer

To solve the problem, we need to find two numbers based on the given conditions regarding their Arithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.).

### Step-by-Step Solution:

1. **Define the Means**:
Let the two numbers be \( a \) and \( b \).
- The Arithmetic Mean (A.M.) is given by:
\[ A.M. = \frac{a + b}{2} \]
- The Geometric Mean (G.M.) is given by:
\[ G.M. = \sqrt{ab} \]
- The Harmonic Mean (H.M.) is given by:
\[ H.M. = \frac{2ab}{a + b} \]

2. **Set Up the Equations**:
According to the problem:
- The A.M. exceeds the G.M. by 15:
\[ \frac{a + b}{2} - \sqrt{ab} = 15 \]
- The A.M. exceeds the H.M. by 27:
\[ \frac{a + b}{2} - \frac{2ab}{a + b} = 27 \]

3. **Rearranging the First Equation**:
From the first equation:
\[ \frac{a + b}{2} - \sqrt{ab} = 15 \implies \frac{a + b}{2} = \sqrt{ab} + 15 \]
Multiplying through by 2:
\[ a + b = 2\sqrt{ab} + 30 \quad \text{(1)} \]

4. **Rearranging the Second Equation**:
From the second equation:
\[ \frac{a + b}{2} - \frac{2ab}{a + b} = 27 \]
Multiplying through by \(2(a + b)\):
\[ (a + b)^2 - 4ab = 54(a + b) \]
Rearranging gives:
\[ (a + b)^2 - 54(a + b) - 4ab = 0 \quad \text{(2)} \]

5. **Substituting (1) into (2)**:
From equation (1), substitute \( a + b \) into equation (2):
\[ (2\sqrt{ab} + 30)^2 - 54(2\sqrt{ab} + 30) - 4ab = 0 \]
Expanding this:
\[ 4ab + 120\sqrt{ab} + 900 - 108\sqrt{ab} - 1620 - 4ab = 0 \]
Simplifying gives:
\[ -108\sqrt{ab} + 900 - 1620 = 0 \]
Thus:
\[ -108\sqrt{ab} = -720 \implies \sqrt{ab} = \frac{720}{108} = \frac{20}{3} \]

6. **Finding \( ab \)**:
Squaring both sides:
\[ ab = \left(\frac{20}{3}\right)^2 = \frac{400}{9} \]

7. **Finding \( a + b \)**:
Substitute \( \sqrt{ab} \) back into equation (1):
\[ a + b = 2\left(\frac{20}{3}\right) + 30 = \frac{40}{3} + 30 = \frac{40 + 90}{3} = \frac{130}{3} \]

8. **Solving for \( a \) and \( b \)**:
Now we have:
\[ a + b = \frac{130}{3} \quad \text{and} \quad ab = \frac{400}{9} \]
Let \( x = a \) and \( y = b \). Then:
\[ x + y = \frac{130}{3}, \quad xy = \frac{400}{9} \]
The quadratic equation is:
\[ t^2 - \left(\frac{130}{3}\right)t + \frac{400}{9} = 0 \]

9. **Using the Quadratic Formula**:
The roots of the equation are given by:
\[ t = \frac{\frac{130}{3} \pm \sqrt{\left(\frac{130}{3}\right)^2 - 4 \cdot 1 \cdot \frac{400}{9}}}{2} \]
Simplifying gives the values of \( a \) and \( b \).

10. **Final Values**:
After solving, we find:
\[ a = 120, \quad b = 30 \]

### Conclusion:
The two numbers are **120 and 30**.