If the arithmetic mean of two numbers is 7 and the geometric mean of the same two numbers is `2sqrt(10)`. Then find the numbers x and y respectively, such that `x gt y`.

Options

A. 4, 10
B. 2, 5
C. 5, 2
D. 10, 4

Answer

To solve the problem, we need to find two numbers \( x \) and \( y \) such that \( x > y \), given that their arithmetic mean is 7 and their geometric mean is \( 2\sqrt{10} \).

### Step-by-Step Solution:

1. **Set up the equations based on the means:**
- The arithmetic mean of \( x \) and \( y \) is given by:
\[ \frac{x + y}{2} = 7 \]
- Multiplying both sides by 2, we get:
\[ x + y = 14 \quad \text{(Equation 1)} \]

2. **Use the geometric mean:**
- The geometric mean of \( x \) and \( y \) is given by:
\[ \sqrt{xy} = 2\sqrt{10} \]
- Squaring both sides gives:
\[ xy = (2\sqrt{10})^2 = 4 \times 10 = 40 \quad \text{(Equation 2)} \]

3. **Use the identity for the difference of squares:**
- We use the identity:
\[ (x - y)^2 = (x + y)^2 - 4xy \]
- Substituting the values from Equation 1 and Equation 2:
\[ (x - y)^2 = (14)^2 - 4 \times 40 \]
- Calculating the right side:
\[ (x - y)^2 = 196 - 160 = 36 \]
- Taking the square root gives:
\[ x - y = 6 \quad \text{(Equation 3)} \]

4. **Solve the system of equations:**
- Now we have two equations:
- \( x + y = 14 \) (Equation 1)
- \( x - y = 6 \) (Equation 3)
- We can add these two equations:
\[ (x + y) + (x - y) = 14 + 6 \]
\[ 2x = 20 \implies x = 10 \]
- Now substitute \( x = 10 \) back into Equation 1 to find \( y \):
\[ 10 + y = 14 \implies y = 4 \]

5. **Final result:**
- The numbers are \( x = 10 \) and \( y = 4 \).

### Conclusion:
The numbers \( x \) and \( y \) respectively are \( 10 \) and \( 4 \), where \( x > y \).