If the arithmetic and geometric means of two numbers are 10, 8 respectively, then one number exceeds the other number by

Options

A. 8
B. 10
C. 12
D. 16

Explanation

To solve the problem, we need to find the two numbers \( A \) and \( B \) given that their arithmetic mean is 10 and their geometric mean is 8. We will then determine how much one number exceeds the other.

### Step-by-Step Solution:

1. **Use the formula for Arithmetic Mean (AM)**:
\[ \text{AM} = \frac{A + B}{2} \]
Given that the arithmetic mean is 10, we can set up the equation:
\[ \frac{A + B}{2} = 10 \]
Multiplying both sides by 2 gives:
\[ A + B = 20 \quad \text{(Equation 1)} \]

2. **Use the formula for Geometric Mean (GM)**:
\[ \text{GM} = \sqrt{AB} \]
Given that the geometric mean is 8, we can set up the equation:
\[ \sqrt{AB} = 8 \]
Squaring both sides gives:
\[ AB = 64 \quad \text{(Equation 2)} \]

3. **Substitute Equation 1 into Equation 2**:
From Equation 1, we can express \( B \) in terms of \( A \):
\[ B = 20 - A \]
Now substitute \( B \) into Equation 2:
\[ A(20 - A) = 64 \]
Expanding this gives:
\[ 20A - A^2 = 64 \]
Rearranging the equation results in:
\[ A^2 - 20A + 64 = 0 \]

4. **Solve the quadratic equation**:
We can use the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -20, c = 64 \).
\[ A = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \]
\[ A = \frac{20 \pm \sqrt{400 - 256}}{2} \]
\[ A = \frac{20 \pm \sqrt{144}}{2} \]
\[ A = \frac{20 \pm 12}{2} \]
This gives us two possible values for \( A \):
\[ A = \frac{32}{2} = 16 \quad \text{or} \quad A = \frac{8}{2} = 4 \]

5. **Find the corresponding values of \( B \)**:
If \( A = 16 \):
\[ B = 20 - 16 = 4 \]
If \( A = 4 \):
\[ B = 20 - 4 = 16 \]

6. **Determine how much one number exceeds the other**:
The difference between the two numbers is:
\[ |A - B| = |16 - 4| = 12 \]

### Final Answer:
One number exceeds the other by **12**.