An ocean depth finder shows the number of feet in the depth of water at a certain place. The difference between d, the actual depth of the water, and the depth finder reading , x, is |d-x| and must be less than or equal to 0.05d. If the depth finder reading is 620 feet, what is the maximum value of the actual depth of the water, to the nearest foot?

Answer

To solve the problem, we need to find the maximum value of the actual depth of water \( d \) given the depth finder reading \( x = 620 \) feet and the condition that the difference between \( d \) and \( x \) must be less than or equal to \( 0.05d \).

### Step-by-Step Solution:

1. **Set up the inequality**:
We know that the difference between the actual depth \( d \) and the depth finder reading \( x \) is given by:
\[ |d - x| \leq 0.05d \]
Substituting \( x = 620 \):
\[ |d - 620| \leq 0.05d \]

2. **Consider the two cases for the absolute value**:
The absolute value inequality can be split into two cases:
- Case 1: \( d - 620 \leq 0.05d \)
- Case 2: \( 620 - d \leq 0.05d \)

3. **Solve Case 1**:
For \( d - 620 \leq 0.05d \):
\[ d - 620 \leq 0.05d \]
Rearranging gives:
\[ d - 0.05d \leq 620 \]
\[ 0.95d \leq 620 \]
Dividing both sides by 0.95:
\[ d \leq \frac{620}{0.95} \approx 652.63 \]

4. **Solve Case 2**:
For \( 620 - d \leq 0.05d \):
\[ 620 - d \leq 0.05d \]
Rearranging gives:
\[ 620 \leq d + 0.05d \]
\[ 620 \leq 1.05d \]
Dividing both sides by 1.05:
\[ d \geq \frac{620}{1.05} \approx 590.48 \]

5. **Combine the results**:
From Case 1, we found that \( d \leq 652.63 \) and from Case 2, we found that \( d \geq 590.48 \). Therefore, the actual depth \( d \) must satisfy:
\[ 590.48 \leq d \leq 652.63 \]

6. **Determine the maximum value of \( d \)**:
The maximum value of \( d \) is approximately \( 652.63 \). Rounding to the nearest foot, we get:
\[ d \approx 653 \text{ feet} \]

### Final Answer:
The maximum value of the actual depth of the water, to the nearest foot, is **653 feet**.