A student walks from point A to point B at 12 miles per hour and reaches 5 minutes late. Had he walked at 16 miles per hour, he would have been 5 minutes early. What is his noNAl rate of walking?

Options

A. `sqrt (12 times 16)`
B. `(2 times 12 times 16)/(12+16)` sec kmph
C. `(2 times 16)/(12+16)` sec kmph
D. `(2 times 16)/2` sec kmph

Answer

To solve the problem, we need to determine the normal speed of the student when he is neither late nor early. We will denote the normal speed as \( T \) miles per hour.

### Step-by-Step Solution:

1. **Understand the Problem**:
- The student walks from point A to point B at 12 miles per hour and is 5 minutes late.
- If he walks at 16 miles per hour, he is 5 minutes early.
- We need to find his normal speed \( T \) miles per hour.

2. **Convert Time into Hours**:
- Since the student is late by 5 minutes when walking at 12 mph, we convert 5 minutes into hours:
\[ 5 \text{ minutes} = \frac{5}{60} \text{ hours} = \frac{1}{12} \text{ hours} \]

3. **Set Up the Equations**:
- Let \( D \) be the distance from A to B.
- The time taken to cover the distance at normal speed \( T \) is \( \frac{D}{T} \).
- When walking at 12 mph, the time taken is:
\[ \frac{D}{12} = \frac{D}{T} + \frac{1}{12} \]
- When walking at 16 mph, the time taken is:
\[ \frac{D}{16} = \frac{D}{T} - \frac{1}{12} \]

4. **Rearranging the Equations**:
- From the first equation:
\[ \frac{D}{12} - \frac{D}{T} = \frac{1}{12} \]
- From the second equation:
\[ \frac{D}{T} - \frac{D}{16} = \frac{1}{12} \]

5. **Combine the Two Equations**:
- We can rearrange both equations:
- First equation becomes:
\[ \frac{D}{T} = \frac{D}{12} - \frac{1}{12} \]
- Second equation becomes:
\[ \frac{D}{T} = \frac{D}{16} + \frac{1}{12} \]

6. **Set the Two Expressions for \( \frac{D}{T} \) Equal**:
\[ \frac{D}{12} - \frac{1}{12} = \frac{D}{16} + \frac{1}{12} \]

7. **Clear the Fractions**:
- Multiply through by \( 48T \) (the common denominator):
\[ 4D - 4T = 3D + 4T \]

8. **Rearranging**:
- Combine like terms:
\[ 4D - 3D = 4T + 4T \]
\[ D = 8T \]

9. **Substituting Back**:
- Substitute \( D = 8T \) back into one of the original equations to find \( T \):
\[ \frac{8T}{12} = \frac{8T}{T} + \frac{1}{12} \]
\[ \frac{2T}{3} = 8 + \frac{1}{12} \]

10. **Solve for \( T \)**:
- Solve the equation to find the value of \( T \):
\[ \frac{2T}{3} = 8 + \frac{1}{12} \]
- Convert 8 into twelfths:
\[ 8 = \frac{96}{12} \]
\[ \frac{2T}{3} = \frac{96 + 1}{12} = \frac{97}{12} \]
- Cross-multiply to solve for \( T \):
\[ 2T \cdot 12 = 3 \cdot 97 \]
\[ 24T = 291 \]
\[ T = \frac{291}{24} = 12.125 \text{ miles per hour} \]

### Final Answer:
The normal speed of the student is approximately **12.125 miles per hour**.