A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/hour and reaches school 5 minutes early. How far is the school from his house?

Options

A. A) `(5)/(8)km`
B. B) 8km
C. C) 5km
D. D) 10km

Answer

To solve the problem, we need to find the distance from the student's house to the school using the information provided about his speed and the time he is late or early.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let the distance from the house to the school be \( d \) km.
- Speed on the first day = 8 km/h.
- Speed on the second day = 10 km/h.
- Time late on the first day = 2.5 minutes.
- Time early on the second day = 5 minutes.

2. **Convert Time to Hours:**
- Convert the time late and early from minutes to hours:
- 2.5 minutes = \( \frac{2.5}{60} \) hours = \( \frac{1}{24} \) hours.
- 5 minutes = \( \frac{5}{60} \) hours = \( \frac{1}{12} \) hours.

3. **Calculate Total Time Difference:**
- The total time difference between the two days is:
\[ \text{Total time difference} = \left(\frac{1}{24} + \frac{1}{12}\right) \text{ hours} \]
- To add these fractions, find a common denominator (which is 24):
\[ \frac{1}{12} = \frac{2}{24} \]
- Thus,
\[ \text{Total time difference} = \frac{1}{24} + \frac{2}{24} = \frac{3}{24} = \frac{1}{8} \text{ hours} \]

4. **Use the Formula for Distance:**
- The formula for distance when two speeds and the time difference are known is:
\[ d = \frac{(v_1 \times v_2) \times \Delta t}{(v_2 - v_1)} \]
- Here, \( v_1 = 8 \) km/h, \( v_2 = 10 \) km/h, and \( \Delta t = \frac{1}{8} \) hours.
- Plugging in the values:
\[ d = \frac{(8 \times 10) \times \frac{1}{8}}{(10 - 8)} \]

5. **Simplify the Expression:**
- Calculate the numerator:
\[ 8 \times 10 = 80 \]
- Now substitute back into the equation:
\[ d = \frac{80 \times \frac{1}{8}}{2} \]
- Simplify:
\[ d = \frac{80/8}{2} = \frac{10}{2} = 5 \text{ km} \]

### Final Answer:
The distance from the student's house to the school is **5 km**.