A school boy, travelling at 3 km/h reaches his school 32 min late, but if he travels at 5 km/h then he reaches his school 28 min early. Find the distance then he travels every day to reach the school ?

Answer

To solve the problem, we need to find the distance the school boy travels every day to reach his school. We will use the information given about his travel times at different speeds.

### Step-by-Step Solution:

1. **Define Variables:**
Let the distance to the school be \( D \) kilometers.

2. **Calculate Time Taken at Different Speeds:**
- When the boy travels at 3 km/h, the time taken to reach school is:
\[ \text{Time}_1 = \frac{D}{3} \text{ hours} \]
- When he travels at 5 km/h, the time taken is:
\[ \text{Time}_2 = \frac{D}{5} \text{ hours} \]

3. **Set Up the Time Difference Equation:**
According to the problem:
- He is 32 minutes late when traveling at 3 km/h.
- He is 28 minutes early when traveling at 5 km/h.

We can convert these times into hours:
- 32 minutes = \( \frac{32}{60} = \frac{8}{15} \) hours
- 28 minutes = \( \frac{28}{60} = \frac{14}{30} = \frac{7}{15} \) hours

The total time difference between the two scenarios is:
\[ \frac{8}{15} + \frac{7}{15} = 1 \text{ hour} \]

4. **Set Up the Equation:**
The difference in time taken at the two speeds can be expressed as:
\[ \frac{D}{3} - \frac{D}{5} = 1 \]

5. **Find a Common Denominator:**
The common denominator for 3 and 5 is 15. Rewrite the equation:
\[ \frac{5D}{15} - \frac{3D}{15} = 1 \]
Simplifying gives:
\[ \frac{2D}{15} = 1 \]

6. **Solve for \( D \):**
Multiply both sides by 15:
\[ 2D = 15 \]
Now, divide by 2:
\[ D = \frac{15}{2} = 7.5 \text{ km} \]

### Conclusion:
The distance the school boy travels every day to reach the school is **7.5 km**.