A boy walking at a speed of 15 km/h reaches his school 20 min late. Next time he increases his speed by 5 km/h but still he late by 5 min. Find the distance of the school from his home.

Options

A. 5 km
B. 10 km
C. 15 km
D. 20 km

Answer

To solve the problem, we need to find the distance from the boy's home to his school based on the information given about his speed and the time he is late.

### Step-by-Step Solution:

1. **Identify the speeds and times:**
- The boy's initial speed (S1) = 15 km/h
- The boy's increased speed (S2) = 15 km/h + 5 km/h = 20 km/h
- Time late at S1 = 20 minutes = 20/60 hours = 1/3 hours
- Time late at S2 = 5 minutes = 5/60 hours = 1/12 hours

2. **Let the distance to school be D km.**
- The time taken to reach school at speed S1 (15 km/h) can be expressed as:
\[ \text{Time at S1} = \frac{D}{15} \text{ hours} \]
- Since he is 20 minutes late, the time he should have taken (T) to reach on time is:
\[ T = \frac{D}{15} - \frac{1}{3} \]

3. **For the second scenario (S2 = 20 km/h):**
- The time taken to reach school at speed S2 (20 km/h) can be expressed as:
\[ \text{Time at S2} = \frac{D}{20} \text{ hours} \]
- Since he is 5 minutes late, the time he should have taken (T) can also be expressed as:
\[ T = \frac{D}{20} - \frac{1}{12} \]

4. **Set the two expressions for T equal to each other:**
\[ \frac{D}{15} - \frac{1}{3} = \frac{D}{20} - \frac{1}{12} \]

5. **Clear the fractions by finding a common denominator:**
- The common denominator for 15, 20, and 12 is 60.
- Multiply the entire equation by 60 to eliminate the denominators:
\[ 60 \left( \frac{D}{15} \right) - 60 \left( \frac{1}{3} \right) = 60 \left( \frac{D}{20} \right) - 60 \left( \frac{1}{12} \right) \]
This simplifies to:
\[ 4D - 20 = 3D - 5 \]

6. **Rearrange the equation:**
\[ 4D - 3D = 20 - 5 \]
\[ D = 15 \]

7. **Conclusion:**
- The distance from the boy's home to his school is **15 km**.