One day a boy walked from his house to his school at the speed of 4 km/hr and he reached ten minutes late to the school. Next day, he ran at the speed of 8 km/hr and was 5 minutes early to the school. Find the distance between his house and school.

Answer

To solve the problem step by step, we will use the information provided about the boy's journey to find the distance between his house and school.

### Step 1: Understand the problem
The boy walks to school at a speed of 4 km/hr and is 10 minutes late. The next day, he runs at a speed of 8 km/hr and arrives 5 minutes early. We need to find the distance between his house and school.

### Step 2: Convert time differences into hours
The total time difference between the two days is:
- 10 minutes late + 5 minutes early = 15 minutes
- Convert 15 minutes into hours:
\[ 15 \text{ minutes} = \frac{15}{60} \text{ hours} = \frac{1}{4} \text{ hours} \]

### Step 3: Set up the equations
Let the distance between the house and school be \( s \) kilometers.

- Time taken when walking:
\[ \text{Time} = \frac{s}{4} \text{ hours} \]
- Time taken when running:
\[ \text{Time} = \frac{s}{8} \text{ hours} \]

### Step 4: Write the equation based on the time difference
Since he is 10 minutes late on the first day and 5 minutes early on the second day, the difference in time taken can be expressed as:
\[ \frac{s}{4} - \frac{s}{8} = \frac{1}{4} \]

### Step 5: Solve the equation
To solve the equation, we first find a common denominator for the left side:
- The common denominator of 4 and 8 is 8.

Rewriting the equation:
\[ \frac{2s}{8} - \frac{s}{8} = \frac{1}{4} \]
This simplifies to:
\[ \frac{s}{8} = \frac{1}{4} \]

Now, we can cross-multiply to solve for \( s \):
\[ s = 8 \times \frac{1}{4} = 2 \]

### Step 6: Conclusion
The distance between the boy's house and school is \( 2 \) kilometers.