Osaka walks from his house at 5 km/h and reaches his office 10 minutes late if this speed had been 6 km/h he would have reached 15 minutes early.The distance of his office from his house is:

Options

A. 15 km
B. 12.5 km
C. 10.5 km
D. 18 km

Answer

To find the distance from Osaka's house to his office, we can set up the problem using the information given about his walking speeds and the times he is late or early.

### Step-by-Step Solution:

1. **Define Variables:**
Let the distance from Osaka's house to his office be \( D \) km.
Let the time he should take to reach his office on time be \( T \) hours.

2. **Calculate Time Taken at 5 km/h:**
If Osaka walks at 5 km/h, the time taken to reach his office is:
\[ \text{Time at 5 km/h} = \frac{D}{5} \]
According to the problem, he reaches 10 minutes late, so:
\[ \frac{D}{5} = T + \frac{10}{60} \quad \text{(convert 10 minutes to hours)} \]
This simplifies to:
\[ \frac{D}{5} = T + \frac{1}{6} \]

3. **Calculate Time Taken at 6 km/h:**
If Osaka walks at 6 km/h, the time taken to reach his office is:
\[ \text{Time at 6 km/h} = \frac{D}{6} \]
According to the problem, he reaches 15 minutes early, so:
\[ \frac{D}{6} = T - \frac{15}{60} \quad \text{(convert 15 minutes to hours)} \]
This simplifies to:
\[ \frac{D}{6} = T - \frac{1}{4} \]

4. **Set Up Equations:**
Now we have two equations:
\[ \frac{D}{5} = T + \frac{1}{6} \quad \text{(1)} \]
\[ \frac{D}{6} = T - \frac{1}{4} \quad \text{(2)} \]

5. **Eliminate T:**
From equation (1), we can express \( T \):
\[ T = \frac{D}{5} - \frac{1}{6} \]
Substitute \( T \) into equation (2):
\[ \frac{D}{6} = \left(\frac{D}{5} - \frac{1}{6}\right) - \frac{1}{4} \]

6. **Simplify the Equation:**
Multiply through by 60 to eliminate the denominators:
\[ 10D = 12D - 10 - 15 \]
Simplifying gives:
\[ 10D = 12D - 25 \]
Rearranging yields:
\[ 2D = 25 \quad \Rightarrow \quad D = \frac{25}{2} = 12.5 \text{ km} \]

### Final Answer:
The distance from Osaka's house to his office is **12.5 km**.