I have to reach a certain place at a certain time and I find that I shall be 15 min too late, if I walk at 4 km an hour, and 10 min too soon, if I walk at 6 km an hour. How far have I to walk?

Options

A. 25 km
B. 5 km
C. 10 km
D. None of these

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Define the variables
Let the distance to be covered be \( d \) kilometers.

### Step 2: Calculate the time taken at different speeds
- When walking at 4 km/h, the time taken is \( \frac{d}{4} \) hours.
- When walking at 6 km/h, the time taken is \( \frac{d}{6} \) hours.

### Step 3: Set up the time difference equation
According to the problem:
- If walking at 4 km/h, I am 15 minutes late. This means the time taken is 15 minutes more than the required time \( T \).
- If walking at 6 km/h, I am 10 minutes early. This means the time taken is 10 minutes less than the required time \( T \).

We can express this as:
\[ \frac{d}{4} = T + \frac{15}{60} \quad \text{(1)} \]
\[ \frac{d}{6} = T - \frac{10}{60} \quad \text{(2)} \]

### Step 4: Rearranging the equations
From equation (1):
\[ T = \frac{d}{4} - \frac{15}{60} \]
From equation (2):
\[ T = \frac{d}{6} + \frac{10}{60} \]

### Step 5: Set the two expressions for T equal to each other
\[ \frac{d}{4} - \frac{15}{60} = \frac{d}{6} + \frac{10}{60} \]

### Step 6: Simplify the equation
To eliminate the fractions, we can multiply the entire equation by 60 (the least common multiple of the denominators):
\[ 60 \left( \frac{d}{4} \right) - 15 = 60 \left( \frac{d}{6} \right) + 10 \]
This simplifies to:
\[ 15d - 15 = 10d + 10 \]

### Step 7: Solve for d
Rearranging gives:
\[ 15d - 10d = 10 + 15 \]
\[ 5d = 25 \]
\[ d = 5 \]

### Conclusion
The distance I have to walk is **5 kilometers**.

---