Walking at `6/7`th of his usual speed a man is 25 minutes too late. His usual time to cover this distance is

Options

A. 2 hours 30 minutes
B. 2 hours 15 minutes
C. 2 hours 25 minutes
D. 2 hours 10 minutes

Answer

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

### Step 1: Understand the relationship between speed, time, and distance.
We know that speed is equal to distance divided by time. Therefore, if speed decreases, time taken to cover the same distance increases.

### Step 2: Define the variables.
Let:
- Usual speed = \( S \)
- Usual time = \( T \)
- Distance = \( D \)

From the formula of speed, we have:
\[ S = \frac{D}{T} \]

### Step 3: Express the new speed and time.
The man walks at \( \frac{6}{7} \) of his usual speed, so the new speed is:
\[ \text{New Speed} = \frac{6}{7} S \]

Let the new time taken to cover the same distance be \( T' \). Since he is 25 minutes late, we can express this as:
\[ T' = T + 25 \text{ minutes} \]

### Step 4: Relate the new speed and time.
Using the speed formula again for the new speed, we have:
\[ \frac{6}{7} S = \frac{D}{T'} \]

### Step 5: Substitute \( T' \) into the equation.
Now we can substitute \( T' \) into the equation:
\[ \frac{6}{7} S = \frac{D}{T + 25} \]

### Step 6: Substitute \( D \) in terms of \( S \) and \( T \).
Since \( D = S \cdot T \), we can substitute this into the equation:
\[ \frac{6}{7} S = \frac{S \cdot T}{T + 25} \]

### Step 7: Cancel \( S \) from both sides.
Assuming \( S \neq 0 \), we can cancel \( S \):
\[ \frac{6}{7} = \frac{T}{T + 25} \]

### Step 8: Cross-multiply to solve for \( T \).
Cross-multiplying gives:
\[ 6(T + 25) = 7T \]
\[ 6T + 150 = 7T \]

### Step 9: Rearrange the equation to isolate \( T \).
Rearranging gives:
\[ 150 = 7T - 6T \]
\[ 150 = T \]

### Step 10: Conclusion.
Thus, the usual time \( T \) to cover the distance is:
\[ T = 150 \text{ minutes} \]

### Step 11: Convert minutes to hours and minutes.
To convert 150 minutes into hours:
\[ 150 \div 60 = 2.5 \text{ hours} \]
This is equivalent to 2 hours and 30 minutes.

### Final Answer:
The usual time to cover this distance is **2 hours and 30 minutes**.

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