By walking at `4/3` of his usual speed, a man reaches his office 25 min earlier than usual. What is his usual time is?

Options

A. 100 min
B. 125 min
C. 25 min
D. 40 min

Answer

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

### Step 1: Define Variables
Let:
- \( D \) = Distance to the office (which remains constant)
- \( S \) = Usual speed of the man
- \( T \) = Usual time taken to reach the office

### Step 2: Express the Usual Time
The usual time taken to reach the office can be expressed as:
\[ T = \frac{D}{S} \]

### Step 3: Calculate the New Speed
The man walks at \( \frac{4}{3} \) of his usual speed, so his new speed is:
\[ \text{New Speed} = \frac{4}{3}S \]

### Step 4: Express the New Time
The time taken to reach the office at the new speed can be expressed as:
\[ \text{New Time} = \frac{D}{\frac{4}{3}S} = \frac{3D}{4S} \]

### Step 5: Relate the New Time to the Usual Time
According to the problem, the man reaches his office 25 minutes earlier than usual. Therefore, we can set up the equation:
\[ T - 25 = \frac{3D}{4S} \]

### Step 6: Substitute \( T \)
Substituting \( T \) from Step 2 into the equation gives:
\[ \frac{D}{S} - 25 = \frac{3D}{4S} \]

### Step 7: Eliminate \( D \) and Solve for \( T \)
To eliminate \( D \), we can multiply the entire equation by \( 4S \):
\[ 4D - 100S = 3D \]

Now, rearranging gives:
\[ 4D - 3D = 100S \]
\[ D = 100S \]

### Step 8: Substitute Back to Find \( T \)
Now substituting \( D \) back into the equation for \( T \):
\[ T = \frac{D}{S} = \frac{100S}{S} = 100 \text{ minutes} \]

### Conclusion
Thus, the usual time taken by the man to reach his office is:
\[ \boxed{100 \text{ minutes}} \]

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