By walking at `(3)/(4)` of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is

Options

A. 75 minutes
B. 60 minutes
C. 40 minutes
D. 30 minutes

Answer

To solve the problem, we need to find the usual time taken by a man to reach his office when he walks at a speed that is \( \frac{3}{4} \) of his usual speed and arrives 20 minutes late.

### Step-by-Step Solution:

1. **Define Variables**:
Let the usual speed of the man be \( V \) (in any unit, e.g., km/h) and the usual time taken to reach his office be \( T \) (in minutes).

2. **Calculate Distance**:
The distance to the office can be expressed using the usual speed and time:
\[ \text{Distance} = \text{Speed} \times \text{Time} = V \times T \]

3. **New Speed**:
When the man walks at \( \frac{3}{4} \) of his usual speed, his new speed becomes:
\[ \text{New Speed} = \frac{3}{4} V \]

4. **New Time**:
Since he arrives 20 minutes later than usual, the time taken at the new speed is:
\[ \text{New Time} = T + 20 \]

5. **Set Up the Distance Equation**:
The distance remains the same in both scenarios. Therefore, we can set up the equation:
\[ V \times T = \left(\frac{3}{4} V\right) \times (T + 20) \]

6. **Simplify the Equation**:
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ T = \frac{3}{4} (T + 20) \]

7. **Distribute and Rearrange**:
Expanding the right side gives:
\[ T = \frac{3}{4}T + 15 \]
Now, rearranging the equation:
\[ T - \frac{3}{4}T = 15 \]
\[ \frac{1}{4}T = 15 \]

8. **Solve for \( T \)**:
Multiplying both sides by 4 gives:
\[ T = 60 \]

9. **Conclusion**:
The usual time taken by the man to reach his office is **60 minutes**.