Walking at `(3)/(4)` of his usual speed, a man is `1(1)/(2)` hours late. His usual time to cover the same distance, (in hours) is

Options

A. `4(1)/(2)`
B. 4
C. `5(1)/(2)`
D. 5

Answer

To solve the problem step by step, we will use the relationship between speed, time, and distance.

### Step 1: Define Variables
Let the usual speed of the man be \( S \) and the usual time to cover the distance be \( T \).

### Step 2: Express the New Speed and Time
When the man walks at \( \frac{3}{4} \) of his usual speed, his new speed becomes:
\[ \text{New Speed} = \frac{3}{4} S \]
Since speed and time are inversely related, the new time taken to cover the same distance will be:
\[ \text{New Time} = \frac{T}{\frac{3}{4}} = \frac{4T}{3} \]

### Step 3: Set Up the Equation for Delay
According to the problem, the man is \( 1\frac{1}{2} \) hours late, which can be expressed as:
\[ 1\frac{1}{2} = \frac{3}{2} \text{ hours} \]
This means that the difference between the new time and the usual time is:
\[ \frac{4T}{3} - T = \frac{3}{2} \]

### Step 4: Solve the Equation
Now, we will simplify the left side of the equation:
\[ \frac{4T}{3} - T = \frac{4T}{3} - \frac{3T}{3} = \frac{4T - 3T}{3} = \frac{T}{3} \]
So, we have:
\[ \frac{T}{3} = \frac{3}{2} \]

### Step 5: Multiply Both Sides by 3
To find \( T \), we multiply both sides by 3:
\[ T = 3 \times \frac{3}{2} = \frac{9}{2} \text{ hours} \]

### Conclusion
Thus, the usual time for the man to cover the same distance is:
\[ \frac{9}{2} \text{ hours} \text{ or } 4.5 \text{ hours} \]

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