Walking at four fifth of his usual speed Vijay Malya reaches his office 15 minutes late on a particular day. The next day, he walked at 5/4 of his usual speed. How early would he be to the office when compared to the previous day?

Options

A. 27 min
B. 32 min
C. 30 min
D. none of these

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understanding the Problem
Vijay Malya walks at four-fifths of his usual speed and is 15 minutes late. We need to find out how early he arrives when he walks at five-fourths of his usual speed the next day.

### Step 2: Establishing Variables
Let:
- Usual speed = \( S \)
- Usual time taken to reach the office = \( T \)
- Distance to the office = \( D \)

From the relationship between speed, distance, and time, we have:
\[ D = S \times T \]

### Step 3: Speed on the First Day
On the first day, his speed is:
\[ \text{Speed on first day} = \frac{4}{5} S \]

### Step 4: Time Taken on the First Day
The time taken on the first day can be calculated as:
\[ \text{Time on first day} = \frac{D}{\text{Speed on first day}} = \frac{D}{\frac{4}{5} S} = \frac{5D}{4S} \]

### Step 5: Relating Time to Usual Time
Since he is 15 minutes late, we can express this as:
\[ \text{Time on first day} = T + 15 \text{ minutes} \]

### Step 6: Setting Up the Equation
We know:
\[ \frac{5D}{4S} = T + 15 \]

Since \( D = S \times T \), we can substitute \( D \):
\[ \frac{5(S \times T)}{4S} = T + 15 \]
This simplifies to:
\[ \frac{5T}{4} = T + 15 \]

### Step 7: Solving for \( T \)
Rearranging gives:
\[ \frac{5T}{4} - T = 15 \]
\[ \frac{5T - 4T}{4} = 15 \]
\[ \frac{T}{4} = 15 \]
Multiplying both sides by 4:
\[ T = 60 \text{ minutes} \]

### Step 8: Time Taken on the First Day
From our earlier equation:
\[ \text{Time on first day} = T + 15 = 60 + 15 = 75 \text{ minutes} \]

### Step 9: Speed on the Second Day
On the second day, his speed is:
\[ \text{Speed on second day} = \frac{5}{4} S \]

### Step 10: Time Taken on the Second Day
The time taken on the second day can be calculated as:
\[ \text{Time on second day} = \frac{D}{\text{Speed on second day}} = \frac{D}{\frac{5}{4} S} = \frac{4D}{5S} \]

### Step 11: Relating Time on the Second Day to Usual Time
Substituting \( D = S \times T \):
\[ \text{Time on second day} = \frac{4(S \times T)}{5S} = \frac{4T}{5} \]

### Step 12: Calculating Time on the Second Day
Substituting \( T = 60 \):
\[ \text{Time on second day} = \frac{4 \times 60}{5} = \frac{240}{5} = 48 \text{ minutes} \]

### Step 13: Finding the Difference in Time
Now, we find how early he is on the second day compared to the first day:
\[ \text{Difference} = \text{Time on first day} - \text{Time on second day} = 75 - 48 = 27 \text{ minutes} \]

### Final Answer
Vijay Malya would be **27 minutes early** on the second day compared to the previous day.

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