If Anuj walks at the speed of 4 km/hr, then he reaches his school 6 minutes late but if he walks at the speed of 5km/hr, then he reaches 6 minutes before the scheduled time. What is the distance (in km) of his school from his house?

Options

A. 4
B. 3
C. 6
D. `3.5`

Answer

To solve the problem step by step, we can follow this approach:

### Step 1: Define Variables
Let the distance from Anuj's house to his school be \( d \) kilometers.

### Step 2: Calculate Time Taken at Different Speeds
1. **Time taken at 4 km/hr**:
\[ \text{Time} = \frac{d}{4} \text{ hours} \]
Since he is 6 minutes late, the scheduled time to reach school is:
\[ \text{Scheduled Time} = \frac{d}{4} - \frac{6}{60} = \frac{d}{4} - \frac{1}{10} \text{ hours} \]

2. **Time taken at 5 km/hr**:
\[ \text{Time} = \frac{d}{5} \text{ hours} \]
Since he is 6 minutes early, the scheduled time to reach school is:
\[ \text{Scheduled Time} = \frac{d}{5} + \frac{6}{60} = \frac{d}{5} + \frac{1}{10} \text{ hours} \]

### Step 3: Set Up the Equation
Since both expressions represent the scheduled time, we can set them equal to each other:
\[ \frac{d}{4} - \frac{1}{10} = \frac{d}{5} + \frac{1}{10} \]

### Step 4: Solve the Equation
1. First, eliminate the fractions by finding a common denominator. The least common multiple of 4 and 5 is 20. Multiply the entire equation by 20:
\[ 20 \left(\frac{d}{4} - \frac{1}{10}\right) = 20 \left(\frac{d}{5} + \frac{1}{10}\right) \]
This simplifies to:
\[ 5d - 2 = 4d + 2 \]

2. Rearranging gives:
\[ 5d - 4d = 2 + 2 \]
\[ d = 4 \]

### Step 5: Conclusion
The distance from Anuj's house to his school is \( \boxed{4} \) kilometers.

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