Rani goes to school at 10 km/h and reaches the school 6 min late. Next day, she covers the distance at 12 km/h and reaches the school 9 min earlier than the scheduled time. What is the distance of her school from her house?

Options

A. 15 km
B. 16 km
C. 60 km
D. None of these

Answer

To solve the problem step by step, let's denote the distance from Rani's house to her school as \( d \) kilometers, and the scheduled time to reach school as \( t \) hours.

### Step 1: Set up the equations based on the information given.

1. **First Day:**
- Speed = 10 km/h
- Rani is 6 minutes late.
- Time taken = \( t + \frac{6}{60} \) hours = \( t + \frac{1}{10} \) hours.
- Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), we have:
\[ t + \frac{1}{10} = \frac{d}{10} \]

2. **Second Day:**
- Speed = 12 km/h
- Rani is 9 minutes early.
- Time taken = \( t - \frac{9}{60} \) hours = \( t - \frac{3}{20} \) hours.
- Again using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), we have:
\[ t - \frac{3}{20} = \frac{d}{12} \]

### Step 2: Rearranging the equations.

From the first equation:
\[ d = 10 \left( t + \frac{1}{10} \right) = 10t + 1 \]

From the second equation:
\[ d = 12 \left( t - \frac{3}{20} \right) = 12t - \frac{36}{20} = 12t - \frac{9}{5} \]

### Step 3: Set the equations for \( d \) equal to each other.

Now we have two expressions for \( d \):
\[ 10t + 1 = 12t - \frac{9}{5} \]

### Step 4: Solve for \( t \).

Rearranging gives:
\[ 1 + \frac{9}{5} = 12t - 10t \]
\[ \frac{5}{5} + \frac{9}{5} = 2t \]
\[ \frac{14}{5} = 2t \]
\[ t = \frac{14}{10} = \frac{7}{5} \text{ hours} \]

### Step 5: Substitute \( t \) back to find \( d \).

Using \( t = \frac{7}{5} \) in the first expression for \( d \):
\[ d = 10t + 1 = 10 \left( \frac{7}{5} \right) + 1 = 14 + 1 = 15 \text{ km} \]

### Conclusion

Thus, the distance from Rani's house to her school is \( \boxed{15} \) kilometers.

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