A student reached his school late by 20 minutes by travelling at a speed of 9 km/hr. Had he travelled at the speed of 12 km/hr, he would have reached his school 20 minutes early. Find the distance between his house and the school?

Options

A. 12 km
B. 6 km
C. 3 km
D. 24 km

Answer

To solve the problem step by step, we will use the information provided about the student's travel times and speeds.

### Step 1: Define Variables
Let the distance between the student's house and the school be \( D \) kilometers.

### Step 2: Convert Time
The student is late by 20 minutes when traveling at 9 km/hr. We need to convert 20 minutes into hours:
\[ 20 \text{ minutes} = \frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hours} \]

### Step 3: Set Up Equations for Both Cases
1. **Case 1 (Speed = 9 km/hr)**:
- The time taken to reach school is \( T + \frac{1}{3} \) hours (since he is late).
- Using the formula \( \text{Distance} = \text{Speed} \times \text{Time} \):
\[ D = 9 \left( T + \frac{1}{3} \right) \]
Expanding this gives:
\[ D = 9T + 3 \]

2. **Case 2 (Speed = 12 km/hr)**:
- The time taken to reach school is \( T - \frac{1}{3} \) hours (since he is early).
- Again using the formula \( \text{Distance} = \text{Speed} \times \text{Time} \):
\[ D = 12 \left( T - \frac{1}{3} \right) \]
Expanding this gives:
\[ D = 12T - 4 \]

### Step 4: Set the Two Equations Equal
Since both expressions equal \( D \), we can set them equal to each other:
\[ 9T + 3 = 12T - 4 \]

### Step 5: Solve for \( T \)
Rearranging the equation to isolate \( T \):
\[ 3 + 4 = 12T - 9T \]
\[ 7 = 3T \]
\[ T = \frac{7}{3} \text{ hours} \]

### Step 6: Substitute \( T \) Back to Find \( D \)
Now we can substitute \( T \) back into either equation to find \( D \). Using the first equation:
\[ D = 9 \left( \frac{7}{3} + \frac{1}{3} \right) \]
\[ D = 9 \left( \frac{8}{3} \right) = 24 \text{ km} \]

### Final Answer
The distance between the student's house and the school is \( \boxed{24} \) kilometers.

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