Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house?

Options

A. 5 km
B. 8 km
C. 3 km
D. 2 km

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Define Variables
Let the distance between the student's house and school be \( D \) km. Let \( T \) be the time (in hours) that the student would take to reach school on time.

### Step 2: Convert Minutes to Hours
Since the student reaches 15 minutes early when walking at 5 km/hr, we convert 15 minutes to hours:
\[ 15 \text{ minutes} = \frac{15}{60} = \frac{1}{4} \text{ hours} \]
When walking at 5 km/hr, the time taken is:
\[ T - \frac{1}{4} \text{ hours} \]

Similarly, when walking at 3 km/hr, the student is late by 9 minutes. We convert 9 minutes to hours:
\[ 9 \text{ minutes} = \frac{9}{60} = \frac{3}{20} \text{ hours} \]
In this case, the time taken is:
\[ T + \frac{3}{20} \text{ hours} \]

### Step 3: Set Up Equations
Using the formula \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \), we can set up two equations based on the two scenarios:

1. For the speed of 5 km/hr:
\[ \frac{D}{T - \frac{1}{4}} = 5 \]
This simplifies to:
\[ D = 5 \left( T - \frac{1}{4} \right) = 5T - \frac{5}{4} \]

2. For the speed of 3 km/hr:
\[ \frac{D}{T + \frac{3}{20}} = 3 \]
This simplifies to:
\[ D = 3 \left( T + \frac{3}{20} \right) = 3T + \frac{9}{20} \]

### Step 4: Equate the Two Expressions for Distance
Since both expressions are equal to \( D \), we can set them equal to each other:
\[ 5T - \frac{5}{4} = 3T + \frac{9}{20} \]

### Step 5: Solve for \( T \)
Rearranging the equation:
\[ 5T - 3T = \frac{9}{20} + \frac{5}{4} \]
Convert \( \frac{5}{4} \) to have a common denominator of 20:
\[ \frac{5}{4} = \frac{25}{20} \]
Now the equation becomes:
\[ 2T = \frac{9}{20} + \frac{25}{20} = \frac{34}{20} \]
Thus,
\[ T = \frac{34}{40} = \frac{17}{20} \text{ hours} \]

### Step 6: Substitute \( T \) Back to Find \( D \)
Now we can substitute \( T \) back into one of the distance equations. Using \( D = 5T - \frac{5}{4} \):
\[ D = 5 \left( \frac{17}{20} \right) - \frac{5}{4} \]
Convert \( \frac{5}{4} \) to have a common denominator of 20:
\[ \frac{5}{4} = \frac{25}{20} \]
Now substituting:
\[ D = \frac{85}{20} - \frac{25}{20} = \frac{60}{20} = 3 \text{ km} \]

### Final Answer
The distance between the student's house and school is \( D = 3 \) km.

---