A student walks from his house at a speed of`2 (1)/( 2)` km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house ?

Options

A. `5/4` km
B. `7/4` km
C. `9/4 km`
D. `(11)/(4)` km

Answer

To solve the problem, we need to determine the distance from the student's house to the school. Let's denote the distance as \( d \) kilometers and the time taken to reach school on time as \( t \) hours.

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

1. The student's speed on the first day is \( 2.5 \) km/h (which is \( 2 \frac{1}{2} \) km/h).
2. He reaches the school 6 minutes late. This means he took \( t + \frac{1}{10} \) hours (since 6 minutes is \( \frac{6}{60} = \frac{1}{10} \) hours) to reach the school.
3. The distance can be expressed as:
\[ d = \text{speed} \times \text{time} \]
For the first day:
\[ d = 2.5 \left(t + \frac{1}{10}\right) \]

### Step 2: Write the equation for the second day.

1. The student's speed on the second day is \( 3.5 \) km/h (since he increased his speed by 1 km/h).
2. He reaches the school 6 minutes early, which means he took \( t - \frac{1}{10} \) hours to reach the school.
3. The distance can also be expressed as:
\[ d = 3.5 \left(t - \frac{1}{10}\right) \]

### Step 3: Set the two equations equal to each other.

Since both expressions equal \( d \), we can set them equal:
\[ 2.5 \left(t + \frac{1}{10}\right) = 3.5 \left(t - \frac{1}{10}\right) \]

### Step 4: Expand both sides.

Expanding both sides gives:
\[ 2.5t + 0.25 = 3.5t - 0.35 \]

### Step 5: Rearrange the equation to solve for \( t \).

Rearranging the equation:
\[ 0.25 + 0.35 = 3.5t - 2.5t \]
\[ 0.60 = 1t \]
\[ t = 0.60 \text{ hours} \]

### Step 6: Substitute \( t \) back into one of the distance equations.

Using the first day's equation to find \( d \):
\[ d = 2.5 \left(0.60 + \frac{1}{10}\right) \]
Calculating \( 0.60 + \frac{1}{10} = 0.60 + 0.10 = 0.70 \):
\[ d = 2.5 \times 0.70 = 1.75 \text{ km} \]

### Final Answer

The distance from the student's house to the school is \( 1.75 \) km.