A students rides on a bicycle at 8 km/hr and reaches his school 2.5 minutes late, The next day he-increases the speed to 10 km/hr and reaches school 5 minutes early. How far is the school from the house?

Options

A. 6 km
B. 4 km
C. 5 km
D. 4.5 km

Answer

To find the distance from the student's house to the school, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = distance from the house to the school (in km)
- \( T \) = the right time to reach the school (in minutes)

### Step 2: Convert Time to Hours
Since speed is given in km/hr, we need to convert the time into hours:
- \( T \) minutes = \( \frac{T}{60} \) hours

### Step 3: Set Up the First Equation
When the student rides at 8 km/hr and is 2.5 minutes late:
- Time taken = \( T + 2.5 \) minutes = \( \frac{T + 2.5}{60} \) hours
- Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):
\[ \frac{x}{8} = \frac{T + 2.5}{60} \]
This can be rearranged to:
\[ x = 8 \cdot \frac{T + 2.5}{60} \]
\[ x = \frac{8(T + 2.5)}{60} \]
\[ x = \frac{2(T + 2.5)}{15} \quad \text{(Equation 1)} \]

### Step 4: Set Up the Second Equation
When the student rides at 10 km/hr and is 5 minutes early:
- Time taken = \( T - 5 \) minutes = \( \frac{T - 5}{60} \) hours
- Using the same formula:
\[ \frac{x}{10} = \frac{T - 5}{60} \]
This can be rearranged to:
\[ x = 10 \cdot \frac{T - 5}{60} \]
\[ x = \frac{10(T - 5)}{60} \]
\[ x = \frac{T - 5}{6} \quad \text{(Equation 2)} \]

### Step 5: Set the Two Equations Equal
Now we have two expressions for \( x \):
\[ \frac{2(T + 2.5)}{15} = \frac{T - 5}{6} \]

### Step 6: Cross Multiply to Solve for \( T \)
Cross multiplying gives:
\[ 2(T + 2.5) \cdot 6 = 15(T - 5) \]
\[ 12(T + 2.5) = 15(T - 5) \]
Expanding both sides:
\[ 12T + 30 = 15T - 75 \]
Rearranging gives:
\[ 30 + 75 = 15T - 12T \]
\[ 105 = 3T \]
\[ T = 35 \text{ minutes} \]

### Step 7: Substitute \( T \) Back to Find \( x \)
Now substitute \( T \) back into either Equation 1 or Equation 2 to find \( x \). Using Equation 2:
\[ x = \frac{T - 5}{6} = \frac{35 - 5}{6} = \frac{30}{6} = 5 \text{ km} \]

### Final Answer
The distance from the student's house to the school is **5 km**.

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