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

Answer

To solve the problem, we need to determine the distance from the student's house to the school. Let's denote this distance as \( x \) kilometers.

### Step 1: Calculate the time taken to reach school at 8 km/hr
The time taken to travel a distance is given by the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
At a speed of 8 km/hr, the time taken to reach the school is:
\[ \text{Time}_{8} = \frac{x}{8} \text{ hours} \]
Since the student is 2.5 minutes late, we need to convert 2.5 minutes into hours:
\[ 2.5 \text{ minutes} = \frac{2.5}{60} \text{ hours} = \frac{1}{24} \text{ hours} \]
Thus, the actual time he should take to reach school is:
\[ \text{Actual Time} = \text{Time}_{8} - \frac{1}{24} \]
So we have:
\[ \text{Actual Time} = \frac{x}{8} - \frac{1}{24} \]

### Step 2: Calculate the time taken to reach school at 10 km/hr
At a speed of 10 km/hr, the time taken to reach the school is:
\[ \text{Time}_{10} = \frac{x}{10} \text{ hours} \]
Since the student is 5 minutes early, we convert 5 minutes into hours:
\[ 5 \text{ minutes} = \frac{5}{60} \text{ hours} = \frac{1}{12} \text{ hours} \]
Thus, the actual time he should take to reach school is:
\[ \text{Actual Time} = \text{Time}_{10} + \frac{1}{12} \]
So we have:
\[ \text{Actual Time} = \frac{x}{10} + \frac{1}{12} \]

### Step 3: Set the two expressions for actual time equal
Since both expressions represent the actual time taken to reach school, we can set them equal to each other:
\[ \frac{x}{8} - \frac{1}{24} = \frac{x}{10} + \frac{1}{12} \]

### Step 4: Solve for \( x \)
To solve this equation, we first find a common denominator for the fractions. The least common multiple of 8, 10, 24, and 12 is 120. We will multiply every term by 120 to eliminate the denominators:
\[ 120 \left(\frac{x}{8}\right) - 120 \left(\frac{1}{24}\right) = 120 \left(\frac{x}{10}\right) + 120 \left(\frac{1}{12}\right) \]
This simplifies to:
\[ 15x - 5 = 12x + 10 \]
Now, rearranging the equation gives:
\[ 15x - 12x = 10 + 5 \]
\[ 3x = 15 \]
Dividing both sides by 3 gives:
\[ x = 5 \]

### Conclusion
The distance from the student's house to the school is \( 5 \) kilometers.

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