A student walks from his house at 5 km/ph, reaches school 10 minutes late. If his speed had been 6 km/ph, he would have reached 15 minutes early. The distance of the school from his house is

Options

A. `5/2` kms
B. `5/22` kms
C. `25/2` kms
D. `25/22` kms

Answer

To solve the problem step by step, we will define the distance from the student's house to the school as \( x \) kilometers.

### Step 1: Define the problem
Let the distance from the house to the school be \( x \) km.

### Step 2: Calculate time taken at different speeds
1. **At 5 km/h**:
- Time taken to reach school = \( \frac{x}{5} \) hours.
- Since he is 10 minutes late, the time he should have taken (on time) = \( \frac{x}{5} - \frac{10}{60} \) hours.

2. **At 6 km/h**:
- Time taken to reach school = \( \frac{x}{6} \) hours.
- Since he is 15 minutes early, the time he should have taken (on time) = \( \frac{x}{6} + \frac{15}{60} \) hours.

### Step 3: Set up the equation
Since both expressions represent the same "on time" duration, we can set them equal to each other:
\[ \frac{x}{5} - \frac{10}{60} = \frac{x}{6} + \frac{15}{60} \]

### Step 4: Simplify the equation
1. Convert the minutes to hours:
- \( \frac{10}{60} = \frac{1}{6} \) hours
- \( \frac{15}{60} = \frac{1}{4} \) hours

2. Substitute these values into the equation:
\[ \frac{x}{5} - \frac{1}{6} = \frac{x}{6} + \frac{1}{4} \]

### Step 5: Find a common denominator
The least common multiple of 5 and 6 is 30. Multiply through by 30 to eliminate the denominators:
\[ 30 \left( \frac{x}{5} \right) - 30 \left( \frac{1}{6} \right) = 30 \left( \frac{x}{6} \right) + 30 \left( \frac{1}{4} \right) \]
This simplifies to:
\[ 6x - 5 = 5x + 7.5 \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ 6x - 5x = 7.5 + 5 \]
\[ x = 12.5 \]

### Step 7: Final answer
The distance from the student's house to the school is \( 12.5 \) km.