A student goes to school at the rate of `(5)/(2)` km/hr and reaches 6 minutes late. If he travels at the speed of 3km/hr, he reaches 10 minutes earlier. The distance of the school is

Options

A. 45km
B. 20km
C. 10km
D. 4km

Answer

To solve the problem step by step, we will use the information provided about the student's travel times at different speeds.

### Step 1: Define the variables
Let the distance to the school be \( d \) km.

### Step 2: Calculate time taken at different speeds
1. When the student travels at a speed of \( \frac{5}{2} \) km/hr:
- Time taken = \( \frac{d}{\frac{5}{2}} = \frac{2d}{5} \) hours.
- Since he is 6 minutes late, the time he should have taken is \( \frac{2d}{5} - \frac{6}{60} \) hours (converting 6 minutes to hours).

2. When the student travels at a speed of \( 3 \) km/hr:
- Time taken = \( \frac{d}{3} \) hours.
- Since he is 10 minutes early, the time he should have taken is \( \frac{d}{3} + \frac{10}{60} \) hours (converting 10 minutes to hours).

### Step 3: Set up the equations
From the above information, we can set up the following equations based on the time taken:

1. From the first scenario:
\[ \frac{2d}{5} - \frac{6}{60} = T \quad \text{(where \( T \) is the actual time taken)} \]

2. From the second scenario:
\[ \frac{d}{3} + \frac{10}{60} = T \]

### Step 4: Equate the two expressions for \( T \)
Now, we can equate the two expressions for \( T \):
\[ \frac{2d}{5} - \frac{6}{60} = \frac{d}{3} + \frac{10}{60} \]

### Step 5: Simplify the equation
1. Convert \( \frac{6}{60} \) and \( \frac{10}{60} \) to a common denominator:
\[ \frac{6}{60} = \frac{1}{10}, \quad \frac{10}{60} = \frac{1}{6} \]

2. Rewrite the equation:
\[ \frac{2d}{5} - \frac{1}{10} = \frac{d}{3} + \frac{1}{6} \]

3. To eliminate fractions, find the least common multiple (LCM) of the denominators (5, 10, 3, 6), which is 30. Multiply the entire equation by 30:
\[ 30 \left(\frac{2d}{5}\right) - 30 \left(\frac{1}{10}\right) = 30 \left(\frac{d}{3}\right) + 30 \left(\frac{1}{6}\right) \]
This simplifies to:
\[ 12d - 3 = 10d + 5 \]

### Step 6: Solve for \( d \)
1. Rearranging gives:
\[ 12d - 10d = 5 + 3 \]
\[ 2d = 8 \]
\[ d = 4 \]

### Conclusion
The distance to the school is \( 4 \) km.