If a scooterist drives at the rate of 24 km/h, he reaches his destination 5 min too late, if he drives at the rate of 30 km/h, he reaches his destination 4 min too soon. Then, the distance of his destination is

Options

A. 12 km
B. 9 km
C. 18 km
D. None of these

Answer

To solve the problem step by step, we will first convert the time differences into hours and then set up equations based on the given speeds and time differences.

### Step-by-Step Solution:

1. **Convert Time Differences to Hours**:
- The scooterist is 5 minutes late when traveling at 24 km/h.
\[ 5 \text{ minutes} = \frac{5}{60} \text{ hours} = \frac{1}{12} \text{ hours} \]
- The scooterist is 4 minutes early when traveling at 30 km/h.
\[ 4 \text{ minutes} = \frac{4}{60} \text{ hours} = \frac{1}{15} \text{ hours} \]

2. **Set Up the Equations**:
- Let \( D \) be the distance to the destination and \( T \) be the time taken to reach the destination at the correct speed.
- When traveling at 24 km/h, the time taken is \( T + \frac{1}{12} \):
\[ \frac{D}{24} = T + \frac{1}{12} \]
- When traveling at 30 km/h, the time taken is \( T - \frac{1}{15} \):
\[ \frac{D}{30} = T - \frac{1}{15} \]

3. **Express \( T \) in Terms of \( D \)**:
- From the first equation:
\[ T = \frac{D}{24} - \frac{1}{12} \]
- From the second equation:
\[ T = \frac{D}{30} + \frac{1}{15} \]

4. **Set the Two Expressions for \( T \) Equal**:
\[ \frac{D}{24} - \frac{1}{12} = \frac{D}{30} + \frac{1}{15} \]

5. **Eliminate the Fractions**:
- To eliminate the fractions, find a common denominator (which is 120):
\[ 5D - 10 = 4D + 8 \]

6. **Solve for \( D \)**:
- Rearranging the equation gives:
\[ 5D - 4D = 8 + 10 \]
\[ D = 18 \text{ km} \]

### Final Answer:
The distance to the destination is **18 kilometers**.

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