A bus at a speed of 40 km/h reaches its destination late by 10 minutes while at the speed of 50 km/h reaches its destination before 5 minutes. Find the distance.

Options

A. 80 km
B. 50 km
C. 40 km
D. 30 km

Answer

To solve the problem, we need to find the distance based on the information provided about the bus's speeds and the time differences. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- The bus travels at two different speeds: 40 km/h and 50 km/h.
- At 40 km/h, the bus is 10 minutes late.
- At 50 km/h, the bus arrives 5 minutes early.
- We need to find the distance to the destination.

2. **Convert Time Differences to Hours:**
- The total time difference between being late and being early is:
\[ 10 \text{ minutes (late)} + 5 \text{ minutes (early)} = 15 \text{ minutes} \]
- Convert 15 minutes into hours:
\[ 15 \text{ minutes} = \frac{15}{60} \text{ hours} = \frac{1}{4} \text{ hours} \]

3. **Define Variables:**
- Let \( D \) be the distance to the destination.
- Let \( T \) be the time taken to reach the destination on time (in hours).

4. **Set Up Equations Based on Speed and Time:**
- When traveling at 40 km/h, the time taken is:
\[ \text{Time at 40 km/h} = \frac{D}{40} \]
Since the bus is 10 minutes late:
\[ \frac{D}{40} = T + \frac{1}{6} \text{ hours} \quad \text{(10 minutes = } \frac{10}{60} = \frac{1}{6} \text{ hours)} \]
- When traveling at 50 km/h, the time taken is:
\[ \text{Time at 50 km/h} = \frac{D}{50} \]
Since the bus is 5 minutes early:
\[ \frac{D}{50} = T - \frac{1}{12} \text{ hours} \quad \text{(5 minutes = } \frac{5}{60} = \frac{1}{12} \text{ hours)} \]

5. **Set Up the Two Equations:**
- From the first equation:
\[ \frac{D}{40} = T + \frac{1}{6} \]
- From the second equation:
\[ \frac{D}{50} = T - \frac{1}{12} \]

6. **Eliminate \( T \) from the Equations:**
- Rearranging both equations gives:
\[ T = \frac{D}{40} - \frac{1}{6} \quad \text{(1)} \]
\[ T = \frac{D}{50} + \frac{1}{12} \quad \text{(2)} \]
- Set the two expressions for \( T \) equal to each other:
\[ \frac{D}{40} - \frac{1}{6} = \frac{D}{50} + \frac{1}{12} \]

7. **Solve for \( D \):**
- To eliminate fractions, find a common denominator (which is 120):
\[ 3D - 20 = 2.4D + 10 \]
- Rearranging gives:
\[ 3D - 2.4D = 30 \]
\[ 0.6D = 30 \]
\[ D = \frac{30}{0.6} = 50 \text{ km} \]

### Final Answer:
The distance to the destination is **50 kilometers**.