Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes?

Options

A. A) 3km/hr
B. B) 4km/hr
C. C) 5km/hr
D. D) 7km/hr

Answer

To solve the problem step by step, we will use the information given about the buses and the man.

### Step 1: Understand the problem
We know that:
- Buses leave the terminal every 10 minutes at a speed of 20 km/hr.
- The man meets the buses every 8 minutes.

### Step 2: Convert time intervals to hours
Since the speeds are given in km/hr, we need to convert the time intervals from minutes to hours:
- 10 minutes = 10/60 hours = 1/6 hours
- 8 minutes = 8/60 hours = 2/15 hours

### Step 3: Calculate the distance traveled by the bus
The distance traveled by the bus in 10 minutes (1/6 hours) is:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 20 \, \text{km/hr} \times \frac{1}{6} \, \text{hr} = \frac{20}{6} \, \text{km} = \frac{10}{3} \, \text{km} \]

### Step 4: Set up the relationship between the bus and the man
Let the speed of the man be \( v \) km/hr. The distance the man travels in 8 minutes (2/15 hours) is:
\[ \text{Distance} = v \times \frac{2}{15} \, \text{hr} \]

### Step 5: Set the distances equal
Since the man meets the bus after traveling the same distance, we can set the distances equal:
\[ \frac{10}{3} = v \times \frac{2}{15} \]

### Step 6: Solve for the man's speed \( v \)
To find \( v \), we rearrange the equation:
\[ v = \frac{10}{3} \times \frac{15}{2} \]
\[ v = \frac{10 \times 15}{3 \times 2} = \frac{150}{6} = 25 \, \text{km/hr} \]

### Step 7: Conclusion
The speed of the man coming from the opposite direction towards the bus terminal is **25 km/hr**.

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