While walking down on the pavements of New York city. I notice that every 20 minute there is a city bus coming in the opposite direction and every 30 minute there is a city bus overtaking me from behind. What is the time gap between one city bus passing a stationary point known as Local Bus Stop beside the route and the immediately next city bus in the same direction passing the same stationary point?

Options

A. 27 min
B. 24 min
C. 25 min
D. can't be determined

Answer

To solve the problem step by step, we will analyze the information given and derive the necessary equations.

### Step 1: Understand the problem
We have two types of buses:
1. Buses coming from the opposite direction every 20 minutes.
2. Buses overtaking the person from behind every 30 minutes.

### Step 2: Define variables
Let:
- Speed of the bus = \( x \) km/h
- Speed of the person = \( y \) km/h
- Distance between successive buses = \( d \) km

### Step 3: Calculate distance for buses coming from opposite direction
Since a bus comes every 20 minutes from the opposite direction, we can express the distance \( d \) as:
\[ d = (20 \text{ minutes}) \times (x + y) \]
Converting 20 minutes to hours:
\[ d = \frac{1}{3} \text{ hours} \times (x + y) = \frac{(x + y)}{3} \]

### Step 4: Calculate distance for buses overtaking from behind
For the buses overtaking from behind every 30 minutes, the distance \( d \) can be expressed as:
\[ d = (30 \text{ minutes}) \times (x - y) \]
Converting 30 minutes to hours:
\[ d = \frac{1}{2} \text{ hours} \times (x - y) = \frac{(x - y)}{2} \]

### Step 5: Set the two equations for distance equal
Since both expressions represent the same distance \( d \):
\[ \frac{(x + y)}{3} = \frac{(x - y)}{2} \]

### Step 6: Cross multiply to eliminate fractions
Cross multiplying gives:
\[ 2(x + y) = 3(x - y) \]

### Step 7: Expand and simplify the equation
Expanding both sides:
\[ 2x + 2y = 3x - 3y \]
Rearranging gives:
\[ 2y + 3y = 3x - 2x \]
So,
\[ 5y = x \]
This means:
\[ x = 5y \]

### Step 8: Substitute back to find distance
Now, substituting \( x \) back into either equation for \( d \). Using the first equation:
\[ d = \frac{(5y + y)}{3} = \frac{6y}{3} = 2y \]

### Step 9: Calculate the time gap between buses
Now, we need to find the time gap between two buses passing the same point:
The time taken for the bus to cover distance \( d \) at speed \( x \):
\[ \text{Time} = \frac{d}{x} = \frac{2y}{5y} = \frac{2}{5} \text{ hours} \]
Converting to minutes:
\[ \frac{2}{5} \text{ hours} = \frac{2}{5} \times 60 = 24 \text{ minutes} \]

### Conclusion
The time gap between one city bus passing a stationary point and the immediately next city bus passing the same point is **24 minutes**.