A bus starts running with some initial speed and its speed increases every hour by 9km/hr. If it takes 11 hours to cover a distance of 572km, then what was the initial speed (in km/hr) of the bus?

Options

A. `3.5`
B. 7
C. `10.5`
D. 14

Answer

To solve the problem, we need to determine the initial speed of the bus, given that its speed increases by 9 km/hr every hour and it covers a distance of 572 km in 11 hours.

### Step-by-Step Solution:

1. **Define Variables**:
Let the initial speed of the bus be \( x \) km/hr.

2. **Determine Speeds for Each Hour**:
- In the first hour, the speed is \( x \) km/hr.
- In the second hour, the speed is \( x + 9 \) km/hr.
- In the third hour, the speed is \( x + 18 \) km/hr.
- Continuing this pattern, the speed in the \( n \)-th hour can be expressed as:
\[ \text{Speed in } n\text{-th hour} = x + 9(n-1) \]

3. **Calculate Total Distance Covered in 11 Hours**:
The total distance covered in 11 hours can be expressed as the sum of the speeds for each hour:
\[ \text{Total Distance} = x + (x + 9) + (x + 18) + \ldots + (x + 90) \]
This can be simplified to:
\[ \text{Total Distance} = 11x + (0 + 9 + 18 + \ldots + 90) \]

4. **Sum of the Arithmetic Series**:
The series \( 0 + 9 + 18 + \ldots + 90 \) is an arithmetic series where:
- First term \( a = 0 \)
- Common difference \( d = 9 \)
- Number of terms \( n = 11 \)

The sum of the first \( n \) terms of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Plugging in the values:
\[ S_{11} = \frac{11}{2} \times (2 \times 0 + (11-1) \times 9) = \frac{11}{2} \times (0 + 90) = \frac{11 \times 90}{2} = 495 \]

5. **Total Distance Equation**:
Now, substituting back into the total distance equation:
\[ 572 = 11x + 495 \]

6. **Solve for \( x \)**:
Rearranging the equation:
\[ 11x = 572 - 495 \]
\[ 11x = 77 \]
\[ x = \frac{77}{11} = 7 \]

7. **Conclusion**:
The initial speed of the bus is \( 7 \) km/hr.