A bus left 60 minutes later than the scheduled time but in order to reach its destination 48 km away in time, it had to increase the speed by 4km/hr from the usual speed. What is usual speed (in km/hr) of the bus?

Options

A. A) 9
B. B) 12
C. C) 15
D. D) 8

Answer

To solve the problem step by step, we will use the information provided in the question to set up equations based on the relationship between distance, speed, and time.

### Step 1: Define Variables
Let the usual speed of the bus be \( V \) km/hr.

### Step 2: Calculate Time Taken at Usual Speed
The time taken to cover 48 km at the usual speed \( V \) is given by:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{48}{V} \quad \text{(Equation 1)} \]

### Step 3: Calculate Time Taken at Increased Speed
If the bus increases its speed by 4 km/hr, the new speed becomes \( V + 4 \) km/hr. The time taken to cover the same distance of 48 km at this increased speed is:
\[ \text{Time} = \frac{48}{V + 4} \quad \text{(Equation 2)} \]

### Step 4: Relate the Two Times
Since the bus left 60 minutes (or 1 hour) late, the time taken at the increased speed must be equal to the time taken at the usual speed minus 1 hour:
\[ \frac{48}{V + 4} = \frac{48}{V} - 1 \]

### Step 5: Clear the Fractions
To eliminate the fractions, we can multiply through by \( V(V + 4) \):
\[ 48V = 48(V + 4) - V(V + 4) \]

### Step 6: Expand and Simplify
Expanding both sides gives:
\[ 48V = 48V + 192 - V^2 - 4V \]
Now, simplifying this equation:
\[ 0 = 192 - V^2 - 4V \]
Rearranging gives:
\[ V^2 + 4V - 192 = 0 \]

### Step 7: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:
\[ V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 4 \), and \( c = -192 \):
\[ V = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-192)}}{2 \cdot 1} \]
\[ V = \frac{-4 \pm \sqrt{16 + 768}}{2} \]
\[ V = \frac{-4 \pm \sqrt{784}}{2} \]
\[ V = \frac{-4 \pm 28}{2} \]

### Step 8: Calculate Possible Values for V
Calculating the two possible values:
1. \( V = \frac{24}{2} = 12 \)
2. \( V = \frac{-32}{2} = -16 \) (not a valid speed)

Thus, the usual speed of the bus is:
\[ V = 12 \text{ km/hr} \]

### Final Answer
The usual speed of the bus is **12 km/hr**.

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