Explanation:

- Let's denote the usual speed of the bus as \( x \) km/hr.

- The usual time to cover 48 km is \( \frac{48}{x} \) hours.

- With the increased speed, the bus travels at \( x + 4 \) km/hr.

- The time taken at the increased speed is \( \frac{48}{x+4} \) hours.

- Since the bus left 60 minutes (or 1 hour) late, the time difference becomes 1 hour.

- We set up the equation: \( \frac{48}{x} - \frac{48}{x+4} = 1 \).

- Solving this equation gives \( x = 12 \).

The usual speed of the bus is: 12 km/hr

Option:2, 12