If the speed of a bus is increased by 5 km/hour from its normal speed, it…

For a fixed distance, time = distance ÷ speed, so time and speed vary inversely of each other. When a speed is increased and the same distance is then covered in a known number of fewer hours, equating the time taken at the original speed to the time taken at the increased speed (with their difference equal to the given time saving) gives one quadratic equation in the original speed. Since speed must be positive, only the positive root of that quadratic is a valid answer; any negative root is rejected.

Let the normal speed of the bus be v km/h.

1. Time taken to cover 300 km at the normal speed v = 300/v hours.

2. After the speed is increased by 5 km/h, the new speed is (v + 5) km/h, and the time taken for the same 300 km is 300/(v + 5) hours.

3. The new journey takes 2 hours less than the normal journey: 300/v − 300/(v + 5) = 2.

4. Combine the two fractions over the common denominator v(v + 5): [300(v + 5) − 300v] / [v(v + 5)] = 2.

5. Simplify the numerator: 300v + 1500 − 300v = 1500, so 1500 / [v(v + 5)] = 2.

6. Cross-multiply: 1500 = 2v(v + 5) = 2v2 + 10v.

7. Divide by 2 and rearrange into standard quadratic form: v2 + 5v − 750 = 0.

8. Factorise: (v − 25)(v + 30) = 0, so v = 25 or v = −30.

9. Since speed cannot be negative, reject v = −30; hence v = 25 km/h.

Check: at v = 25 km/h, the normal time is 300/25 = 12 hours; at the increased speed of 30 km/h, the time is 300/30 = 10 hours. The difference is 12 − 10 = 2 hours, exactly as given — confirming the value.

Hence, the normal speed of the bus is 25 km/hour.