A bus covers a distance of 180 kilometres. If its speed is increased by 15 kilometres per hour, the travel time decreases by 1 hour. What was the original speed of the bus?

45 kilometres per hour

Let the original speed be \(x\) kilometres per hour. The original travel time is \(\frac{180}{x}\) hours, while the time at the increased speed \(x+15\) is \(\frac{180}{x+15}\) hours. Hence, \(\frac{180}{x}-\frac{180}{x+15}=1\). On simplifying, \(x^2+15x-2700=0\), or \((x-45)(x+60)=0\). Thus, \(x=45\) or \(x=-60\); since speed cannot be negative, the original speed was 45 kilometres per hour. Exam tip: In distance–speed–time problems, use \(\text{time}=\frac{\text{distance}}{\text{speed}}\) before forming the quadratic equation.

This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Word Problems and Applications.