A train covers a distance of 360 kilometres. If its speed is increased by 15 kilometres per hour, the travel time decreases by 2 hours. What was the train’s original speed?

45 km/h

Let the original speed be \(x\) km/h. Then \(\frac{360}{x}-\frac{360}{x+15}=2\). On simplifying, \(x(x+15)=2700\), so \(x^2+15x-2700=0\). Factoring gives \((x-45)(x+60)=0\), hence \(x=45\) or \(x=-60\). Since speed cannot be negative, the original speed is \(45\) km/h. Check: at 45 km/h the time is 8 hours, and at 60 km/h it is 6 hours, a decrease of exactly 2 hours. Exam tip: reject any negative root when the variable represents speed, distance, or time.

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