A person travels \(300\) kilometres. If the person's speed is reduced by \(10\) kilometres per hour, the travel time increases by \(1\) hour. What was the original speed?

60 किलोमीटर प्रति घंटा (60 km/h)

Let the original speed be \(x\) km/h. The original travel time is \(\frac{300}{x}\) hours, while the time at the reduced speed is \(\frac{300}{x-10}\) hours. Therefore, \(\frac{300}{x-10}-\frac{300}{x}=1\). On simplifying, \(x^2-10x-3000=0\), giving \(x=60\) or \(x=-50\). Since speed cannot be negative, the original speed is \(60\) km/h. Check: at 60 km/h the time is 5 hours, and at 50 km/h it is 6 hours, an increase of exactly 1 hour. Exam tip: reject physically impossible negative roots and verify the remaining value in the original condition.

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