Given, passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed.
Let the usual speed be ‘a’ km/hr
Increased speed = (a + 5) km/hrwe know, \(time=\frac{distance}{speed}\)
⇒ time taken by train to cover with usual speed to travel 300 km \(\frac{300}{a}\) km/h
⇒ time taken by train to cover with increased speed to travel 300 km \(=\frac{300}{a+5}\) km/h
Therefore,
According to question
\(\Rightarrow \frac{300}{a}-\frac{300}{a+5}=2\)
⇒ 300 ×(a + 5 –a) = 2(a2 + 5a)
⇒ a2 + 5a – 750 = 0
⇒ a2 + 30a – 25a – 750 = 0
⇒ a(a + 30) – 25(a + 30) = 0
⇒ (a + 30)(a – 25) = 0
⇒ a = 25 km/hr
