Answer:
Its usual speed is 25 kilometre/hour
Explanation:
Let the usual time duration for a journey of 300 km. = t
Let the usual speed = v
The parameters given are;
The time duration for the 300 km. journey at increased speed = t - 2
The increased speed of the passenger train = v + 5
Distance, d = Speed, v × Time, t
Therefore, we have;
v × t = 300
∴ t = 300/v
Also (v + 5)×(t - 2) = 300
Substituting the value of t = 300/v, we have;
(v + 5)×(300/v - 2) = 300
![- (2 \cdot v^2 - 290 \cdot v - 1500)/(v) = 300](https://img.qammunity.org/2021/formulas/mathematics/high-school/uoeo0ddy64xy2fs5iibfqdqethvnhgsz90.png)
Which gives;
2·v² + 10·v - 1500 = 0
Which is equivalent to v² + 5·v - 750 = 0
Therefore we have;
(v + 30)·(v - 25) = 0 whereby v = -35 or 25 km/h
v = Natural number = 25 km/hour
Therefore its usual speed is 25 kilometre/hour.