A train travels a distance of 300 km at a constant speed. If the speed had been 10 km/hr faster, the train would have taken 1 hour less to cover the same distance. What is the original speed of the train?
Let the original speed of the train be s' km / hr and the original time taken be 't' hours.
We know that distance = speed × time.
Therefore: 300 = s×t ______ (i)
If the speed had been 10 km/hr. faster, the new speed would be (s + 10) km / hr and the time taken would be (t - 1) hours. The distance remains the same:
Therefore: 300 = (s + 10)×(t - 1)
⇒
Now substitute this value of 't' back into equation (i)
[Multiply both sides by 10]
3000 = s2 + 10s
s2 + 10s - 3000 = 0
s2 + 60s - 10s - 3000 = 0
∴ (s + 60)(s - 50) = 0
This gives us two possible solutions for 's': s = - 60 or s = 50
Since speed cannot be negative, the original speed of the train is 50 km/hr.
Answer: The original speed of the train was 50 km/hr.