Let the original speed of the train be s'  km / hr and the original time taken be 't' hours. 

We know that distance = speed $\times$ time. 

Therefore: 300 = s$\times$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)$\times$(t - 1) 

$\Rightarrow 300 = st - s + 10t - 10$

$\Rightarrow 300 = 300 - s + 10t - 10$

$\Rightarrow 300 - 300 = - s + 10t - 10$

$\Rightarrow s = 10t - 10$

$\therefore t=\frac{\left(s+10\right)}{10}$

Now substitute this value of 't' back into equation (i) 

$300 = s \times \frac{(s + 10)}{10}$

$\Rightarrow 3000 = s(s + 10)$[Multiply both sides by 10] 

$\Rightarrow$ 3000 = s² + 10s 

$\Rightarrow$ s² + 10s - 3000 = 0 

$\Rightarrow$ 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.