A plane left 30 minutes later than the scheduled time and in order to reach the destination 1500 km away in time, it had to increase the speed by 250 km/hr from the usual speed. Find its usual speed.

Let usual speed of the plane be 's' km/hr and new speed be 'x + 250' km/hr

As the plane left 30 minutes late, the difference between this time and usual time will be equal to 30 minutes. 

$$\frac{1500}{s} - \frac{1500}{x + 250} = \frac{30}{60}$$

$$\frac{1500s + 1500 \times 250 - 1500s}{s(s + 250)} = \frac{1}{2}$$

$$1500(s + 250 - s) \times 2 = s(s+250)$$

$$75000 = s^{2} + 250s$$

 $$s^{2} + 250s - 7,50,000 = 0$$

Roots for the above equation are $$750$$ and '$$-1000$$'

As '$$-1000$$' is a negative value, consider only $$750$$.

Hence, option D is the correct answer.