A plane left 30 min later than its scheduled time to reach its destination 1500 km away. In order to reach in time it increases its speed by 250 km/h. What is its original speed?

Let the usual speed of the plane be v kmph and the time taken by the plane to cover 1500m = t hrs

To reach the destination on time, the increased speed of the plane = v+250 kmph

Since he started 30 min later, the time taken should be t-$$\ \frac{\ 1}{2}$$ hr

v*t = (v+250)(t-$$\ \frac{\ 1}{2}$$) 

v = (250t-125)2

$$\ \frac{\ 1500}{v+250}=t-\ \frac{\ 1}{2}$$

$$\ \frac{\ 1500}{500t-250+250}=t-\ \frac{\ 1}{2}$$

$$\ \frac{\ 3}{t}=t-\ \frac{\ 1}{2}$$

On solving for t, we get

t= 2 hrs

His original speed = $$\ \frac{\ 1500}{2}$$ = 750 kmph

B is the correct answer.