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

Let usual speed be x km/hr

Time taken = $\dfrac{1500}{x}$ hours

If speed is increased by 250 km/hr then time taken = $\dfrac{1500}{x + 250}$ hours

Given, on increasing speed it takes 30 minutes less,

$\therefore \dfrac{1500}{x} - \dfrac{1500}{x + 250} = \dfrac{30}{60} \\[1em] \Rightarrow \dfrac{1500(x + 250) - 1500x}{x(x + 250)} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1500x + 375000 - 1500x}{x^2 + 250x} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{375000}{x^2 + 250x} = \dfrac{1}{2} \\[1em] \Rightarrow 750000 = x^2 + 250x \\[1em] \Rightarrow x^2 + 250x - 750000 = 0 \\[1em] \Rightarrow x^2 + 1000x - 750x - 750000 = 0 \\[1em] \Rightarrow x(x + 1000) - 750(x + 1000) = 0 \\[1em] \Rightarrow (x - 750)(x + 1000) = 0 \\[1em] \Rightarrow x - 750 = 0 \text{ or } x + 1000 = 0 \\[1em] \Rightarrow x = 750 \text{ or } x = -1000.$

Since, speed cannot be negative

∴ x ≠ -1000.

Hence, usual speed = 750 km/hr.