A plane left 30 minutes later than its scheduled time to reach its destination…

Concept: For a fixed distance, Time = Distance / Speed. If departing later still allows on-time arrival by increasing speed, the extra speed must exactly recover the time already lost -- so the new travel time equals the original travel time minus the delay.

Application: Let the original speed be s kmph and t hrs the time it would normally take to cover 1500 km.

1. At the original speed, s x t = 1500, so t = 1500/s, where s is the original speed (kmph) and t is the normal travel time (hrs).

2. Departing 30 minutes (0.5 hr) late but still arriving on time means the faster leg must take 0.5 hr less: new time = t - 0.5.

3. The new speed is s + 250, and the distance is unchanged, so (s + 250)(t - 0.5) = 1500.

4. Substituting t = 1500/s: (s + 250)(1500/s - 0.5) = 1500.

5. Expanding: 1500 + 375000/s - 0.5s - 125 = 1500. Subtracting 1500 from both sides: 375000/s - 0.5s - 125 = 0.

6. Multiplying throughout by s: 375000 - 0.5s2 - 125s = 0. Multiplying by -2 to clear the fraction: s2 + 250s - 750000 = 0.

7. Solving this quadratic (discriminant = 2502 + 4 x 750000 = 3062500, whose square root is 1750) gives s = (-250 + 1750)/2 = 750; the negative root is rejected since speed cannot be negative.

Cross-check: At 750 kmph, the 1500 km trip takes 2 hours. Raising the speed by 250 kmph to 1000 kmph cuts this to 1.5 hours -- a saving of exactly 30 minutes, matching the delay in departure.

Hence, the original speed is 750 kmph.