An aeroplane started 30 minutes later than scheduled from a place 1500 km from…

Concept

For a fixed distance, travel time equals distance divided by speed. If a journey starts late but must finish at the original time, the increase in speed must reduce the travel time by exactly the starting delay.

Application

1. Let x km/h be the scheduled speed. The actual journey speed was therefore (x + 250) km/h.

2. The scheduled travel time was 1500/x hours, while the actual travel time was 1500/(x + 250) hours.

3. The 30-minute delay equals 1/2 hour, so 1500/x − 1500/(x + 250) = 1/2.

4. Combining the fractions gives 375000/[x(x + 250)] = 1/2, hence x(x + 250) = 750000.

5. Thus x2 + 250x − 750000 = 0 = (x − 750)(x + 1000). Speed must be positive, so x = 750 km/h.

6. The speed during the actual journey was x + 250 = 750 + 250 = 1000 km/h.

Cross-check

At 750 km/h, the scheduled travel time is 1500/750 = 2 hours. At 1000 km/h, the actual travel time is 1500/1000 = 1.5 hours, exactly 30 minutes less.

Result

Therefore, the aeroplane travelled at 1000 km/h during the journey.