A flight has to travel between 2 cities A and B, 1200 km apart. The flight was…

Concept

For a fixed distance d, speed and time are linked by d = speed × time, so time = d ÷ speed. If the speed changes, the new time can be written for the same distance, and the difference between the two times equals the stated delay. Setting up that single time-difference equation in one unknown (the original speed) lets us solve the problem.

Application

1. Let the original speed be v km/hr. The distance is fixed at 1200 km, so the original time is 1200 ÷ v hours.

2. The reduced speed is (v − 200) km/hr, so the slower time is 1200 ÷ (v − 200) hours.

3. The delay is 30 min = 0.5 hr, so: 1200 ÷ (v − 200) − 1200 ÷ v = 0.5.

4. Multiply through by v(v − 200): 1200v − 1200(v − 200) = 0.5 · v(v − 200), giving 1200 × 200 = 0.5(v² − 200v), i.e. 240000 = 0.5v² − 100v.

5. Rearrange: v² − 200v − 480000 = 0. Factoring, (v − 800)(v + 600) = 0, so v = 800 km/hr (reject the negative root).

6. Original time = 1200 ÷ 800 = 1.5 hr.

Cross-check

At the reduced speed 800 − 200 = 600 km/hr, the time is 1200 ÷ 600 = 2 hr. The difference 2 − 1.5 = 0.5 hr = 30 min, matching the stated delay, so the original duration of 1.5 hr is consistent.