For a journey of 600 km, an aircraft was slowed down due to bad weather. Its…

Concept: For a fixed route, distance = speed × time stays constant. When the speed is reduced by a fixed amount and, as a result, the time taken increases by a fixed amount, writing the usual speed and usual time as two unknowns and forming two equations (before and after the change) lets you eliminate one unknown and solve the resulting quadratic in the other — the physically valid (positive) root is the answer.

Application:

1. Let the usual speed be x km/hr and the usual time be t hours. Since distance = speed × time, xt = 600.

2. After the slowdown, speed = (x − 200) km/hr and time = (t + 0.5) hours, covering the same 600 km: (x − 200)(t + 0.5) = 600.

3. Expanding: xt + 0.5x − 200t − 100 = 600. Since xt = 600, this reduces to 0.5x − 200t − 100 = 0.

4. Substitute x = 600/t (from step 1) into this equation: 0.5(600/t) − 200t − 100 = 0, i.e. 300/t − 200t − 100 = 0.

5. Multiply through by t and divide by 100: 3 − 2t2 − t = 0, i.e. 2t2 + t − 3 = 0.

6. Factor the quadratic: (2t + 3)(t − 1) = 0, giving t = 1 or t = −3/2.

7. Time cannot be negative, so the usual duration is t = 1 hour.

Cross-check: With t = 1 hour, usual speed x = 600/1 = 600 km/hr. After the slowdown, speed = 400 km/hr and time = 1.5 hours, giving distance = 400 × 1.5 = 600 km — matching the given distance, confirming the result.