A flight has to travel between 2 cities A and B, 3000 km apart. The flight was…

20172017

A flight has to travel between 2 cities A and B, 3000 km apart. The flight was slowed down due to bad weather. Its average speed for the trip reduced by 200 km/hr and the time of flight increased by 30 min. The duration of the original speed is:

  1. A.

    2.6 hrs

  2. B.

    2.75 hrs

  3. C.

    2.25 hrs

  4. D.

    2.5 hrs

Attempted by 37 students.

Show answer & explanation

Correct answer: D

Concept

For a fixed distance, time = distance ÷ speed. If the same distance is covered at two different speeds, the difference between the two travel times equals the stated change. Writing that time difference as an equation in the unknown original speed lets you solve for it, after which the original duration follows directly.

Application

  1. Let the original speed be x km/hr. Original time = 3000/x hours.

  2. Reduced speed = (x − 200) km/hr, so the new time = 3000/(x − 200) hours.

  3. The journey takes 30 min = 0.5 hr longer, so 3000/(x − 200) − 3000/x = 0.5.

  4. Combine over a common denominator: [3000x − 3000(x − 200)] / [x(x − 200)] = 0.5, i.e. 600000 / [x(x − 200)] = 0.5.

  5. Cross-multiply: x(x − 200) = 1200000, giving x2 − 200x − 1200000 = 0.

  6. Factorise: (x − 1200)(x + 1000) = 0. Speed must be positive, so x = 1200 km/hr.

  7. Original duration = distance ÷ speed = 3000 ÷ 1200 = 2.5 hours.

Cross-check

At 1200 km/hr the trip takes 2.5 hr; at 1000 km/hr it takes 3000/1000 = 3 hr. The difference is exactly 0.5 hr = 30 min, matching the given condition.

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