Question

In a flight of 2800km an aircraft was slowed down due to bad weather and its average speed was reduced to 100km/hr and time increased 30min.Find original duration of the flight.

Answer 1

Let the speed of aircraft be x km/h

Distance travelled = 2800 km

Time taken = Distance / Speed = (2800/x ) hours

Given that when the speed is reduced by 100 km/h, time is reduced by 30 min

New speed = (x – 100) km/h

That is, 30 min�= 30/60 hr�= (1/2) hr�

Time taken after the speed was reduce = [2800/(x – 100)] hr

Hence, � [2800/(x – 100)] =�(2800/x) + (1/2)

[2800/(x-100)]� -�[2800/x]� =� 1/2

Hence, x2�- 100x - 560000� =�0

x2�- 800x + 700x - 560000� =� 0

x(x - 800) + 700(x - 800)� =� 0

(x + 700) (x - 800)� =� 0

x� =� - 700 or 800

Since speed cannot be negative, x = 800

Thus�original duration of the flight�=� 2800/800 = 7/2 hours

� = 3 hr 30 mins

Answer 2

Let original duration of flight be x hrs.
Distance =2800km
Speed = distance/time
(2800/x) km/hrs

New duration of flight = x hrs +30 min.
= x+0.5 hrs.
Distance = 2800 km
Speed = d/t
= 2800/x+0.5 km/hr.

Since, New speed = 100 km/hr.
2800/x+0.5 = 100
2800 = 100x + 50
100x = 2800-50
100x = 2750
x = 27. 5 hrs.
= 27 hrs. 30 mins.