In a flight of 2800 km, an aircraft was slowed down due to bad weather. its average speed is reduced by 100km/h and time increased by 30 mins. find the original duration.
Answers
speed of aircraft= x
Distance = 2800 km
Time = Distance / Speed = 2800/x
Given:
The speed has reduced by 100 and time is reduced by 30 min
New speed = (x – 100)
That is, 30 min = 30/60 hr = (1/2) hr
Time taken= 2800/(x – 100) hr
=2800/(x – 100)
= (2800/x) + (1/2)
2800/(x-100) -` (2800/x) = 1/2
x² - 100x - 560000 = 0
x² - 800x + 700x - 560000 = 0
x(x - 800) + 700(x - 800) = 0
(x + 700) (x - 800) = 0
SO,
x = - 700 or 800
Thus original duration of the flight = 2800/800 = 7/2 hours
OR = 3 hr 30 mins
▶ Answer :-
→ The original duration of the flight = 3 hours 30 minutes .
▶ Step-by-step explanation :-
→ Let the original speed of the aircraft be x km/hr .
→ Time taken to cover 2800 km = hours .
→ Reduced speed = ( x - 100 ) km/hr .
→ Time taken to cover 2800 km at this speed = hours .
▶ Now,
•°• Original speed of the aircraft = 800 km/hr .
And, original duration of the flight = hours = 3 hours 30 minutes .
✔✔ Hence, it is solved ✅✅.