In a flight of 3000 km, an aircraft was slowed down due to bad weather.Its average speed for the trip was reduced by 100 km/hr and the time increased by one hour.Find the original duration of flight.
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Solution:-
Let the usual speed of the aircraft be 'x' km/hr
So, the reduced speed will be (x - 100) km/hr
Time taken to reach the destination at usual speed
T1 = (3000/x) hr
Time taken to reach the destination at reduced speed
T2 = (3000/x-100) hr
Given : T2 - T1 = 1 hr
∴ (3000/x-100) hr - (3000/x) hr = 1 hr
⇒ (3000x - 3000x + 300000)/(x)(x-100) = 1 hr
⇒ x² - 100x - 300000 = 0
⇒ x² - 600x + 500x - 300000 = 0
⇒ x(x - 600) + 500(x - 600) = 0
⇒ (x - 600)(x + 500) = 0
⇒ x = 600 or x = - 500
Speed cannot be negative. Hence x = 600 km/hr is correct speed.
The speed was reduced by 100 km/hr, so 600 - 100 = 500 km/hr was the original speed.
Original duration of flight = 3000/500 = 6 hours
Answer.
Let the usual speed of the aircraft be 'x' km/hr
So, the reduced speed will be (x - 100) km/hr
Time taken to reach the destination at usual speed
T1 = (3000/x) hr
Time taken to reach the destination at reduced speed
T2 = (3000/x-100) hr
Given : T2 - T1 = 1 hr
∴ (3000/x-100) hr - (3000/x) hr = 1 hr
⇒ (3000x - 3000x + 300000)/(x)(x-100) = 1 hr
⇒ x² - 100x - 300000 = 0
⇒ x² - 600x + 500x - 300000 = 0
⇒ x(x - 600) + 500(x - 600) = 0
⇒ (x - 600)(x + 500) = 0
⇒ x = 600 or x = - 500
Speed cannot be negative. Hence x = 600 km/hr is correct speed.
The speed was reduced by 100 km/hr, so 600 - 100 = 500 km/hr was the original speed.
Original duration of flight = 3000/500 = 6 hours
Answer.
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