Question

In a flight for 3000 kilometre and aircraft was slowed down due to bad weather its average speed for the trip was reduced by hundred kilometre per hour and consequently time of flight increased by 1 hour find the original duration of flight

Answer 1

The original duration of the flight is of 5 hours.

Explanation:

Given:

Total distance covered by the flight =
`3000` km

Let the speed of the flight in normal condition be
`x` km/hr

And the average speed in bad weather = (
`x-100`) km/hr

Time taken increases when the speed of the flight decreases.

Time taken in normal condition,
`t_1` =
`(distance)/(speed)` =
`(3000)/(x)` hr

Time taken in bad weather,
`t_2=(3000)/(x-100) +1` hr

Here we can observe the difference of time =
`1` hr

So,

⇒
`t_2-t_1 =1`

⇒
`(3000)/((x-100))-(3000)/((x)) =1`

⇒
`(3000(x)-3000(x-100))/((x)(x-100)) =1`

⇒
`3000(x)-3000(x-100)=(x)(x-100)`

⇒
`3000x-3000x+300,000 =x^2-100x`

⇒
`x^2-100x-30,000=0`

⇒
`x^2-600x+500x-300,000=0` using middle term splitting method.

⇒
`x(x-600)+500(x-600)=0`

⇒
`(x+500)(x-600)=0`

⇒
`x=600 ,-500`

Note: We can also use quadratic formula directly to find 'x' values.

Discarding the negative value as speed can't be negative.

Speed of the flight in normal condition =
`600` km\hr

Speed of the flight in bad weather =
`500` km\hr

Original duration of the flight = Ratio of distance and speed in normal condition.

⇒ Original duration =
`(3000)/(600)= 5` hr

But due to bad weather the duration of the flight will be an hour more that is
`6` hrs.

The original duration of the flight is 5 hours.