Q. In a fight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of the flight.

Ans: 

Step 1: Let’s consider that the aircraft’s original average speed was X km/kr and that it takes Y hours to cover the given distance of 600 km.

Since Speed =

∴ X = ……..(i)

Step 2: By given condition, when speed is reduced by 200 km/hr, then the flight time increased by 30 mins

Here, 30 mins = hr

∴ X – 200 =

∴ 600 = (X – 200) (Y + ) …… (ii)

Step 3: By substituting the value of X from equation (i) in equation (ii), we get:

600 = ( – 200) (Y + )

∴ 600 = () ()

∴ (2) (Y) (600) = (600 – 200 Y)(2 Y + 1)

∴ 200 x 6 Y = 200 (3 – Y)(2 Y + 1)

∴ x 6 Y = (3 – Y) (2 Y + 1)

∴ 6 Y = (3 – Y) (2 Y + 1)

∴ 6 Y = 6 Y – 2 Y ² + 3 – Y

∴ = – 2 Y ² + 3 – Y

∴ 0 = – 2 Y ² – Y + 3 = 0

∴ 2 Y ² + Y – 3 = 0

Step 4: We solve the above quadratic equation by mid term splitting:

∴ 2 Y ² + 3 Y – 2 Y – 3 = 0

∴ Y (2 Y + 3) – (2 Y + 3) = 0

∴ (2 Y + 3) (Y – 1) = 0

∴ Y = and Y = 1

Here we reject, Y = because Y is the time and it can not be a negative value

∴ Y = 1

Therefore, The duration of the flight is 1 hour.

(Note: Here since we were asked to calculate flight time i.e. Y, therefore in equation (i), we calculated value of X in terms of Y. This helps us to dissolve unwanted variables and hence saves time to calculate. e.g. when we substituted the value of x in equation (ii), X was dissolved and we received Quad equation only in terms of y.) 

Check:
If Y = 1, then X = 600 /1 = 600 km/hr (from equation i)
Time taken in normal course = 600/ 600 = 1 hr
Time taken when the aircraft slows down = 600/(600 – 200) = 600/400 = 1.5 hrs
Extra time taken in slowed down journey = 1.5 – 1 = 0.5 hrs = 30 mins
Since it matches with the given condition on the question, hence our solution is correct.

Please press the “Heart” button if you like the solution.