Question

An aeroplane left 40 minutes late due to heavy rains and in order to reach its destination. 1,600 km away in time, it had to increase its speed by 400 km / hr from its original speed. Find the original speed of the aeroplane.
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Answer 1

distance is 1600 let the usual speed by X km / hr we know that speed is equal to distance / time and time is equal to distance / speed usual time is 1600/ X now due to bad weather the speed is Inc B 400km/ hr which means new speed is = 1600/ X
+ 400 now it's given than the plane left 40 mins late which means new time =1600/x + 400 + 2/3 ( 40/60 = 2/3hrs) 1600/X = 1600/X + 400 +2/3 1600/X = 1600 /X +400=2/3 1600 ( X + 400 ) _ 1600X) × ( X+ 400) = 2/3 1600 X + 640000_ 1600 X / X2 + 400 X = 2/3 640000 = 2/3 (X2+ 400 X) 640000× 3/2 = X2 + 400 X 960000 = X2 + 400 X 0 = X 2 + 400 X _ 960000 0 = X2 + 1200 X _ 80O X _ 960000 0- X ( X + 1200) _ 800 (X+ 1200) 0 = (X+ 1200) (X-800) X = _1200 or X= 800 speed cannot be negative because usual speed = 800 km / hr

Answer 2

Let the usual speed of the aeroplane be x km/hr.

Given that Distance = 1600km.

We know that Time(t1) = Distance/Speed

                                     = 1600/x


Given that it has to increase its speed by 400km/hr.

We know that Time(t2) = 1600/x + 400

Given that Time t2 - t1 = 40 minutes.

                                       = 40/60

                                       = 4/6



= > (1600/x) - (1600/x + 400) = 4/6

= > (3840000/x^2 + 400x) = 4

= > 3840000 = 4(x^2 + 400x)

= > 4x^2 + 1600x - 3840000 = 0 

= > x^2 + 400x - 960000 = 0

= > x^2 - 800x + 1200x - 960000 = 0

= > x(x - 800) + 1200(x - 800) = 0

= > (x + 1200)(x - 800) = 0

= > x = 800,-1200.


The speed cannot be negative.


Therefore the original speed of the aeroplane = 800km/hr.



Hope this helps