Question

An aeropale start late by 30 min. To reach the destination which is 1500 km away.the pilot ase the speed by 250km/hour just to reach on time. Find orignal speed of aeroplane

Answer 1

let the speed of the plane be x km per hour 

case 1 
speed=x
distance=1500

time=1500/x



case 2 
speed=x+250
distance=1500

time=1500/x+250

According to problem 

1500/x - 1500/x+250 = 30/60

1500 multiplied into (250 + x) - 1500x / x(250+x)= 1/2


375000 + 1500x - 1500x/ x square + 250x = 1/2
(1500x-1500x=0)

750000 = x square + 250x

x square + 250x - 750000=0 

x square + 1000x-750x -750000= 0

x(x+1000) - 750(X+1000)=0
(x-750)(x+1000)=0

x-750=0 and x+1000=0

x=750 km per hour and x= -1000 km per hour 

x=-1000 km per hour gets rejected because speed can't be in negative 

So, the original speed pf aeroplane is 750 km per hour.

Answer 2

Solution :-

Let the original speed of train be x km/hr

New speed = (x + 250) km/hr

We know that,

Time = Distance / Speed

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

According to the question,

=> 1500/x - 1500/(x + 250) = 1/2

=> (1500x + 37500 - 1500x)/x(x + 250) = 1/2

=> 2(37500) = x(x + 250)

=> 75000 = x² + 250x

=> x² + 250x - 75000 = 0

=> x² + 1000x - 750x - 75000 = 0

=> x(x + 1000) - 750(x + 1000) = 0

=> (x - 750) (x + 1000) = 0

=> x = 750 or x = - 1000

∴ x ≠ - 1000 (Because speed can't be negative)

Hence,

Its usual speed = 750 km/hr