Question

A plane left 30 min late than its scheduled time.in order to reach the distination 1500kn in time, the speed has to be increase 250km/h from the usual speed.find the usual speed

Answer 1

Given: distance=1500km 

Let the original speed of a plane =x km/h

New speed= x+250                            ( speed increase by 250 km/h)

Time taken at original speed = distance/ speed= 1500/x h

Time taken at new speed = distance/ speed= 1500/x+250 h

∴ 1500/x = 1500/x+250 + 30/60                    (30 min= 30 /60= 1/2 hr)

1500/x = 1500/x+250 + 1/2
 
 1500/x - 1500/x+250= 1/2

  (1500 (x + 250) - 150 x) / x(x+250)= 1/2  

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

1500 × 250 / x² +250 x = 1/2

 2(1500 × 250) =  x² +250 x

2(375000) =  x² +250 x

750000 =  x² +250 x

 x² +250 x - 750000=0

x² +1000x - 750x -750000=0

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

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

(x+1000) =0

x= -1000           ( speed cannot be negative)
 
(x-750) =0

x= 750 

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Hence the original speed (usual speed) = 750 km /h

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Hope this will help you.....

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