Question

A plane left 30 minutes later than the schedule time and in order to reach the destination 1500 km away in time it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

Answer 1

Let the usual speed of the plane be x km/h.

Time taken to cover 1500 km with the usual speed of x km/h = 1500 / x hrs
Time taken to cover 1500 km with the increase speed of (x+250) km/h = 1500 / (x+250) hrs

ATQ
1500 / x = 1500 / (x+250) + ½
1500 / x -  1500 / (x+250) = ½
1500(x+250) -1500x / x(x+250)= ½
1500x + 1500 × 250 -1500x / x² +250x = ½
1500 × 250 / x² +250x = ½
2(1500 × 250 ) = x² +250x
750000 = x² +250x
x² +250x - 750000 =0
x² +1000x -750x - 750000= 0
x(x+1000) -750(x + 1000)= 0
(x+1000)  (x - 750) = 0
(x+1000)  = 0  or  (x - 750) = 0

x = -1000  or  x = 750

Speed cannot be negative, so x = 750

Hence,  the usual speed of the plane is 750 km/h.

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Answer 2

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Answer,


Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km
Case (i)
Speed = Distance / Time = (1500 / x) Hrs
 
Case (iI)
Time taken by the aeroplane = (x - 1/2) Hrs
Distance to the destination = 1500 km
Speed = Distance / Time = 1500 / (x - 1/2) Hrs
 
Increased speed = 250 km/hr
 
⇒ [1500 / (x - 1/2)] - [1500 / x] = 250
⇒ 1/(2x2 - x) = 1/6
⇒ 2x2 - x = 6
⇒ (x - 2)(2x + 3) = 0
⇒ x = 2 or -3/2
Since, the time can not be negative,
The usual time taken by the aeroplane = 2 hrs
and the usual speed = (1500 / 2) = 750 km/hr.



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