A plane left 30 min late than its scheduled time and in order to reach the destination 1500km away in time,it had to increase its speed by 100km/h from the usual speed.Find the usual speed
Answers
Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km
--------Speed = Distance / Time = (1500 / x) Hrs
------------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 = 100 km/hr
[1500 / (x - 1/2)] - [1500 / x] = 100
150((10/x-1/2)-10/x0=100
15 /x -1/2- 15/x =1
30x -30x +15=2x^2-x
15=2x^2-x
2x^2-x-15=0
2x^2 - 6x+5x-15=0
2x(x-3)+5(x-3)
2x+5=0 or x-3 =0
x=-5/2 or x=3
Since, the time can not be negative,
The usual time taken by the aeroplane = 3hrs
and the usual speed = (1500 / 3) = 500km/hr
Solution :-
Let the original speed of train be x km/hr
New speed = (x + 100) 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 + 100) = 1/2
=> (1500x + 15000 - 1500x)/x(x + 100) = 1/2
=> 2(15000) = x(x + 100)
=> 30000 = x² + 100x
=> x² + 100x - 30000 = 0
=> x² + 600x - 500x - 30000 = 0
=> x(x + 600) - 500(x + 600) = 0
=> (x - 500) (x + 600) = 0
=> x = 500 or x = - 600
∴ x ≠ - 600 (Because speed can't be negative)
Hence,
Its usual speed = 500 km/hr