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Question
An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
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Solution
Let the usual speed of aero plane be x km/hr. Then,
Increased speed of the aero plane = (x + 250) km/hr
Time taken by the aero plane under usual speed to cover 1250 km = `1250/x`hr
Time taken by the aero plane under increased speed to cover 1250 km = `1250/(x +250)`hr
Therefore,
`1250/x-1250/(x+250)=50/60`
`(1250(x+250)-1250x)/(x(x+250))=5/6`
`(1250x+312500-1250x)/(x^2+250x)=5/6`
`312500/(x^2+250x)=5/6`
312500(6) = 5(x2 + 250x)
1875000 = 5x2 + 1250x
5x2 + 1250x - 1875000 = 0
5(x2 + 250x - 375000) = 0
x2 + 250x - 375000 = 0
x2 - 500x + 750x - 375000 = 0
x(x - 500) + 750(x - 500) = 0
(x - 500)(x + 750) = 0
So, either
x - 500 = 0
x = 500
Or
x + 750 = 0
x = -750
But, the speed of the aero plane can never be negative.
Hence, the usual speed of train is x = 500 km/hr.
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