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Question
A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
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Solution
Let the usual speed of train be x km/hr then
Increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed to cover 150km = `150/x`hr
Time taken by the train under increased speed to cover 150km = `150/(x + 5)`hr
Therefore,
`150/x-150/(x+5)=1`
`(150(x+5)-150x)/(x(x+5))=1`
`(150x+750-150)/(x^2+5x)=1`
`750/(x^2+5x)=1`
750 = x2 + 5x
x2 + 5x - 750 = 0
x2 - 25x + 30x - 750 = 0
x(x - 25) + 30(x - 25) = 0
(x - 25)(x + 30) = 0
So, either
x - 25 = 0
x = 25
Or
x + 30 = 0
x = -30
But, the speed of the train can never be negative.
Hence, the usual speed of train is x = 25km/hr
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