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Questions
A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.
A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/h more, it would have taken 30 minutes less for the same journey. Find the original speed of the train.
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Solution
Let the original speed of train be x km/hr.
Then, Increased speed of the train = (x + 15) km/hr
Time taken by the train under usual speed to cover 90 km = `90/x`hr
Time taken by the train under increased speed to cover 90 km = `90/(x + 15)`hr
Therefore,
`90/x - 90/(x + 15) = 30/60`
`(90(x + 15) - 90x)/(x(x + 15)) = 1/2`
`(90x + 1350 - 90x)/(x^2 + 15x) = 1/2`
`1350/(x^2 + 15x) = 1/2`
1350(2) = x2 + 15x
2700 = x2 + 15x
x2 + 15x – 2700 = 0
x2 – 45x + 60x – 2700 = 0
x(x – 45) + 60(x – 45) = 0
(x – 45)(x + 60) = 0
So, either
x – 45 = 0
x = 45
Or
x + 60 = 0
x = –60
But, the speed of the train can never be negative.
Hence, the original speed of train is x = 45 km/hr
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