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Question
A passenger train takes 2 hours less for a journey of 300 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 the passenger train be x km/hr. Then,
Increased speed of the passenger train = (x + 5) km/hr
Time taken by the train under usual speed to cover 300 km = `300/x`hr
Time taken by the train under increased speed to cover 300 k m = `300/(x + 5)`hr
Therefore,
`300/x-300/(x+5)=2`
`(300(x+5)-300x)/(x(x+5))=2`
`(300x+1500-300x)/(x^2+5x)=2`
`1500/(x^2+5x)=2`
1500 = 2(x2 + 5x)
1500 = 2x2 + 10x
2x2 + 10 - 1500 = 0
2(x2 + 5x - 750) = 0
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 passenger train can never be negative.
Hence, the usual speed of passenger train is x = 25 km/hr
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