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Question
The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?
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Solution
Let the ongoing speed of person be x km/hr. Then,
Returning speed of the person is = (x + 10)km/hr.
Time taken by the person in going direction to cover 150km = `150/x`hr
Time taken by the person in returning direction to cover 150km = `150/(x+10)`hr
Therefore,
`150/x-150/(x+10)=5/2`
`(150(x+10)-150x)/(x(x+10))=5/2`
`(150x+1500-150x)/(x^2+10x)=5/2`
`1500/(x^2+10)=5/2`
1500(2)=5(x2+10x)
3000 = 5x2 + 50x
5x2 + 50x - 3000 = 0
5(x2 + 10x - 600) = 0
x2 + 10x - 600 = 0
x2 - 20x + 30x - 600 = 0
x(x - 20) + 30(x - 20) = 0
(x - 20)(x + 30) = 0
So, either
x - 20 = 0
x = 20
Or
x + 30 = 0
x = -30
But, the speed of the train can never be negative.
Thus, when x = 20 then
= x + 10
= 20 + 10
= 30
Hence, ongoing speed of person is x = 20 km/hr
and returning speed of the person is x = 30 km/hr respectively.
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