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Question
A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Solution
Let the original speed of the train be x km/hr.
According to the question:
`90/x-90/(x+15)=1/2`
⇒ `(90(x+15)-90x)/(x(x+15))=1/2`
⇒`( 90x+1350-90x)/(x^2+15x)=1/2`
⇒`1350/(x^2+15x)=1/2`
⇒`2700=x^2+15x`
⇒`x^2+(60-45)x-2700=0`
⇒`x^2+60x-45x-2700=0`
⇒`x(x+60)-45x(x+60)=0`
⇒`(x+60)(x-45)=0`
⇒`x=-60 or x=45`
x cannot be negative; therefore, the original speed of train is 45 km/hr.
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