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. 

Answer 1

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.