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:

\(\frac{90}x\) -\(\frac{90}{(x+15)}\) = \(\frac{1}2\)

⇒ \(\frac{90(x+15)-90x}{x(x+15)}\) = \(\frac{1}2\)

⇒ \(\frac{90x+1350-90x}{x^2+15x}\) = \(\frac{1}2\)

x cannot be negative; therefore, the original speed of train is 45 km/hr.