A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Given:

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for the journey.

To do:

We have to find the original speed of the train.  

Solution:

Let the original speed of the train be $x$ km/hr.

This implies,

Time taken by the train to travel 90 km at original speed$=\frac{90}{x}$ hours

Time taken by the train to travel 90 km when the speed is 15 km/hr more than the original speed$=\frac{90}{x+15}$ hours

$30$ minutes in hours$=\frac{30}{60}$ hours.    (since 1 hour = 60 minutes)

According to the question,

$\frac{90}{x}-\frac{90}{x+15}=\frac{30}{60}$

$\frac{90(x+15)-90(x)}{(x)(x+15)}=\frac{30\times1}{30\times2}$

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

$2(90)(15)=1(x^2+15x)$   (On cross multiplication)

$30(90)=x^2+15x$

$x^2+15x-2700=0$

Solving for $x$ by factorization method, we get,

$x^2+60x-45x-2700=0$

$x(x+60)-45(x+60)=0$

$(x+60)(x-45)=0$

$x+60=0$ or $x-45=0$

$x=-60$ or $x=45$

Speed cannot be negative. Therefore, the value of $x$ is $45$ km/hr.

The original speed of the train is $45$ km/hr.