A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Given:

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same 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 360 km at original speed$=\frac{360}{x}$ hours

Time taken by the train to travel 360 km when the speed is 5 km/hr more than the original speed$=\frac{360}{x+5}$ hours

According to the question,

$\frac{360}{x}-\frac{360}{x+5}=1$

$\frac{360(x+5)-360(x)}{(x)(x+5)}=1$

$\frac{360(x+5-x)}{x^2+5x}=1$

$(360)(5)=1(x^2+5x)$   (On cross multiplication)

$1800=x^2+5x$

$x^2+5x-1800=0$

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

$x^2+45x-40x-1800=0$

$x(x+45)-40(x+45)=0$

$(x+45)(x-40)=0$

$x+45=0$ or $x-40=0$

$x=-45$ or $x=40$

Speed cannot be negative. Therefore, the value of $x$ is $40$ km/hr.

The original speed of the train is $40$ km/hr.