Question

A train travelling at a uniform speed $360$ km would have taken $48$ minutes less to travel the same distance if its speed were $5$ km/h more. Find the original speed of the train.

Answer 1

Let the original speed of train $=xkm/hr$

So, the new increased speed of train $=(x+5)km/hr$

Time taken by train in covering $360km$ with original speed $=\frac{Distance}{Speed}=\frac{360}{x}hr$

Time taken by train in covering $360km$ with new speed $=\frac{360}{x+5}hr$

According to the question,

$\frac{360}{x}-\frac{360}{x+5}=\frac{48}{60}hr$

$\Rightarrow 360[\frac{1}{x}-\frac{1}{(x+5)}]=\frac{4}{5}$

$\Rightarrow 360\left[\frac{x+5-x}{x(x+5)}\right]=\frac{4}{5}$

$\Rightarrow \frac{90\left[5\right]}{x^2+5x}=\frac{1}{5}$

$\Rightarrow x^2+5x=90\times 25$

$\Rightarrow x^2+5x-90\times 25=0$

$\Rightarrow x^2+50x-45x-90\times 25=0$

$\Rightarrow x(x+50)-45[x+10\times 5]=0$

$\Rightarrow (x+50)(x-45)=0$

$\Rightarrow x+50=0$ or $x-45=0$

$\Rightarrow x=-50$ or $x=45$

$x=-50$ is rejected as it is negative.

Hence, the original speed of train is $45km/hr$