Question

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

Answer 1

Step 1: Construct the equation based on given condition.
let the original speed of the train $=xkm/h$.
Then the increased speed of the train
$=(x+5)km/h.$
and distance $=360km.$
According to the question,
$\frac{360}{x}-\frac{360}{x+5}=\frac{4}{5}$
$(\therefore \text{speed}=\frac{\text{distance}}{\text{time}})$

$[\because 48\text{minutes=}\frac{48}{60}\text{hours}=\frac{4}{5}\text{hours}]$
$\Rightarrow \frac{360(x+5)-360x}{x(x+5)}=\frac{4}{5}$
$\Rightarrow \frac{360x+1800-360x}{x(x+5)}=\frac{4}{5}$
$\Rightarrow \frac{1800}{x^2+5x}=\frac{4}{5}$

$\Rightarrow \frac{1800\times 5}{4}=x^2+5x$
$\Rightarrow x^2+5x=\frac{1800\times 5}{4}=2250$
$\Rightarrow x^2+5x-2250=0.$

Step 2: Find the roots using factorization.
By splitting the middle term, we get :
$x^2+(50x-45x)-2250=0$
$\Rightarrow x^2+50x-45x-2250=0$
$\Rightarrow x(x+50)-45(x+50)=0$
$\Rightarrow x(x+50)(x+45)=0$
$\Rightarrow x+50=0$ or $x-45=0$
$x\ne -50$ because speed can not be negative
So, $x-45=0$
$\Rightarrow x=45$
Hence, the original speed of the train is $45km/h.$