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 $5km/hr$ more. Find the original speed of the train.

Answer 1

Let the speed of train be $xkm/hr$

Distance to be travelled $=360km$
Since,$\text{Speed}=\frac{\text{distance}}{\text{time}}$
$\Rightarrow$ $\text{time}=\frac{\text{Distance}}{\text{speed}}$
Time for cover $360km$ with speed $xkm/hr$ is $\frac{360}{x}hr$.....(i)
In case if the speed is $5km/hr$ is more then the time required to cover the same distance of $360km$ is $\frac{360}{x+5}hr$.....(ii)
Then times difference between (i) and (ii) is $48$min$=\frac{48}{60}$ hr.
$\Rightarrow \frac{360}{x}-\frac{360}{x+5}=\frac{48}{60}$
Dividing both sides by $360$, we get
$\Rightarrow \frac{1}{x}-\frac{1}{x+5}=\frac{48}{60\times 360}$
$\Rightarrow \frac{x+5-x}{(x+5)\left(x\right)}=\frac{1}{450}$
$\Rightarrow \frac{5}{x^2+5x}=\frac{1}{450}$
$\Rightarrow x^2+5x=2250$
$\Rightarrow x^2+5x-2250=0$
$\Rightarrow x^2-45x+50x-2250=0$
$\Rightarrow x(x-45)+50(x-45)=0$
$\Rightarrow (x-45)(x+50)=0$
$x=45,-50$
Since, speed can't be a negative quantity.
Therefore, required speed of the train is $=45km/hr$.