Question

# The speed of the express train is $x\text{ }km/hr$ and the speed of ordinary train is $12\text{ km/hr}$ less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of $240\text{ }km$ .Find the speed of the express train.

Hint: Here we apply the concept of speed and distance.
The speed of the express train is $x\text{ }km/hr$.
The speed of an ordinary train is $12\text{ km/hr}$ less than that of the express train.

Required Formula:
\begin{align} & dis\tan ce=speed\times time \\ & time=\dfrac{dis\tan ce}{speed} \\ \end{align}
We need to find the speed of the express train with the given information.

${{T}_{ordinary}}\text{ and }{{T}_{\exp ress}}\text{ }$be the times taken by express train and ordinary train respectively.
The speed of the express train is $x\text{ }km/hr$.
The speed of an ordinary train is $12\text{ km/hr}$ less than that of the express train.
Therefore, the speed of an ordinary train is $(x-12)\text{ km/hr}$.

According to the question,
${{T}_{ordinary}}-{{T}_{express}}=1$
\begin{align} & \Rightarrow \dfrac{dis\tan ce\text{ }travelled}{speed\text{ }of\text{ }ordinary\text{ }train}-\dfrac{dis\tan ce\text{ }travelled}{speed\text{ }of\text{ }\exp ress\text{ }train}=1 \\ & \Rightarrow \dfrac{240}{x-12}-\dfrac{240}{x}=1 \\ & \Rightarrow 240(\dfrac{x-x+12}{x(x-12)})=1 \\ & \Rightarrow 240\times 12=x(x-12) \\ & \Rightarrow 2880={{x}^{2}}-12x \\ & \Rightarrow {{x}^{2}}-12x-2880=0 \\ & \Rightarrow {{x}^{2}}-60x+48x-2880=0 \\ & \Rightarrow x(x-60)+48(x-60)=0 \\ & \Rightarrow x=60 \\ \end{align}
As the speed of a train cannot be negative, we omit $x=48$ .

Therefore, speed of the express train $=60km/hr$.

Note: In such types of questions which involve concepts of speed and distance, the knowledge about the related formula is needed. Accordingly, we apply the known values and solve them to get the unknown value by satisfying the condition involved in the question.