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# 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.

Last updated date: 15th Sep 2024
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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 $(x-12)\text{ km/hr}$.
${{T}_{ordinary}}-{{T}_{express}}=1$
\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 $=60km/hr$.