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# A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hr less for the same journey. Find the speed of the train.

Last updated date: 20th Jun 2024
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Hint: We assume the velocity of the train as variable. We use the formula of $time=\dfrac{dis\tan ce}{speed}$ to find the two different types of time required for different speeds. We put the relation between times into a quadratic equation. We solve the equation to get the speed of the train.

Let the uniform speed of the train be x km/h. The train travels 360 km at that uniform speed.
We know the relation between time required, speed and the distance are $time=\dfrac{dis\tan ce}{speed}$.
We now try to find the time required by the train when it’s traveling 360 km with a speed of x km/h. let it be t hour.
So, the time required is $t=\dfrac{360}{x}$.
Now the given condition is if the speed had been 5 km/h more, it would have taken 1 hr less for the same journey.
This means if the speed had been $\left( x+5 \right)$ km/h, it would have taken 1 hr less which is $\left( t-1 \right)$ hour for the same journey.
We try to again express the time $\left( t-1 \right)$ according to the formula and get $t-1=\dfrac{360}{x+5}$.
We have two equations of two unknowns.
\begin{align} & t-1=\dfrac{360}{x+5} \\ & \Rightarrow \dfrac{360}{x}-1=\dfrac{360}{x+5} \\ & \Rightarrow \dfrac{360}{x}-\dfrac{360}{x+5}=1 \\ \end{align}
We get the quadratic form and get $360\left( x+5 \right)-360x=x\left( x+5 \right)$.
The equation becomes ${{x}^{2}}+5x-1800=0$.
\begin{align} & {{x}^{2}}+5x-1800=0 \\ & \Rightarrow {{x}^{2}}+45x-40x-1800=0 \\ & \Rightarrow \left( x+45 \right)x-40\left( x+45 \right)=0 \\ & \Rightarrow \left( x+45 \right)\left( x-40 \right)=0 \\ \end{align}
The solution of the factorisation is $x=40$. It can’t be -45 as it’s velocity.

Therefore, the speed of the train is 40 km/h.

Note: We also could have used the fixed distance value to find the relation which means $\left( t-1 \right)\left( x+5 \right)=360=tx$. The relation comes from the corollary formula of $time=\dfrac{dis\tan ce}{speed}$ which is $dis\tan ce=time\times speed$. The equation would have been given $\left( t-1 \right)\left( x+5 \right)=tx$.
Although the rest of the solution process would have been the same.