Question

A train travels 360 km at a uniform speed. If the speed had been 5 km/ hr more, then it would have taken 1 hour less for the same journey. From a quadratic equation to find the speed of the train.

Answer 1

Hint: First of all we will suppose the speed of the train to be a variable and we will use the basic motion formula which is as follows:
\[time=\dfrac{distance}{speed}\]

Complete step-by-step answer:
By using the formula and the given condition in the question we will get the required quadratic equation.
Let us suppose the speed of the train to be V km/ hr.
We know that \[time=\dfrac{distance}{speed}\]
So time taken by the train to cover 360 km, distance is as follows:
\[time=\dfrac{360}{V}\]
We have been told that if the speed had been 5 km/ hr more, then it would take 1 hour less for the same journey.
So, the new speed \[=(V+5)km/hr\]
Distance = 360 km
\[time=\dfrac{360}{V+5}\]
Now, according to the question,
\[\dfrac{360}{V+5}=\dfrac{360}{V}-1\]
On taking \[\dfrac{360}{V+5}\] to the right hand side and the term ‘1’ to the left hand side, we get as follows:
\[\begin{align}
  & 1=\dfrac{360}{V}-\dfrac{360}{V+5} \\
 & 1=360\left( \dfrac{1}{V}-\dfrac{1}{V+5} \right) \\
 & 1=360\left( \dfrac{V+5-V}{V(V+5)} \right) \\
 & 1=360\left( \dfrac{5}{V(V+S)} \right) \\
\end{align}\]
On multiplying the equation with \[V(V+5)\], we get as follows:
\[\begin{align}
  & V(V+5)=360\left( \dfrac{5}{V(V+5)} \right)\times V(V+5) \\
 & {{V}^{2}}+5V=1800 \\
 & {{V}^{2}}+5V-1800=0 \\
\end{align}\]
Therefore, the quadratic equation to find the speed of the train is \[{{V}^{2}}+5V-1800=0\].

Note: The first step is to form the right equation, if we formulate it as \[\dfrac{360}{V+5}-1=\dfrac{360}{V}\] then we might end up completely different quadratic equation. Be careful while simplifying the equation and also take care of the sign after you take the LCM. Do not suppose the variable speed in m/ s unit since also the other measurement is in km or hour. So, it is better to take speed in km/ hr.