QUESTION

# A train travels at a certain average speed for a distance of 63km and then travels a distance of 72km at an average speed of 6km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed?

Hint: In this question, the train travels a certain distance with one speed and some other distance with another speed. However, as the distances travelled would each take some time to complete and the events of travelling follow one after another, the total time taken would be the sum of the time taken for the first 63km and the time taken to complete the next 72 km.

As the train travels the first 63km with its original average speed and the next 72km with 6km/hr more than its average speed, we can take the original average speed to be x km/hr . Thus,
Speed at which the train travels the first 63km = x km/hr
Speed at which the train travels the next 72km = x+6 km/hr
As the time taken to cover a distance is given by the formula
$\text{Time taken=}\dfrac{\text{Distance Travelled}}{\text{Speed}}$
And $\text{Total time taken= time taken to travel first }$
We can obtain the time taken to travel the distances as
$\begin{array}{*{35}{l}} \text{Time taken by the train to travel the first 63km }=\text{ }\dfrac{63km}{\text{x km/hr}} \\ ~ \\ \end{array}.....................\text{(1}\text{.1)}$$\begin{array}{*{35}{l}} \text{Time taken by the train to travel the next 72km }=\text{ }\dfrac{72km}{\text{x+6 km/hr}} \\ ~ \\ \end{array}.....................(1.2)$
Therefore, the total time taken would be the sum of the time taken for the first 63km and the next 72km. However, the total time taken is given to be 3hours.
Thus, from equations (1.1) and (1.2), we obtain
\begin{align} & \dfrac{63}{x}+\dfrac{72}{x+6}=3\Rightarrow 63(x+6)+72x=3x(x+6) \\ & \Rightarrow \left( 63+72 \right)x+378=3{{x}^{2}}+18x \\ & \Rightarrow 3{{x}^{2}}-117x-378=0 \\ & \Rightarrow {{x}^{2}}-39x-126=0..............(1.3) \\ \end{align}
The solution of a quadratic equation $a{{x}^{2}}+bx+c=0$ is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Using this formula for equation (1.3), we get
\begin{align} & x=\dfrac{39\pm \sqrt{{{39}^{2}}+4\times 1\times 126}}{2}=\dfrac{39\pm 45}{2} \\ & \Rightarrow x=\dfrac{39-45}{2}=-3\text{ or }x=\dfrac{39+45}{2}=42 \\ \end{align}
As the velocity cannot be negative, the solution x=-3 should be rejected. Thus, the solution should be x = 42km/hr.
Therefore, the original average speed of the train is 42km/hr.
Note: We should note that, in this question, the equation comes out to be a quadratic equation. As there are two solutions of a quadratic equation, we can reject one solution only if it does not correspond to the correct physical situation.