Question

A train travels a distance of 300km at a constant speed. If the speed of the train is increased by 5km an hour, the journey would have taken 2 hours less. Find the original speed of the train.

Answer 1

Hint: In this problem, first we need to find the quadratic equation using the time and distance formula. Next, use the middle term factorization to obtain the original speed.
Let the constant speed of the train be ${v} \dfrac{km}{h}$.

Complete step-by-step solution -
Now, the time $t_1$ taken by train to cover 300 km is shown below.
$({t_1} = \dfrac{{300}}{v})$
The speed of the train is increased by 5km/h. Now, the time $({t_2})$ taken by the train to cover 300 km is shown below.
$({t_2} = \dfrac{{300}}{{v + 5}}$
Now, as per the question,
$ {t_1} - {t_2} = 2 \\$
$\Rightarrow \dfrac{{300}}{v} - \dfrac{{300}}{{v + 5}} = 2 \\$
$\Rightarrow 300\left( {\dfrac{1}{v} - \dfrac{1}{{v + 5}}} \right) = 2 \\$
$\Rightarrow 300\left( {\dfrac{{v + 5 - v}}{{{v^2} + 5v}}} \right) = 2 \\$
$\Rightarrow 150\left( 5 \right) = {v^2} + 5v \\$
$\Rightarrow {v^2} + 5v - 750 = 0 \\ $
Further, simplify the above equation.
${v^2} + \left( {30 - 25} \right)v - 750 = 0 \\$
 $\Rightarrow {v^2} + 30v - 25v - 750 = 0 \\$
 $\Rightarrow v\left( {v + 30} \right) - 25\left( {v + 30} \right) = 0 \\$
$ \Rightarrow \left( {v - 25} \right)\left( {v + 30} \right) = 0 \\$
$ \Rightarrow v = 25\,\,{\text{or}}\,\, - 30 \\ $
Since the speed of the train cannot be negative, hence the original speed of the train is 25km/h.

Note: In this problem, find the quadratic equation in terms of the original speed of the train.
Here we can also use the quadratic formula for solving the quadratic equation as an alternate approach. For solving these types of problems there is more than one approach.