QUESTION

# The distance between two places is 900 km. An ordinary express train takes 5 hours more than a superfast express train to cover this distance. If the speed of the superfast express train is 15 km/hr more than that of the ordinary express train, find the speed of the trains.

Hint: In order to find the speed of two trains first we will assume ordinary train speed as some variable that we will find time taken by both the trains. By using a given statement we will make a quadratic equation and after solving it we will get the answer.

Let the speed of ordinary train be x km/hr
And time taken by ordinary train $= \dfrac{{900}}{x}hr\;{\text{ }}\left[ {\therefore {\text{time}} = \dfrac{{{\text{distance}}}}{{{\text{speed}}}}} \right]$
So, according to the given statement, the speed of superfast train $= x + 15km/hr$
And the time taken by superfast train $= \dfrac{{900}}{{x + 15}}hr$
Therefore, an ordinary train takes 5 hours more than a superfast train to cover 900 km distance.
According to given statement, we get
$\Rightarrow \dfrac{{900}}{x} - \dfrac{{900}}{{x + 15}} = 5 \\ \Rightarrow 900\left( {\dfrac{1}{x} - \dfrac{1}{{x + 15}}} \right) = 5 \\ \Rightarrow 900\left( {\dfrac{{x + 15 - x}}{{x\left( {x + 15} \right)}}} \right) = 5 \\ \Rightarrow 900\left( {\dfrac{{15}}{{{x^2} + 15x}}} \right) = 5 \\ \Rightarrow 900 \times \dfrac{{15}}{5} = {x^2} + 15x{\text{ }}\left\{ {{\text{by cross multiplication}}} \right\} \\ \Rightarrow 2700 = {x^2} + 15 \\ \Rightarrow {x^2} + 15x - 2700 = 0 \\$
Now, we will solve this quadratic equation to get the speed of an ordinary train.
$\Rightarrow {x^2} + 15x - 2700 = 0 \\ \Rightarrow x\left( {x + 60} \right) - 45\left( {x + 60} \right) = 0 \\ \Rightarrow \left( {x + 60} \right)\left( {x - 45} \right) = 0 \\ \Rightarrow x = - 60{\text{ or }}x = 45 \\$
Since, the speed cannot be negative therefore neglecting the negative term.
$\therefore x = 45$
The speed of ordinary train $x = 45km/hr$
And the speed of superfast train
$= x + 15 \\ = (45 + 15)km/hr \\ = 60km/hr \\$

Note: In order to solve problems related to solving basic algebraic and quadratic equations. The first problem is find the conditions given in the problem. Second step is to frame the equations with the given conditions, remembering how to solve quadratic equations whether by factoring or using formulas. These problems are statement based so read the statement carefully.