Question

A fast train takes $3$ hours less than a slow train for a journey of $600\mathrm{km}$. If the speed of the slow train is $10\mathrm{km}/\mathrm{hr}$ less than that of the first train, find the speeds of the two trains.

Answer 1

Step 1: Determining the quadratic equation in $x$

Let the speed of the slow train be $x\mathrm{km}/\mathrm{hr}$.

Then, the speed of the fast train will be $(x+10)\mathrm{km}/\mathrm{hr}$.

The time taken by the slow train to cover a distance of $600\mathrm{km}$ is $\frac{600}{x}$ hours.

And the time taken by the fast train to cover a distance of $600\mathrm{km}$ is $\frac{600}{x+10}$ hours.

It is given that the fast train takes $3$ hours less than the slow train.

$$\begin{gathered} \therefore \frac{600}{x}-\frac{600}{x+10}=3 \\ \Rightarrow 600\left[\right(x+10)-x]=3x(x+10) \\ \Rightarrow 6000=3x^2+30x \\ \Rightarrow x^2+10x-2000=0 \end{gathered}$$

Step 2: Determining the speed of trains

By factorization method,

$x^2+10x-2000=0$

For splitting the middle term, i.e. $10$, we need to find two numbers such that their sum is $10$ and the product is $2000$.

Such numbers can be $50$ and $-40$.

$$\begin{gathered} \Rightarrow x^2+50x-40x-2000=0 \\ \Rightarrow x(x+50)-40(x+50)=0 \\ \Rightarrow (x+50)(x-40)=0 \\ \therefore x=-50,40 \end{gathered}$$

Since the speed of the train cannot be negative,

$\therefore x=40$ and $(x+10)=40+10=50$

Therefore, the speeds of the slow train and fast train are $40\mathrm{km}/\mathrm{hr}$ and $50\mathrm{km}/\mathrm{hr}$, respectively.