Question

A train travels a distance of \[480km\] at a uniform speed. If the speed had been $8km/h$ less, then it would have taken $3$ hours more to cover the same distance. We need to find the speed of the train.

Answer 1

Hint: First, we shall analyze the given information. It is given that the distance is\[480km\].
We are asked to find the speed of the train. Also, we are given a condition that if the speed is decreased by $\left( {x - 8} \right)km/h$, then it would have taken $3$ hours more to cover the same distance.
Using this condition, we are able to calculate the required speed.
Formula used:
 The formula to obtain the speed is as follows.
\[Speed = \dfrac{{Distance}}{{Time}}\]
The formula to calculate the train is obtained by applying simple changes to the above formula.
$Time = \dfrac{{Distance}}{{Speed}}$

Complete step by step answer:
Let us consider the required speed of the train $xkm/h$.
It is given that the distance is\[480km\].
We shall calculate the time taken by the train to cover the distance \[480km\]at a speed of $xkm/h$.
We know that $Time = \dfrac{{Distance}}{{Speed}}$
Thus the time taken by the train to cover the distance of \[480km\]at a speed of $xkm/h$$ = \dfrac{{480}}{x}$ hrs
It is also given that the required speed is decreased by $\left( {x - 8} \right)km/h$
We shall calculate the time taken by the train to cover the distance \[480km\] at a speed of $\left( {x - 8} \right)km/h$.
Thus the time taken by the train to cover \[480km\] at a speed of$\left( {x - 8} \right)km/h$$ = \dfrac{{480}}{{\left( {x - 8} \right)}}$hrs
We are given a condition that if the speed is decreased by $\left( {x - 8} \right)km/h$, then it would have taken$3$hours more to cover the same distance.
Now, according to the condition, we have,
$\dfrac{{480}}{{\left( {x - 8} \right)}} = \dfrac{{480}}{x} + 3$
$ \Rightarrow \dfrac{{480}}{{\left( {x - 8} \right)}} - \dfrac{{480}}{x} = 3$
 $ \Rightarrow 480\left( {\dfrac{1}{{\left( {x - 8} \right)}} - \dfrac{1}{x}} \right) = 3$
We shall take LCM inside the brackets.
$ \Rightarrow 480\left( {\dfrac{{x - x + 8}}{{x - 8 \times x}}} \right) = 3$
$ \Rightarrow 480\left( {\dfrac{8}{{{x^2} - 8x}}} \right) = 3$
\[ \Rightarrow \dfrac{8}{{{x^2} - 8x}} = 3 \times \dfrac{1}{{480}}\]
\[ \Rightarrow \dfrac{8}{{{x^2} - 8x}} = \dfrac{1}{{160}}\]
Now, we shall cross multiply both sides.
\[ \Rightarrow 8 \times 160 = {x^2} - 8x\]
\[ \Rightarrow 1280 = {x^2} - 8x\]
\[ \Rightarrow {x^2} - 8x - 1280 = 0\]
The above equation is quadratic. Here, $ - 8x$ can be written as $ - 40x + 32x$ .
\[ \Rightarrow {x^2} - 40x + 32x - 1280 = 0\]
We need to pick the common terms as follows.
\[ \Rightarrow x\left( {x - 40} \right) + 32\left( {x - 40} \right) = 0\]
\[ \Rightarrow \left( {x - 40} \right)\left( {x + 32} \right) = 0\]
Hence, $x - 40 = 0$ or$x + 32 = 0$
We get, $x = 40$ or$x = - 32$
Since the speed of the train cannot be negative, $x = - 32$is not possible.
Therefore, the required speed of the train is $x = 40$ (i.e.) $40km/h$.

Note: The value of the speed is always positive and it cannot be negative. So, we have rejected $x = - 32$.
We have been given a condition that if the speed is decreased by $\left( {x - 8} \right)km/h$, then it would have taken $3$ hours more to cover the same distance. This is the required condition to solve this problem.