Question

An express train takes 2 hours less time than a passenger train to travel 400 km between two stations (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 10 km/h more than that of the passenger train, find the average speed of the two trains.

Answer 1

Hint: Use the relationship between time, distance and speed i.e. \[\text{Time}=\dfrac{\text{distance}}{\text{speed}}\]. Assume that the average speed of the express train is x km/h and the passenger train is y km/h. Use the relationship given in question to form equations in x and y and solve for them.

Complete step-by-step answer:
Let the average speed of the express train be x km/h and the average speed of passenger train is y km/h.
We know the relationship between time, speed and distance i.e. \[\text{Time}=\dfrac{\text{distance}}{\text{speed}}\].
Considering constant speed throughout the given time, the acceleration is zero. Here, we know that the distance covered by both trains is 400 km between two stations. So, let the time taken by passenger train be \[{{t}_{1}}\] hour and time taken by express train be \[{{t}_{2}}\] hour.
Therefore, using the formula mentioned above
${{t}_{1}}=\dfrac{400}{y}$
${{t}_{2}}=\dfrac{400}{x}$
And we know that the express train takes 2 hours less time than a passenger train to travel 400 km between two stations without taking into consideration the time they stop at intermediate stations.
So,
${{t}_{1}}-{{t}_{2}}=2$
$\begin{align}
  & \dfrac{400}{y}-\dfrac{400}{x}=2 \\
 & 400\left( \dfrac{1}{y}-\dfrac{1}{x} \right)=2 \\
\end{align}$
Now,
$\dfrac{1}{y}-\dfrac{1}{x}=\dfrac{1}{200}....\left( i \right)$
Now, it’s given in the question that an express train has a speed greater than passenger train by 10 km/h.
So,
$x=10+y.....\left( ii \right)$
Using (i) and (ii)
Substituting the value of x from (ii) in (i)
$\begin{align}
  & \dfrac{1}{y}-\dfrac{1}{(10+y)}=\dfrac{1}{200} \\
 & \dfrac{10+y-y}{y(10+y)}=\dfrac{1}{200} \\
\end{align}$
$2000={{y}^{2}}+10y$
${{y}^{2}}+10y-2000=0$
Solving the equation using the quadratic formula and we know that when a quadratic equation is given as $a{{x}^{2}}+bx+c=0$ then its roots are given by the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
So using the quadratic formula to determine the value of y.
$\begin{align}
  & y=\dfrac{-10\pm \sqrt{100+8000}}{2} \\
 & y=\dfrac{-10\pm \sqrt{8100}}{2} \\
 & y=\dfrac{-10\pm 90}{2} \\
\end{align}$
Now, we have two values of y.
$y=\dfrac{-10-90}{2}$; $y=\dfrac{-100}{2}=-50$
Hence it is rejected as speed cannot be negative.
And, $y=\dfrac{-10+90}{2}=\dfrac{80}{2}=40$
Putting the value of y in equation (ii) we will get, x = 50.
So, the average speed of the passenger train is 40 km/h and the speed of the express train is 50 km/h.

Note: We can use the middle term splitting method to determine the roots of the equation as an alternative to the quadratic equation formula. We can use middle term splitting in the equation ${{y}^{2}}+10y-2000=0$ as
${{y}^{2}}+10y-2000=0$
$\begin{align}
  & {{y}^{2}}+50y-40y-2000=0 \\
 & y\left( y+50 \right)-40\left( y+50 \right)=0 \\
 & \left( y+50 \right)\left( y-40 \right)=0 \\
\end{align}$
Now we get the value of by comparing both factors with zero.
We get y = – 50 and y = 40. we will accept y = 40.