Question

The distance between Akola and Bhusawal is 168 km. An express train takes 1 hour less than a passenger train to cover the distance. Find the average speed of each train if the average speed of the express train is more by 14 km/hr than the speed of the passenger train.

Answer 1

Hint: We solve this question by first assuming the time taken by the passenger train as t and the speed of passenger train as s and find the time taken by the express train and speed of the express train in terms of t and s respectively. Then we use the formula, $\text{Speed}\times \text{Time}=\text{Distance}$ and find the relation between s and t in both trains. Then we solve those equations to find the value of the speeds of both the trains.

Complete step-by-step solution:
We are given that the distance between Akola and Bhusawal is 168km. So, the distance the trains need to travel is 168km.
$\text{Distance}=168km..........\left( 1 \right)$
We are given that the express train takes 1 hour less than the passenger train to cover the distance. So, let us assume that the time taken by the passenger train as $t\text{ }hrs$, then the time taken by the express train is $\left( t-1 \right)hrs$.
So, we have
\[\begin{align}
  & \text{Time taken by passenger train}=t\ hrs....................\left( 2 \right) \\
 & \text{Time taken by express train}=\left( t-1 \right)\ hrs................\left( 3 \right) \\
\end{align}\]
We are also given that average speed of express train is 14 km/hr more than the average speed of the passenger train. So, let us assume that the speed of the passenger train as $s\ km/hr$. So, we have
\[\begin{align}
  & \text{Average speed of passenger train}=s\ km/hr......................\left( 4 \right) \\
 & \text{Average speed of express train}=\left( s+14 \right)\ km/hr................\left( 5 \right) \\
\end{align}\]
Now let us consider the formula for distance.
$\text{Speed}\times \text{Time}=\text{Distance}$
Applying this formula for the passenger train and using the values obtained in equations (1), (2) and (4) we get,
$st=168..........\left( 6 \right)$
Applying the same formula for express train and using the values obtained in equations (1), (3) and (5) we get,
$\begin{align}
  & \Rightarrow \left( s+14 \right)\left( t-1 \right)=168 \\
 & \Rightarrow st+14t-s-14=168 \\
\end{align}$
Substituting the value of $st$ from equation (6) in the above equation, we get
$\begin{align}
  & \Rightarrow 168+14t-s-14=168 \\
 & \Rightarrow 14t-s-14=0 \\
 & \Rightarrow s=14t-14.................\left( 7 \right) \\
\end{align}$
Substituting this value in equation (6) we get,
$\begin{align}
  & \Rightarrow \left( 14t-14 \right)t=168 \\
 & \Rightarrow 14\left( t-1 \right)t=168 \\
 & \Rightarrow t\left( t-1 \right)=12 \\
 & \Rightarrow {{t}^{2}}-t-12=0 \\
 & \Rightarrow \left( t+3 \right)\left( t-4 \right)=0 \\
 & \Rightarrow t=4,-3 \\
\end{align}$
As time is always positive, $t=-3$ is not possible. So, we get $t=4hrs$
Now let us substitute this value in equation (7). Then we get,
$\begin{align}
  & \Rightarrow s=14\left( 4 \right)-14 \\
 & \Rightarrow s=56-14 \\
 & \Rightarrow s=42 \\
\end{align}$
Substituting these values in equations (4) and (5) we get,
\[\begin{align}
  & \text{Average speed of passenger train}=s\ =42km/hr \\
 & \text{Average speed of express train}=\left( s+14 \right)\ =\left( 42+14 \right)=56km/hr \\
\end{align}\]
Hence answer is 42 km/hr and 56 km/hr.

Note: The commonly made mistake in this question is one might take the formula for distance wrongly as $\text{Speed}=\text{Distance}\times \text{Time}$. The other mistake possible is one might interpret the given information wrongly as,
\[\begin{align}
  & \text{Average speed of passenger train}=\left( s+14 \right)\ km/hr \\
 & \text{Average speed of express train}=s\ km/hr \\
\end{align}\]
But we are given that the speed of express train is 14km/hr more than the passenger train not the reverse.