Question
Answers

An express train takes 1 hours less than a passenger train to travel 132km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/hr more than that of the passenger train, find the average speed of the two trains.

Answer Verified Verified
Hint: First of all consider the average speed of express and passenger train as Se and Sp respectively. Also consider time taken by them as te and tp respectively. Then use,
$\text{average speed =}\dfrac{\text{Total distance travelled}}{\text{Total time taken}}$ . Use equations (Se - Sp) =11 and (tp- te) =1 to get the values of Se and Sp.

Complete step-by-step answer:
Here we are given that the average speed of an express train is 11km/hr more than that of passenger trains. Also this express train takes 1 hour less than passenger train to travel 132km between Mysore and Bangalore. We have to find the average speed of two trains.
First of all, let us consider the average speed of passenger trains to be Sp km/hr.
Also, the average speed of the express train is Se km/hr.
As we are given that the average speed of the express train is 11km/hr more than passenger train. Therefore we get,
Se = Sp +11 km/hr…………. (1)
Now we know that, $\text{average speed =}\dfrac{\text{Total distance travelled}}{\text{Total time taken}}$.
Let us consider that the passenger train takes tp hour to travel 132km.
Therefore we get,
Average speed of passenger train$\text{=}\dfrac{132}{tp}\dfrac{km}{hours}$
Or we get, $Sp=\dfrac{132}{tp}$
By cross multiplying above equation, we get
$tp=\dfrac{132}{Sp}$…………….. (2)
Also, let us consider that the express train takes te hours to travel 132km.
Therefore we get,
Average speed of express train$\text{=}\dfrac{132}{te}\dfrac{km}{hours}$
Or we get, $Se=\dfrac{132}{te}$
By cross multiplying above equation, we get
$te=\dfrac{132}{Se}$…………….. (3)
As we are given that express train takes 1 hour less than passenger train to travel 132km. Therefore we get,
tp = te +1
By putting the value of tp and te from equation (2) and (3) we get,
$\dfrac{132}{Sp}=\dfrac{132}{Se}+1$
Now from equation (1), we put Se = Sp +11,
We get,
$\begin{align}
  & \dfrac{132}{\left( Sp \right)}=\dfrac{132}{\left( Sp+11 \right)}+1 \\
 & or \\
 & \dfrac{132}{\left( Sp \right)}-\dfrac{132}{\left( Sp+11 \right)}=1 \\
\end{align}$
By taking 132 common and solving above equation, we get
\[\begin{align}
  & =132\left[ \dfrac{\left( Sp+11 \right)-Sp}{\left( Sp \right)\left( Sp+11 \right)} \right]=1 \\
 & =\dfrac{132\times 11}{\left( Sp \right)\left( Sp+11 \right)}=1 \\
\end{align}\]
By cross multiplying above equation, we get,
$\begin{align}
  & =\left( Sp \right)\left( Sp+11 \right)=132\times 11 \\
 & or\ {{\left( Sp \right)}^{2}}+11Sp-\left( 132\times 11 \right)=0 \\
 & \Rightarrow {{\left( Sp \right)}^{2}}+11Sp-1452=0 \\
\end{align}$
We can split 11Sp as, 11Sp = 44Sp – 33Sp. Therefore, we get,
$\Rightarrow {{\left( Sp \right)}^{2}}+44Sp-33Sp-1452=0$
As we know that 1452 = 33 x 44, we get above equation as,
$\Rightarrow {{\left( Sp \right)}^{2}}+44Sp-33Sp-\left( 33\times 44 \right)=0$
We can also write above equation as,
$\Rightarrow Sp\left( Sp+44 \right)-33\left( Sp+44 \right)=0$
By taking $\left( Sp+44 \right)$common, we get
$\Rightarrow \left( Sp+44 \right)\left( Sp-33 \right)=0$
Hence we get Sp = -44 and Sp = 33.
As Sp is average speed of passenger train, it can’t be negative, so Sp = 33km/hr.
By putting the value of Sp in equation (1), we get
Se = Sp + 11
= 33 +11
Se = 44km/hr
Therefore we get average speed of passenger and express trains as 33km/hr and 44km/hr respectively.

Note: Here students must take care while writing the equation. Many students make this mistake of writing Sp = Se +11 instead of Se = Sp +11 or te = tp +1 instead tp = te +1. So this mistake must be taken care of. Also students can cross check their answers by putting the values of Sp and Se in different equations and checking if they are satisfying the equations or not.
Bookmark added to your notes.
View Notes
×