Question

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop in intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
A). Speed of passenger train = 31 km/hr, speed of express train = 42 km/hr
B). Speed of passenger train = 33 km/hr, speed of express train = 44 km/hr
C). Speed of passenger train = 43 km/hr, speed of express train = 54 km/hr
D). None of these

Answer 1

Hint: In this question use the given information to make the equations and remember to use time taken by passenger train as $\dfrac{{132}}{x}$ hr, using these instructions will help you to approach closer towards the solution of the problem.

Complete step-by-step solution -
According to the given information the total distance between passenger train and express train is 132km and the average speed of the express train is 11 km/hr more than of passenger train
So let x km/hr be the average speed of passenger train
Thus the average speed of express train will be (x + 11) km/hr
So we know that distance = speed $ \times $ time
Time taken by passenger train = distance/speed
Substituting the value in the above condition
Time taken by passenger train = $\dfrac{{132}}{x}$ hr
And time taken by express train = $\dfrac{{132}}{{x + 11}}$ hr
According to the given information express train takes 1 hr less than the passenger train
So time taken by express train = time taken by passenger train - 1hr
Substituting the values in the above statement we get
$\dfrac{{132}}{{x + 11}}$ hr = $\left( {\dfrac{{132}}{x} - 1} \right)$hr
$ \Rightarrow $$\dfrac{{132}}{{x + 11}}$= $\dfrac{{132 - x}}{x}$
$ \Rightarrow $$132x = \left( {x + 11} \right)\left( {132 - x} \right)$
$ \Rightarrow $$132x = 132x - {x^2} + 1452 - 11x$
$ \Rightarrow $$132x = 121x - {x^2} + 1452$
$ \Rightarrow $$132x - 121x = 1452 - {x^2}$
$ \Rightarrow $$11x = 1452 - {x^2}$
$ \Rightarrow $\[{x^2} + 11x - 1452 = 0\]
By the method of splitting the middle term
\[{x^2} + \left( {44 - 33} \right)x - 1452 = 0\]
$ \Rightarrow $\[{x^2} + 44x - 33x - 1452 = 0\]
$ \Rightarrow $\[x\left( {x + 44} \right) - 33\left( {x + 44} \right) = 0\]
$ \Rightarrow $\[\left( {x - 33} \right)\left( {x + 44} \right) = 0\]
So x = 33, -44
Since x can’t be negative
Therefore x = 33
So the average speed of passenger train is 33 km/hr
For average speed of express train is (x+11) km/hr
Putting the value of x in above equation
Average speed of express train = 33 + 11 = 44 km/hr
Hence option B is the correct option.

Note: In the above solution we used the given information which says that express train takes 1 hr less than passengers train so it means that the time taken by express train = time taken by passengers train – 1hr. Then we got the quadratic equation in which we generally use the splitting the middle term method to solve it but in some cases we use quadratic formula i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. After finding the value of x which is the average speed of the passenger train we will find the value of the express train by using the value of x.