Question

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

Answer 1

Hint: Take the speed of the passenger train as $ x $ . The speed of the express train will be the sum of passenger trains and $ 11{\rm{ km/h}} $ . Time taken will be $ \dfrac{{132}}{x}{\rm{ h}} $ . Time taken by the express train is $ \dfrac{{132}}{{x + 11}}{\rm{ h}} $ .

Complete step-by-step answer:
An express train takes $ 1{\rm{ hours}} $ less than a passenger train to travel $ 132{\rm{ km}} $ between Mysore and Bangalore.
Let us assume that the average speed of passenger trains is $ x{\rm{ km/h}} $ .
The passenger train has a distance of $ 132{\rm{ km}} $ and speed is $ x{\rm{ km/h}} $ .
We know the formula for speed which is ratio of distance and time, we get,
 $ {\rm{speed = }}\dfrac{{{\rm{distance}}}}{{{\rm{time}}}} $
Substitute the distance and speed values we obtain,
 $ x = \dfrac{{132}}{{{\rm{time}}}} $
Then, after rearranging the equation,
 $ {\rm{time = }}\dfrac{{132}}{x} $
It is given that the speed of the express train is $ 11{\rm{ km/h}} $ more than the passenger. So, the speed is, $ x + 11 $ .
We know the formula to find speed is,
 $ {\rm{speed = }}\dfrac{{{\rm{distance}}}}{{{\rm{time express}}}} $
On substitute the values of speed and distance we obtain,
 $ \begin{array}{c}
x + 11 = \dfrac{{132}}{{{\rm{time express}}}}\\
{\rm{time express}} = \dfrac{{132}}{{x + 11}}
\end{array} $
It is given that an express train takes $ 1{\rm{ hours}} $ less than a passenger train to travel $ 132{\rm{ km}} $ between Mysore and Bangalore.

So, time taken by passenger train minus time taken by express is equal to $ 1 $ .
So, we will get the equation,
 $ \begin{array}{c}
\dfrac{{132}}{x} - \dfrac{{132}}{{x + 11}} = 1\\
132\left( {x + 11} \right) - 132x = x\left( {x + 11} \right)
\end{array} $
Then, after rearranging the equation, we get,
 $ \begin{array}{c}
132x + 132 \times 11 - 132x = {x^2} + 11x\\
132 \times 11 = {x^2} + 11x\\
{x^2} + 11x - 1452 = 0
\end{array} $
We know that the above equation is in the form of a quadratic equation. So, the formula to find the roots of is,
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Where, $ a = 1 $ , $ b = 11 $ , $ c = - 1452 $ .
On putting values of $ a $ , $ b $ , and $ c $ , we obtain,
 $ \begin{array}{l}
x = \dfrac{{ - 11 \pm \sqrt {5929} }}{2}\\
x = \dfrac{{ - 11 \pm 77}}{2}
\end{array} $
So, the values we obtained are,
  $ x = \dfrac{{ - 11 + 77}}{2} = 33 $ , $ x = \dfrac{{ - 11 - 77}}{2} = - 44 $ .
Therefore, the average speed cannot be negative; we take $ x = 33 $ .

Note: The speed can never be negative, and speed can be zero and positive. Also, acceleration can be negative. The formula for speed, distance and time is simple but still most students get confused in this relation.