Ex.4.3 Q10 Quadratic Equations Solutions - NCERT Maths Class 10

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Question

An express train takes \(1\) hour 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 express train is \(11\, \rm{km /hr}\) more than that of passenger train, find the average speed of the two trains.

 Video Solution
Quadratic Equations
Ex 4.3 | Question 10

Text Solution

What is Known?

i) Express train takes \(1\) hour less than a passenger train to travel \(132\,\rm{ km.}\)

ii) Average speed of express train is \(11\,\rm{km/hr}\) more than that of passenger train.

What is Unknown?

Average speed of express train and the passenger train.

Reasoning:

Let the average speed of passenger train \(= x\,\rm{km/hr}\)

Average speed of express train \(= (x + 11)\, \rm{km /hr}\)

\[\begin{align}\rm{Distance} &= \rm{Speed} \times \rm{Time}\\\rm{Time}& = \frac{{{\rm{Distance}}}}{{\rm{Speed}}}\end{align}\]

Time taken by passenger train to travel \( 132\,{\rm{km}}= \frac{{132}}{x}\)

Time taken by express train to travel \(132 \,{\rm{km}} =\frac{{132}}{{x + 11}}\)

Difference between the time taken by the passenger and the express train is \(1\) hour. Therefore, we can write:

\[\frac{{132}}{x} - \frac{{132}}{{x + 11}} = 1\]

Steps:

Solving \(\begin{align} \frac{{132}}{x} - \frac{{132}}{{x + 11}} = 1 \end{align}\) by taking the LCM on the LHS:

\[\begin{align}\frac{{132\left( {x + 11} \right) - 132x}}{{x\left( {x + 11} \right)}}& = 1\\\frac{{132x + 1452 - 132x}}{{{x^2} + 11x}} &= 1\\1452 &= {x^2} + 11x\\{x^2} + 11x - 1452 &= 0\end{align}\]

By comparing \({x^2} + 11x - 1452 = 0\) with the general form of a quadratic equation \(ax² + bx + c = 0:\)

\[a = 1,\; b = 11,\; c = - 1452\]

\[\begin{align} {{b}^{2}}-4ac&={{11}^{2}}-4\left( 1 \right)\left( -1452 \right) \\ & =121+5808 \\
 & =5929>0 \\ b{}^\text{2}\text{ }-\text{ }4ac &>0 \\\end{align}\]

\(\therefore\) Real roots exist.

\[\begin{align}{\rm{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\
&= \frac{{ - 11 \pm \sqrt {5929} }}{{2(1)}}\\&= \frac{{ - 11 \pm 77}}{2}\\
x &= \frac{{ - 11 + 77}}{2} \qquad x = \frac{{ - 11 - 77}}{2}\\&= \frac{{66}}{2} \qquad \qquad \quad\;\;  = \frac{{ - 88}}{2}\\&= 33 \qquad \qquad \quad\;\;\;\;= - 44\\\\&x=33 \qquad \qquad \qquad -x=44\end{align}\]

\(x\) can’t be a negative value as it represents the speed of the train.

Speed of passenger train \(= 33\,\rm{ km/hr}\)

Speed of express train \(=x+11 = 33 + 11 = 44\;\rm{km/hr.}\)