Ex.4.3 Q8 Quadratic Equations Solutions - NCERT Maths Class 10

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Question

A train travels \(360\,\rm{ km}\) at a uniform speed. If the speed had been \(5\,\rm{ km /hr}\) more, it would have taken \(1\) hour less for the same journey. Find the speed of the train.

 Video Solution
Quadratic Equations
Ex 4.3 | Question 8

Text Solution

What is known?

i) Distance covered by the train at a uniform speed \(= 360\,\rm{ km} \)

ii) If the speed had been \(5 \,\rm{km/hr}\) more, it would have taken \(1\) hour less for the same journey.

What is Unknown?

Speed of the train.

Reasoning:

Let the speed of the train be \(s\; \rm{km/hr}\) and the time taken be \(t\) hours.

\[\begin{align}\rm{Distance} &= \rm{Speed} \times \rm{Time}\\360&=s\times t\\360 &= s \times t\\t &= \left( {\frac{{360}}{s}} \right)\end{align}\]

Increased speed of the train: \(s + 5\)

New time to cover the same distance: \(t – 1\)

\[(s + 5)(t - 1) = 360\,\,\,\,\,\,\,\,\,\, \ldots (2)\]

Steps:

\[\begin{align}(s + 5)(t - 1) &= 360\\st - s + 5t - 5 &= 360\\360 - s + 5\left( {\frac{{360}}{s}} \right) - 5 &= 360\\ - s + \frac{{1800}}{s} - 5 &= 0 - {s^2} + 1800 - 5s = 0\\{s^2} + 5s - 1800 &= 0\end{align}\]

Solving by quadratic formula:

Comparing with \(ax^\text{2}+bx+c=0\)

\[a{\rm{ }} = {\rm{ }}1,{\rm{ }}b{\rm{ }} = {\rm{ }}5,{\rm{ }}c{\rm{ }} = {\rm{ }} - {\rm{ }}1800\]

\[\begin{align}{{{b}}^2} - 4{\rm{ac}}& = {{(5)}^2} - 4(1)( - 1800)\\&= 25 + 7200\\& = 7225 > 0\end{align}\]

\(\therefore\) Real roots exist.

\[\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\{{s}} &= \frac{{ - 5 \pm \sqrt {7225} }}{2}\\{{s}} &= \frac{{ - 5 \pm 85}}{2}\\{{s}} &= \frac{{ - 5 + 85}}{2},\quad {{s}} = \frac{{ - 5 - 85}}{2}\\{{s}}& = \frac{{80}}{2}\qquad\qquad  {{s}} = \frac{{ - 90}}{2}\\{{s}} &= 40 \qquad\qquad\;  {{s}} = - 45\end{align}\]

Speed of the train cannot be a negative value.

\(\therefore\) Speed of the train is \(40 \,\rm{km /hr.}\)