# 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
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}

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 $$40 \,\rm{km /hr.}$$

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