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Let us assume the original speed is ${\text{x}}$.

Now, we construct a bus that travels at a ${\text{x}}$ speed for a distance of \[75\] km.

As we know,

${\text{time = }}\dfrac{{{\text{distance }}}}{{{\text{speed}}}}$

So, time taken to travel $75$ km at a speed of${\text{x}}$km/hr = ${\text{t = }}\dfrac{{{\text{75}}}}{{\text{x}}}....{\text{(1)}}$

Again, the bus travels a distance of $90$ km at an average speed of $10$ km/hr more than the first speed.

It means the new speed is $10$ km/hr more than the first speed i.e., ${\text{x + 10}}$

Then time taken to travel $90$ km at a speed of ${\text{x + 10}}$ km/hr = ${\text{t = }}\dfrac{{{\text{90}}}}{{{\text{x + 10}}}}{\text{ }}....{\text{(2)}}$

It is given that the total time taken to complete the journey was $3$ hours. It means, it takes $3$ hours to complete both the distance i.e, $(1)$ and $(2)$.

Therefore, \[3{\text{ = }}\dfrac{{{\text{75}}}}{{\text{x}}} + \dfrac{{{\text{90}}}}{{{\text{x + 10}}}}\]

Taking LCM on both sides, we get,

$ \Rightarrow 3 = \dfrac{{{\text{75}}\left( {{\text{x + 10}}} \right){\text{ + 90}}\left( {\text{x}} \right)}}{{{\text{x}}\left( {{\text{x + 10}}} \right)}}$

On multiply the bracket term and we get,

$ \Rightarrow 3 = \dfrac{{{\text{75x + 750 + 90x}}}}{{{{\text{x}}^2} + 10{\text{x}}}}$

On adding the numerator term

$ \Rightarrow 3 = \dfrac{{{\text{165x + 750 }}}}{{{{\text{x}}^2} + 10{\text{x}}}}$

Taking cross multiply we get,

$ \Rightarrow 3\left( {{{\text{x}}^2} + 10{\text{x}}} \right) = 16{\text{5x + 750 }}$

Let us multiply the bracket term and we get,

$ \Rightarrow 3{{\text{x}}^2} + 30{\text{x}} = 16{\text{5x + 750 }}$

Here we can write it as equating form and we get,

$ \Rightarrow 3{{\text{x}}^2} + 30{\text{x - }}16{\text{5x - 750 = 0}}$

Dividing $3$ on both sides,

$ \Rightarrow 3{{\text{x}}^2} + 10{\text{x}} - 55{\text{x - 750 = 0}}$

On adding the same variable term and we get,

$ \Rightarrow {{\text{x}}^2}{\text{ - }}4{\text{5x - 250 = 0}}$

On splitting the \[x\]term and we get

$ \Rightarrow {{\text{x}}^2} - {\text{50x + 5x - 250 = 0}}$

Taking common variables out,

$ \Rightarrow {\text{x}}\left( {{\text{x - 50}}} \right){\text{ + 5}}\left( {{\text{x - 50}}} \right) = 0$

Again we take a common term and we get,

$ \Rightarrow \left( {{\text{x - 50}}} \right)\left( {{\text{x + 5}}} \right){\text{ = }}0$

On equate the term and we get,

$\left( {{\text{x - 50}}} \right){\text{ = 0 or }}\left( {{\text{x + 5}}} \right){\text{ = 0}}$

We can rewrite it as,

${\text{x = 50 or x = - 5}}$

Since speed cannot be in negative, ${\text{x}} \ne - 5$

Therefore ${\text{x = 50km/hr}}$

Formula to find root value, ${\text{x = }}\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$

Now we have a quadratic equation, ${{\text{x}}^2}{\text{ - }}4{\text{5x - 250 = 0}}$

Here, we have given values,

${\text{a}} = 1$

${\text{b}} = - 45$

${\text{c}} = - 250$

Now, we putting the values and we get,

${\text{x}} = \dfrac{{ - \left( { - 45} \right) \pm \sqrt {{{45}^2} - 4(1)( - 250)} }}{{2(1)}}$

On squaring the values and multiply we get,

${\text{x}} = \dfrac{{45 \pm \sqrt {2025 + 1000} }}{2}$

We split the equation as follows:

${\text{x}} = \dfrac{{45 + \sqrt {3025} }}{2},{\text{x}} = \dfrac{{45 - \sqrt {3025} }}{2}$

On squaring the root value and we get,

${\text{x}} = \dfrac{{45 + 55}}{2},{\text{x}} = \dfrac{{45 - 55}}{2}$

Let us add the term and we get,

${\text{x}} = \dfrac{{100}}{2},{\text{x}} = \dfrac{{ - 10}}{2}$

Let us divide the term and we get,

${\text{x = 50 / x = - 5}}$

Since speed cannot be in negative, ${\text{x}} \ne - 5$

Therefore ${\text{x = 50km/hr}}$

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