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A train travels at a certain average speed for a distance of $54$ km and then travels a distance of $63$ km at an average speed of $6$ km/h more than the first speed. If it takes $3$ hours to complete the journey, what was its first average speed ?
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Let first average speed of the train be $x$ km/hr.
$\frac{54}{x} + \frac{63}{x + 6} = 3$
$\Rightarrow 54(x + 6) + 63x = 3x^2 + 18x$
$\Rightarrow 3x^2 - 99x - 324 = 0$ or $x^2 - 33x - 108 = 0$
$\Rightarrow (x - 36) (x + 3) = 0$
$\Rightarrow x = 36, -3$ (rejected)
Therefore, first average speed of the train was $36$ km/hr.
$\frac{54}{x} + \frac{63}{x + 6} = 3$
$\Rightarrow 54(x + 6) + 63x = 3x^2 + 18x$
$\Rightarrow 3x^2 - 99x - 324 = 0$ or $x^2 - 33x - 108 = 0$
$\Rightarrow (x - 36) (x + 3) = 0$
$\Rightarrow x = 36, -3$ (rejected)
Therefore, first average speed of the train was $36$ km/hr.