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A train travelling at a uniform speed for $360$ km would have taken $48$ minutes less to travel the same distance if its speed were $5$ km/h more. Find the original speed of the train.
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Let the original speed of train be 'x' km/h
ATQ
$\frac{360}{x} - \frac{360}{x + 5} = \frac{48}{60}$ ($2$)
$\Rightarrow x^2 + 5x - 2250 = 0$ ($1$)
$\Rightarrow (x + 50)(x - 45) = 0$ ($1$)
So, $x = 45$
$\therefore$ Original speed of the train is $45$ km/h. ($1$)
ATQ
$\frac{360}{x} - \frac{360}{x + 5} = \frac{48}{60}$ ($2$)
$\Rightarrow x^2 + 5x - 2250 = 0$ ($1$)
$\Rightarrow (x + 50)(x - 45) = 0$ ($1$)
So, $x = 45$
$\therefore$ Original speed of the train is $45$ km/h. ($1$)