A bus travels at a certain speed for a distance of 75 km and then travels a distance of 90 km at a speed of 10 km/hr more than the first speed. If it takes 3 hours to complete the journey, find the original speed.

Let's assume the bus's original speed is x km/hr.

Using the formula; Time = $\dfrac{\text{Distance}}{\text{Speed}}$

The time taken to travel 75 km at the original speed x km/hr = $\dfrac{75}{x}$

The time taken to travel 90 km at a speed of (x + 10) km/hr (since the second speed is 10 km/h more than the original speed) = $\dfrac{90}{x + 10}$

The total time taken for the journey is 3 hours. Therefore, we can write the equation:

$\Rightarrow \dfrac{75}{x} + \dfrac{90}{x + 10} = 3\\[1em] \Rightarrow \dfrac{75(x + 10) + 90x}{x(x + 10)} = 3\\[1em] \Rightarrow \dfrac{75x + 750 + 90x}{x(x + 10)} = 3\\[1em] \Rightarrow \dfrac{165x + 750}{x(x + 10)} = 3\\[1em] \Rightarrow 165x + 750 = 3x(x + 10)\\[1em] \Rightarrow 165x + 750 = 3x^2 + 30x\\[1em] \Rightarrow 3x^2 + 30x - 165x - 750 = 0\\[1em] \Rightarrow 3x^2 - 135x - 750 = 0\\[1em] \Rightarrow 3(x^2 - 45x - 250) = 0\\[1em] \Rightarrow x^2 - 45x - 250 = 0\\[1em] \Rightarrow x^2 - 50x + 5x - 250 = 0\\[1em] \Rightarrow x(x - 50) + 5(x - 50) = 0\\[1em] \Rightarrow (x - 50)(x + 5) = 0\\[1em] \Rightarrow (x - 50) = 0 \text{ or }(x + 5) = 0\\[1em] \Rightarrow x = 50 \text{ or }x = -5\\[1em]$

Since, speed of bus cannot be negative.

Hence, speed of bus = 50 km/hr.