A man covers a distance of 100 km, travelling with a uniform speed of x km/hr. Had the speed been 5 km/hr more it would have taken 1 hour less. Find x the original speed.

In first case :

Distance = 100 km

Speed = x km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x}$

In second case :

Distance = 100 km

Speed = (x + 5) km/hr

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{100}{x + 5}$

According to question,

The difference between the time taken between first and second case is 1 hour.

$\therefore \dfrac{100}{x} - \dfrac{100}{x + 5} = 1 \\[1em] \Rightarrow \dfrac{100(x + 5) - 100x}{x(x + 5)} = 1 \\[1em] \Rightarrow \dfrac{100x + 500 - 100x}{x^2 + 5x} = 1 \\[1em] \Rightarrow \dfrac{500}{x^2 + 5x} = 1 \\[1em] \Rightarrow 500 = x^2 + 5x \\[1em] \Rightarrow x^2 + 5x - 500 = 0 \\[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \\[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \\[1em] \Rightarrow (x - 20)(x + 25) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 25 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -25.$

Since, speed cannot be negative.

∴ x = 20 km/hr.

Hence, original speed = 20 km/hr.