A man walks 1 km/hr faster than his usual speed and covers a distance of 3 km in 15 minutes less time. Find his usual speed.

Let the usual speed of man be x km/hr.

Given that when he walks 1 km/h faster than usual, his speed is (x + 1) km/h, and he covers a distance of 3 km in 15 minutes less time than usual.

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

The time it takes for him to cover 3 km at his usual speed x km/h is given by:

Usual time = $\dfrac{3}{x}$ hr

The time it takes for him to cover 3 km at the faster speed x + 1 km/h is given by:

Faster time = $\dfrac{3}{x + 1}$ hr

The faster time is 15 minutes ($\dfrac{15}{60} = \dfrac{1}{4}$ hours) less than the usual time. Therefore, we can write:

$\Rightarrow \dfrac{3}{x} - \dfrac{3}{x + 1} = \dfrac{1}{4}\\[1em] \Rightarrow \dfrac{3(x + 1) - 3x}{x(x + 1)} = \dfrac{1}{4}\\[1em] \Rightarrow \dfrac{3x + 3 - 3x}{x(x + 1)} = \dfrac{1}{4}\\[1em] \Rightarrow \dfrac{3}{x(x + 1)} = \dfrac{1}{4}\\[1em] \Rightarrow 3 \times 4 = x(x + 1)\\[1em] \Rightarrow 12 = x^2 + x\\[1em] \Rightarrow x^2 + x - 12 = 0\\[1em] \Rightarrow x^2 + 4x - 3x - 12 = 0\\[1em] \Rightarrow x(x + 4) - 3(x + 4) = 0\\[1em] \Rightarrow (x + 4)(x - 3) = 0\\[1em] \Rightarrow (x + 4) = 0 \text{ or } (x - 3) = 0\\[1em] \Rightarrow x = -4 \text{ or } x = 3$

Since, speed cannot be negative,

∴ Usual speed of man = 3 km/hr

Hence, usual speed of man = 3 km/hr.