Let one number be \(x\). \[ \therefore \text{The other number is } (50 - x) \] According to the question: \[ \frac{1}{x} + \frac{1}{50 - x} = \frac{1}{12} \] \[ \Rightarrow \frac{50 - x + x}{x(50 - x)} = \frac{1}{12} \Rightarrow \frac{50}{50x - x^2} = \frac{1}{12} \Rightarrow 50x - x^2 = 600 \Rightarrow 50x - x^2 - 600 = 0 \Rightarrow x^2 - 50x + 600 = 0 \Rightarrow x^2 - (30 + 20)x + 600 = 0 \Rightarrow x^2 - 30x - 20x + 600 = 0 \Rightarrow x(x - 30) - 20(x - 30) = 0 \Rightarrow (x - 30)(x - 20) = 0 \] \[ \therefore x = 30 \text{ or } x = 20 \] \[ \therefore \text{The two numbers are } 30 \text{ and } 20 \]