Work Completion Days

Let efficiencies of John, Jack, and Jim be $E_1, E_2, E_3$ respectively. John takes twice as much time as Jack $\implies E_2 = 2 E_1$. Jack and Jim take $\frac{1}{3}$ the time of John alone $\implies E_2 + E_3 = 3 E_1$. Since $E_2 = 2 E_1$, $2 E_1 + E_3 = 3 E_1 \implies E_3 = E_1$. Total efficiency of all three $= E_1 + E_2 + E_3 = E_1 + 2E_1 + E_1 = 4 E_1$. Time taken by all three working together $t_{\text{all}} = \frac{W}{4 E_1}$. Time taken by John alone $t_{\text{John}} = \frac{W}{E_1}$. Given $t_{\text{John}} - t_{\text{all}} = 3$ days: $\frac{W}{E_1} - \frac{W}{4 E_1} = 3 \implies \frac{3W}{4 E_1} = 3 \implies \frac{W}{E_1} = 4$ Since $E_3 = E_1$, time taken by Jim alone working = $\frac{W}{E_3} = \frac{W}{E_1} = 4\text{ days}$.