A and B together can do a piece of work in 30 days. B and C together can do it in 20 days. A starts the work and works on it for 5 days, then B takes up and works for 15 days. Finally C finishes the work in 18 days. The number of days in which C alone can do the work when doing it separately is

Let C complete the work in x days. $\therefore$ B’s 1 day’s work $=\frac{1}{20}-\frac{1}{x}$ and, A’s 1 day’s work $=\frac{2-3}{60}+\frac{1}{x}=\frac{1}{x}-\frac{1}{60}$ According to the question, $5\left(\frac{1}{x}-\frac{1}{60}\right)+15\left(\frac{1}{20}-\frac{1}{x}\right)+\frac{18}{x}=1$ $\Rightarrow \frac{5}{x}-\frac{1}{12}+\frac{15}{20}-\frac{15}{x}+\frac{18}{x}=1$ $\Rightarrow \frac{5}{x}-\frac{15}{x}+\frac{18}{x}=1+\frac{1}{12}-\frac{3}{4}$ $\Rightarrow\left(\frac{5-15+18}{x}\right)=\frac{12+1-9}{12}$ $\Rightarrow \quad \frac{8}{x}=\frac{1}{3}$ $\Rightarrow \quad x=8 \times 3=24$ days Alternative: According to the question, (A + B) × 30 = (B + C) × 20 = 5A + 15B + 18C Now, $\Rightarrow(\mathrm{A}+\mathrm{B}) \times 30=(\mathrm{B}+\mathrm{C}) \times 20$ $3 \mathrm{~A}+\mathrm{B}=2 \quad \ldots \ldots(\mathrm{i})$ $\Rightarrow(\mathrm{B}+\mathrm{C}) \times 20=5 \mathrm{~A}+15 \mathrm{~B}+18 \mathrm{C}$ $5 \mathrm{~A}-5 \mathrm{~B}=2 \mathrm{C} \quad \ldots \ldots$ (ii) $\Rightarrow(\mathrm{A}+\mathrm{B}) \times 30=5 \mathrm{~A}+15 \mathrm{~B}+18 \mathrm{C}$ $25 \mathrm{~A}+15 \mathrm{~B}=18 \mathrm{C} \quad \ldots \ldots$ (iii) From equation (i) and (ii) 2A = 6B $\frac{\mathrm{A}}{\mathrm{B}}=\frac{3}{1}$ Now equation (i) 3 × 3 + 1 = 2C

∴ Total work = (3 +1) × 30 = 120

Time taken by $\mathrm{C}$ alone to complete the work

$=\frac{\text { Total work }}{\text { Efficiency of } \mathrm{C}}$

$=\frac{120}{5}=24$ days