A and B together can complete a work in 15 days. They started together but after 5 days A left the work. If the remaining work is completed by B in 15 more days, then A alone can complete the entire work in how many days?

$(\mathrm{A}+\mathrm{B})$ ’s 1 day’s work $=\frac{1}{15}$ $\therefore(\mathrm{A}+\mathrm{B})$ ’s 5 days’ work $=\frac{5}{15}=\frac{1}{3}$ Remaining work $=1-\frac{1}{3}=\frac{2}{3}$ $\because$ Time taken by B alone in doing $\frac{2}{3}$ work $=15$ days $\therefore$ Time taken by B alone in in doing 1 work $=15 \times \frac{3}{2}=\frac{45}{2} \text { days }$ $\therefore$ A’s 1 day’s work $=\frac{1}{15}-\frac{2}{45}$ $=\frac{3-2}{45}=\frac{1}{45}$ $\therefore$ Required time $=45$ days

Alternative:

(A + B) × 15 = (A + B) + 15B

3A + 3B = A + B + 3B

2A = B

$\frac{\mathrm{A}}{\mathrm{B}}=\frac{1}{2}$

$\therefore$ Total work $=(2+1) \times 15$

$=45$

Time taken by A alone to complete the work

$=\frac{45}{1}=45 \text { days }$