A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work ?

### Concept & Pairwise Efficiencies When given times for pairs of workers, calculate their combined pair efficiencies using LCM. Summing these pairwise efficiencies gives exactly twice the efficiency of all workers combined. $$ \text{Total Efficiency } (A+B+C) = \frac{(A+B) + (B+C) + (C+A)}{2} $$ ### Step-by-Step Solution * **Given:** (A+B) take 30 days, (B+C) take 24 days, (C+A) take 20 days. All work for 10 days, then B and C leave. * **Calculation:** Let Total Work = LCM(30, 24, 20) = 120 units. * Efficiency of (A + B) = $120 / 30 = 4$ units/day. * Efficiency of (B + C) = $120 / 24 = 5$ units/day. * Efficiency of (C + A) = $120 / 20 = 6$ units/day. * Adding them all together: $2(A + B + C) = 4 + 5 + 6 = 15$ units/day. * Combined efficiency of (A + B + C) = $15 / 2 = 7.5$ units/day. * They all work together for 10 days. Work done = $7.5 \times 10 = 75$ units. * Remaining work = $120 - 75 = 45$ units. * Now, B and C leave, meaning only A is working. We need A's individual efficiency. * A's efficiency = (A + B + C)'s efficiency - (B + C)'s efficiency. * A's efficiency = $7.5 - 5 = 2.5$ units/day. * Days required by A to finish the remaining work = Remaining Work / A's Efficiency = $45 / 2.5 = 450 / 25 = 18$ days. ### Exam Strategy & Shortcut Find the total work as 120. Pair efficiencies are 4, 5, 6. Total efficiency of all three is half the sum: $15/2 = 7.5$. After 10 days, $75$ work is done, $45$ is left. A's efficiency is Total $(7.5)$ minus B+C $(5)$, which is $2.5$. $45 / 2.5 = 18$ days. ### Common Pitfall Students often forget to divide the sum of the pairwise efficiencies by 2, mistakenly using 15 as the combined efficiency of A, B, and C, which leads to incorrect downstream calculations. ### Final Answer Therefore, the correct answer is **18 days**.