X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last ?

### Concept & Sequential Work Track the amount of work completed in chronological phases. Subtract the initial work completed by the solo worker from the total work, then divide the remainder by the combined efficiency of all current workers to find the remaining time. ### Step-by-Step Solution * **Given:** X can do the work in 20 days, Y in 12 days. X works alone for 4 days, then Y joins. * **Calculation:** Let Total Work = LCM(20, 12) = 60 units. * Efficiency of X = $60 / 20 = 3$ units/day. * Efficiency of Y = $60 / 12 = 5$ units/day. * X starts alone and works for 4 days. Work done by X = $3 \times 4 = 12$ units. * Remaining work = $60 - 12 = 48$ units. * Y joins X, so they now work together. * Combined efficiency of (X + Y) = $3 + 5 = 8$ units/day. * Time taken by (X + Y) to finish the remaining work = $48 / 8 = 6$ days. * The question asks how long the work lasted in total. * Total time = Time X worked alone + Time X and Y worked together = $4 + 6 = 10$ days. ### Exam Strategy & Shortcut Use the LCM (60). X does 3 units/day, Y does 5 units/day. X does 12 units in 4 days. 48 units remain. Together they do 8 units/day. $48 / 8 = 6$ days. Total time is $4 + 6 = 10$ days. ### Common Pitfall A frequent error is calculating the time for the second phase (6 days) and choosing that as the final answer. The question specifically asks "How long did the work last?", implying the total duration from start to finish. ### Final Answer Therefore, the correct answer is **10 days**.