A, B and C can complete a piece of work in 12 days, 18 days and 36 days respectively (working alone). They all start working together. After 2 days, C leaves the work. After another x days, B also leaves the work, and the remaining work is completed by A alone in 3 more days. In how many days in total was the work completed (from the start)?

Solution

Let total work = 1 unit. Rates per day: A: 1/12 B: 1/18 C: 1/36 First 2 days: A + B + C Combined rate = 1/12 + 1/18 + 1/36 LCM = 36 = (3 + 2 + 1)/36 = 6/36 = 1/6 Work in first 2 days = 2 × 1/6 = 1/3 Remaining work = 1 − 1/3 = 2/3 Next x days: only A + B Rate (A + B) = 1/12 + 1/18 LCM = 36 → (3 + 2)/36 = 5/36 Work in these x days = x × 5/36 Last 3 days: only A Work by A alone in last 3 days = 3 × 1/12 = 3/12 = 1/4 Now total work: (First part) + (Second part) + (Third part) = 1 1/3 + (5x/36) + 1/4 = 1 Take LCM 36: 1/3 = 12/36 1/4 = 9/36 So: (12/36) + (5x/36) + (9/36) = 1 (21 + 5x)/36 = 1 21 + 5x = 36 5x = 15 x = 3 Total time = 2 (first phase) + 3 (second phase) + 3 (last phase) = 8 days.