A can complete a work in 20 days, B in 30 days and C in 60 days. All three start working together. After 2 days, C leaves. Four days before the work is finished, B also leaves and only A works in those last 4 days. In how many days is the work completed?

A’s 1-day work = 1/20 B’s 1-day work = 1/30 C’s 1-day work = 1/60 First 2 days: all three work Daily work = 1/20 + 1/30 + 1/60 = (3 + 2 + 1)/60 = 6/60 = 1/10 Work done in 2 days = 2 × 1/10 = 1/5 Last 4 days: only A works Work done = 4 × 1/20 = 4/20 = 1/5 So work done in first 2 days + last 4 days = 1/5 + 1/5 = 2/5 Remaining work = 1 − 2/5 = 3/5 This remaining work is done by A and B together. A+B daily work = 1/20 + 1/30 = (3 + 2)/60 = 5/60 = 1/12 Let total time = T days. Then time when both A and B work together = (T − 2 − 4) = T − 6 So: (T − 6) × 1/12 = 3/5 T − 6 = 36/5 T = 36/5 + 6 = 36/5 + 30/5 = 66/5 = 13.2 days