P can complete a piece of work in 15 days. Q in 20 days and R in 30 days. P and R worked together for three days and then P was replaced by Q. In how many days, altogether, was the work completed (in days)?

Work done by ( P + R ) in 3 days = 3(1/15+1/30) = 3((2+1)/30)=3/10 Remaining work = 1-3/10=7/10 ( Q + R )’s 1 day’s work = 1/20+1/30 = (3+2)/60 = 5/60 = 1/12 Time taken by ( Q + R ) to finish 7/10 part of the work = 12 × (7/10)  = 42/5 days = 8(2/5)  days Total time = 3 + 8  = 11`2/5`  days. Alternate Method:      P            Q            R     15         20            30 LCM = 60 units Eff  4   :      3   :      2 (P+R)'s 3 days work = (4+2)×3 = 18 units So remaining is done by P & R in = (60-18)/(3+2) = 42/5 days So total days = 3 + (42/5) = 57/5 days