A can complete a piece of work in 10 days. B in 15 days and C in 20 days. A and C worked together for two days and then A was replaced by B. In how many days, altogether, was the work completed (in days)?

Solution

Work done by ( A + C ) in 2 days = 2 (1/10 + 1/20 ) = 2 ((2 + 1)/20 ) = 6/20 = 3/10 Remaining work = 1 - 3/10 = 7/10 ( B + C )’s 1 day’s work = 1/15 + 1/20 = (4 + 3)/60 = 7/60 Time taken by ( B + C ) to finish 7/10 part of the work = 60/7 × 7/10 = 6 days Total time = 2 + 6 = 8 days Alternate method: Let total work = LCM (10,15,20) = 60 units So A’s 1 day work  = 60/10 = 6 units B’s 1 day work  = 60/15 = 4 units C’s 1 day work  = 60/20 = 3 units A & C’s work for 2 days = (6+3)×2 = 18 units So remaining work = 60 – 18 = 42 units Now A is replaced by B So B & C together will do work in 1 day = 3 + 4 = 7 units So they can do the remaining job in = 42/ 7 = 6 days So total days = 6 + 2 =  8 days