A, B and C can complete a work in 20, 24 and 30 days respec tively. All of them starts to gether but after 4 days A leaves the job and B left the job 6 days before the work was completed. C completed the remaining work alone. In how many days was the total work completed?

14

Total work = LCM(20,24,30) = 120 units A @ 6units/day B @ 5 un/day C @ 4un/day Let's say the total work i.e. 120units is completed in x days. A worked for 4 days = 6*4 = 24units by A B worked for x-6 days C worked for all the days ==> work done by A+B+C = 120 ==> 24+5(x-6)+4x = 120 ==> 24+5x-30+4x = 120 ==> 9x = 126 ==> x = 14 days

Take LCM of their respective days as total work. implies 20,24,30's LCM Is 120.. (As we know the efficiency is total work divided by no:of days of work), efficiency of A is 6, B's efficiency 5 and C's 4. As per the qn, all of three together works for 4 days,after that A leaves. So 3 of them works for 4 days completing 60 works out of 120.( efficiency of A+B+C=15. 15×4days=60) B leaves the job 6 days before completing, that means only C works for last 6 days that is 4×6 days =24 unit work of 120 so total 60+24= 84 units of work is completed The rest 36 units work is done by both B AND C. 36÷9=4 DAYS SO TOTAL DAYS IS 4+6+4= 14 DAYS (Never bother about lengthy description.The method is soo simple.There is no option to attach image other wise I would have done it)