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?

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)

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

14