Given:

A can complete the work in 10 days, B in

20 days and C can do it in 40 days.

First day A works then on second day he also works 

and on the third day B and C joins and the cycle then repeats on.

Concept Used:

We will always try to find efficiencies in this type of question.

Total Work = Efficiency × Time

Calculation:

LCM of (10, 20 and 40) = 40

So, we assume our Total work = 40 units.

Now, efficiency of A = Total work/Time taken by A

= 40/10 = 4 units/day

Efficiency of B = 2 unit/day and 

Efficiency of C = 1 unit/day.

First day A works = 4 units

Second day A works = 4 units

Third day A + B + C works = 7 units/day.

So, In 3 days total 15 units will be completed

Accordingly in 6 days total 30 units will be completed.

Then in next 2 days, work done by A = 8 units

Total work done in 8 days = (30 + 8) units

⇒ 38 units

Now, remaining 2 units (40 – 38) = 2units will be completed by A + B + C.

Time taken to complete 2 units work = 2/7 days

Total time taken to complete the work = 8 + (2/7) days

⇒ 58/7 days

∴ The total time taken to complete the work is 58/7 days