Correct Answer - Option 4 : 58/7 days
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