M, N, and O can do work in 22, 24, and 16 days respectively. If they work for a day alternately such that M works on the first day, N works on the second day, O works on the third day and then again M works on the fourth day and so on, then in how many days approximately, the work will be completed?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e.20 days
Given that
M can do a piece of work in 22 days
N can do the same work in 24 days
O can do the same work in 16 days
Total work = LCM (22, 24, 16) = 528
The efficiency of M = 528/22 = 24
The efficiency of N = 528/24 = 22
The efficiency of O = 528/16 = 33
If they work for a day alternately such that M works on the first day, N works on the second day, O works on the third day and so on.
In 3 days, (24 + 22 + 33 = 79) work will be completed
In 20 days, (79 × 6 + 24 + 22 = 520) work will be completed
The remaining (528 - 520 = 8) will be completed by O in 8/33 days i.e. 0.24 days
Hence, the whole work will be completed in 20.24 days ≈ 20 days