A, B, C and D can do a piece of work separately in 30 days, 33 days, 36 days and 44 days, respectively. In how many days can A and B together complete the work, if A and B together are assisted by C and D on alternate two-two days starting with C for first two days, then D on next two days and again C on next two days and so on?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 11 days.
Total work = LCM of (30, 33, 36 and 44) = 1980
A efficiency = 1980 / 30 = 66 units/day
B efficiency = 1980 / 33 = 60 units/day
C efficiency = 1980 / 36 = 55 units/day
D efficiency = 1980 / 44 = 45 units/day
A + B = 66 + 60 = 126 units/day
A + B + C = 126 + 55 = 181 units/day
A + B + D = 126 + 45 = 171 units/day
Work in first 2 days with C = 2 × 181 = 362 units
Work in next 2 days with D = 2 × 171 = 342 units
Work in one 4-day cycle = 362 + 342 = 704 units
Work in two full cycles (8 days) = 704 × 2 = 1408 units
Remaining work = 1980 - 1408 = 572 units
Work in next 2 days (with C) = 2 × 181 = 362 units
Cumulative after 10 days = 1408 + 362 = 1770 units
Remaining work = 1980 - 1770 = 210 units
Work in 1 day (with D) = 171 units
Remaining after that = 210 - 171 = 39 units
Time = Work / Efficiency
Time to finish remaining 39 units at 171 units/day = 39 / 171 = 13 / 57 days
Total time = 8 days + 2 days + 1 day + 13/57 day = 11 + 13/57 = 11 days