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?

Correct option is C

Given:
Time taken by A alone = 30 days
Time taken by B alone = 33 days
Time taken by C alone = 36 days
Time taken by D alone = 44 days
Formula Used:
1-day work = 1 / Total time
Total Work = LCM of given days
Solution:
LCM of 30, 33, 36 and 44 = 1980 units
1-day work:
A = $$\frac{1980}{30}$$​ = 66 units

B = $$\frac{1980}{33}$$​ = 60 units

C = $$\frac{1980}{36}$$​= 55 units

D =$$\frac{1980}{44}$$​= 45 units
A + B = 66 + 60 = 126 units per day
Work Pattern:
First 2 days (A + B + C):
(126 + 55) × 2 = 362 units
Next 2 days (A + B + D):
(126 + 45) × 2 = 342 units
Work in 4 days = 704 units
In 8 days: 704 × 2 = 1408 units
Remaining work = 1980 − 1408 = 572 units
Next Turn: A+B+C in 2 days = 362 units 
Remaininng work done by A, B and D = 572-362 = 210 unit
So, Time taken by A, B and D to complete the remaining work = $$\frac{210}{171} = 1 \frac {39}{171}$$ 
Total Number of Days = 10 + $$1 \frac {39}{171}$$ = $$11 \frac {13}{57}$$ days​