A, B and C can do a piece of work in 22 days, 24 days and 33 days, respectively. In how many days will the work be finished if, A works on the 1st day, C works on the 2nd day, B works on the 3rd day, A works on the 4th day and so on?

Correct option is C

Given:

A can complete the work in 22 days

B can complete the work in 24 days

C can complete the work in 33 days

Formula Used:​

Efficiency = $$\frac{\text{Total Work}}{\text{Days taken individually}}$$​​

Solution:

Let Total Work = 264 units(LCM of 22, 24, 33)

A’s efficiency = $$\frac{264}{22}$$​ = 12 units/day

B’s efficiency = $$\frac{264}{24}$$​ = 11 units/day

C’s efficiency = $$\frac{264}{33}$$​ = 8 units/day

Work in 3-day cycle

Day 1: A works = 12 units

Day 2: C works = 8 units

Day 3: B works = 11 units

Total in 3 days = 12 + 8 + 11 = 31 units

Now, 

​$$\left\lfloor \frac{264}{31} \right\rfloor = 8 \text{ cycles} => 8 × 31 = 248 \text{ units}$$​​

Time taken =  8 $$\times$$​ 3 = 24 days

Remaining work = 264 - 248 = 16 units

Day 25 (A’s turn): does 12 units

Remaining: 16 - 12 = 4 units

Day 26 (C’s turn): does 8 units/day

To do 4 units => $$\frac{4}{8}$$​ = 0.5 day

Total Time:  24 + 1 + 0.5 = 25.5 days