A, B and C can complete a piece of work in 12, 20 and 15 days, respectively. A, B and C will complete the work while working on alternate days. C works on the 1st day, B works on the 2nd day and A works on the 3rd day and so on. In how many days will the work get complete?

Correct option is B

Given:

A can complete the work in 12 days 

B can complete the work in 20 days 

C can complete the work in 15 days 

Working order: C (Day 1), B (Day 2), A (Day 3), C (Day 4), ...

Formula Used:

Work per cycle = C’s 1 day work + B’s 1 day work + A’s 1 day work

Number of full cycles = $$\frac{\text{Total work}}{\text{Work per cycle}}$$​​

Solution:

A’s 1 day work = $$\frac{1}{12}$$​​

B’s 1 day work = $$\frac{1}{20}$$​​

C’s 1 day work = $$\frac{1}{15}$$​​

Work done in 3 days:

​$$= \frac{1}{15} + \frac{1}{20} + \frac{1}{12}$$​​

​$$= \frac{4 + 3 + 5}{60}$$​​

​$$= \frac{12}{60}$$​​

​$$= \frac{1}{5}$$​​

So, work done in 3 days = $$\frac{1}5$$​​

Therefore, in 15 days (5 cycles), work done = $$\frac{5}{5}$$​ = 1 

Alternate Solution: 

LCM of 12, 20, 15 = 60 units (assumed total work)

Work rates:

A = $$\frac{60}{12}$$​ = 5 units/day

B = $$\frac{60}{20}$$​ = 3 units/day

C = $$\frac{60}{15}$$​ = 4 units/day

3-day cycle work:

Day 1 (C) = 4 units

Day 2 (B) = 3 units

Day 3 (A) = 5 units

Total per cycle = 4 + 3 + 5 = 12 units

Number of full 3-day cycles:

$$\left\lfloor \frac{60}{12} \right\rfloor$$​ = 5 cycles

Total days = 5 × 3 = 15 days