​A, B and C can do a piece of work in 20, 30 and 60 days, respectively. A works every day. B assists A on all odd-numbered days, while C assists A on all even-numbered days. In how many days will the work be completed?

Correct option is C

Given:

A can do the work in 20 days.

B can do the work in 30 days.

C can do the work in 60 days.

Solution:

A's work rate: $$\frac{1}{20}$$ work per day.

​B's work rate: $$\frac{1}{30}$$ work per day.

​C's work rate: $$\frac{1}{60}$$ work per day.

​Work Done in Two Days: 

On odd-numbered days (A and B work together):

​Combined work rate = $$\frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12}$$ 

​On even-numbered days (A and C work together):

​Combined work rate = $$\frac{1}{20} + \frac{1}{60} = \frac{3 + 1}{60} = \frac{4}{60} = \frac{1}{15}$$ 

​Total work in 2 days: 

​$$\frac{1}{12} + \frac{1}{15} = \frac{5}{60} + \frac{4}{60} = \frac{9}{60} = \frac{3}{20}$$ 

Work completed in 12 days (6 full 2-day cycles): 

​$$6 \times \frac{3}{20} = \frac{18}{20} = 0.9.$$

​​Remaining work after 12 days:

1−0.9=0.1

​Work on the 13th day (odd, A + B):

Work done = $$\frac{1}{12}$$ =0.0833

Remaining work = 0.1 - 0.0833 = 0.0167

​Time on the 14th day (even, A + C):

A + C work rate = $$\frac{1}{15}$$ 

​Time required = $$\frac{0.0167}{\frac{1}{15}} = 0.25$$ days

​Total Time: 

12days + 1day + 0.25day =13.25days , $$13 \frac{1}{4}$$

Thus, correct answer is (c) $$13 \frac{1}{4}$$ 

Alternate Method: ​

Total work = Efficiency $$\times$$ Time

LCM of (20, 30, 60) = total work = 60  

Individual Efficiency of A = 3, B = 2 , C = 1 

On Day 1 A+B work = 3 + 2 = 5 unit 

On Day 2 A + C work = 3 + 1 = 4 unit 

Total work on two days = 5 + 4 = 9 unit 

So in 12 day work completed = 54 unit 

on 13th day A + B work , 

Work completed = 54 + 5 = 59 unit 

Remaining work = 60 - 59 = 1 unit 

time taken by A + C = $$\frac{1}{4}$$ 

Now, total time to complete the work = 13$$\frac 14$$ days​