​C is twice as efficient as A, while B takes thrice as many days as C to do me same work. A takes 10 days to do the work alone. If they work in pairs [like (A, B) , (B, C) and (C, A) ]. with (A, B) working on the first day, then (B, C) working on the second day and (C, A) working on the third day and continuing the cycle till the work gets completed. then how many days will be required to complete this work?

Correct option is D

​Given:
C is twice as efficient as A,

B takes thrice as many days as C.

A alone completes in 10 days. 

Formula Used: ​

Total work = Efficiency $$\times$$ Time​

Solution: 

​​C is twice as efficient as A 

So , Efficiency Ratio of A : B = 1 : 2 

B takes trice as many days as C 

So, Efficiency ratios of B : C = 1 : 3 

Combining the Efficiency ratio ;

A : B : C = 3 : 2 : 6 

A alone completes the work in 10 days, So

Total work = $$3 \times 10 = 30$$ 

Work is done in pairs ( A, B ) , (B, C), ( C , A ) 

So, A + B = 3 + 2 = 5 ,  B + C = 2 + 6 = 8 , C + A = 6 + 3 = 9 

Total work completed in three days working in pair,

= 5 + 8 + 9 = 22 unit 

Now, pair (A, B ) work again, on 4th day 

So, total work completed = 22 + 5 = 27 unit 

Remaining work = 30 - 27 = 3 unit 

3 unit work is completed by next pair(B , C) 

So, time = $$\frac{3}{8}$$​ 

Thus, total time taken to complete the work = 4 + $$\frac 3 8$$ = $$4\frac 3 8 \ days$$​​