C is twice efficient as A. B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (i.e., AB, BC, CA) starting with AB on the first day, BC on the second day and AC on the third day and so on, then how many days are required to finish the work?

Explanation:

According to question,

A B C

Efficiency   3 2 6

Number of day   2 3 1

Number of days taken by A = 12.

∴Number of days taken by B = 18 and Number of days taken by C = 6

1 day’s work of (A + B) = 5 \36

1 day’s work of (B + C) = 8\ 36

1 day’s work of (C + A) = 9\ 36

Day work done

In 5 days total work done = 35\36

Now, the rest of the work

(i.e, 1\36) is done by AC

Number of days taken by AC the rest of the work

=1\36 \ 9\36 = 1\9

Therefore, total time taken complete the work

= 5+1\9 = 5 1\9