A can do piece of work in 4 days, B in 12 days and C in 6 days. If A is assisted by both B and C on every third day, the total work can be done in

Let total work is L.C.M. (4,12,6) = 12 units

A alone can do the work in 4 days, => A's efficiency = $$\frac{12}{4}=3$$ units/day

Similarly B's efficiency = $$\frac{12}{12}=1$$ unit/day

and C's efficiency = $$\frac{12}{6}=2$$ units/day

Work done by A in 2 days = $$3\times2=6$$ units

Now, (A+B+C)'s 1 day's work = $$3+1+2=6$$ units/day

$$\therefore$$ Total work, i.e. $$6+6=12$$ units is done in = 3 days

=> Ans - (B)