A, B and C can complete a work alone in 400, 600 and 900 days respectively. In how many days can the work be completed if it is started by A and he is assisted by B and C on every second and third day respectively?

A, B and C can complete a work alone in 400, 600 and 900 days respectively.

LCM of 400, 600 and 900 is 3600 which can be assumed as the total work.

Efficiency of A = $$\frac{3600}{400}$$ = 9 units/day

Efficiency of B = $$\frac{3600}{600}$$ = 6 units/day

Efficiency of C = $$\frac{3600}{900}$$ = 4 units/day

if it is started by A and he is assisted by B and C on every second and third day respectively?

FIrst day + second day + third day = Efficiency of A + (Efficiency of A+Efficiency of B) + (Efficiency of A+Efficiency of C)

= 9 + (9+6) + (9+4)

= 9 + 15 + 13

= 37

In three days 37 units is done.

$$3\times97$$ days = $$37\times97$$ units

291 days = 3589 units

Now 11 units of work is remaining. So on 292nd day 9 units will be completed and the remaining 2 units will be completed in $$\frac{2}{15}\ days$$. 

Number of days required to complete the whole work as per the given condition = $$291+1+\frac{2}{15}$$

= $$292+\frac{2}{15}$$

= 292.13

= 292 days (approx.)