$$N_{1}$$ alone can do 20% of a piece of work in 6 days, $$N_{2}$$ alone can do 20% of the same work in 4 days, and $$N_{3}$$ alone can do 50% of the same work in 12 days. In how many days can they complete the work, if all of them work together?

$$N_{1}$$ alone can do 20% of a piece of work in 6 days.

Time taken by $$N_{1}$$ to complete the whole work = $$\frac{6}{20}\times\ 100$$ = 30 days

$$N_{2}$$ alone can do 20% of the same work in 4 days.

Time taken by $$N_{2}$$ to complete the whole work = $$\frac{4}{20}\times\ 100$$ = 20 days

$$N_{3}$$ alone can do 50% of the same work in 12 days.

Time taken by $$N_{3}$$ to complete the whole work = $$\frac{12}{50}\times\ 100$$ = 24 days

LCM of 30, 20 and 24 is 120 which can be assumed as the total work.

Efficiency of $$N_{1}$$ = $$\frac{120}{30}$$ =  4 units/day
Efficiency of $$N_{2}$$ = $$\frac{120}{20}$$ = 6 units/day

Efficiency of $$N_{3}$$ = $$\frac{120}{24}$$ = 5 units/day

Time taken by all of them to complete the work = $$\frac{120}{4+6+5}$$

= $$\frac{120}{15}$$

= 8 days