The efficiencies of A, B and C are in the ratio of 2 : 3: 5. Working together, they can complete a task in 6 days. In how many days will A alone complete 20% of that task?

As per the question,

The efficiencies of A, B and C are in the ratio of 2 :3: 5

So the ratio in the time taken by them will be $$\dfrac{1}{2}:\dfrac{1}{3}:\dfrac{1}{5}$$

So, A can finish the work in $$=\dfrac{x}{2}$$ days

A can finish the work in one day $$=\dfrac{2}{x}$$

B can finish the work in $$=\dfrac{x}{3}$$ days

B can finish the work in one day $$=\dfrac{3}{x}$$

and C can finish the work in $$=\dfrac{x}{5}$$ days

C can finish the work in one day $$=\dfrac{5}{x}$$

It is given that together they can finish the work in 6 days,

Hence together they can finish the work in one day $$=\dfrac{1}{6}$$

So, if they are working together, then they can finish the work in one day $$\dfrac{2}{x}+\dfrac{3}{x}+\dfrac{5}{x}=\dfrac{10}{x}$$

$$\Rightarrow \dfrac{10}{x}=\dfrac{1}{6}$$

$$\Rightarrow x=60$$

Hence, A can finish the $$20\%$$ of the work in $$=30 \times \dfrac{20}{100}=6$$days.