A is 40% more efficient than B and C is 20% less efficient than B. Working together, they can finish a task in 15 days. In how many days, will B alone complete 75% of the task?

Let B can finish 100% of work in  X days.

Therefore B's one day work $$\dfrac{100}{x}\%$$

A can finish $$140\%$$ of work in x days,

Therefore A's one day work $$\dfrac{140}{x}\%$$

C can finish$$80\%$$ of work in x days,

Therefore, C's one day work$$=\dfrac{80}{x}\%$$
So, If A, B and C are working together, then there one day's combine work $$=(\dfrac{100}{x}+\dfrac{140}{x}+\dfrac{80}{x})\%$$
together they can finish 100% of work $$=(\dfrac{100}{x}+\dfrac{140}{x}+\dfrac{80}{x})\%\times 15$$
$$\Rightarrow (\dfrac{100}{x}+\dfrac{140}{x}+\dfrac{80}{x})\%\times 15=100\%$$

$$\Rightarrow x= (\dfrac{100}{100}+\dfrac{140}{100}+\dfrac{80}{100})\times 15$$
$$\Rightarrow x=1+1.4+0.8\times 15$$
$$\Rightarrow x=3.2\times 15$$
$$\Rightarrow x=48.0$$ days
Hence, B alone can finish $$75\%$$work in$$=\dfrac{48\times 75}{100}=36$$ days.