A works twice as fast as B and B works twice as fast as C. All three working together can finish a task in 4 days with the help of D. If D alone can finish the same task in 16 days, then in how many days will A alone finish 75% of the same task?

 we let A = 4, B= 2 , C = 1 

let D = $$x$$

total work = $$(x+7)$$days 

According to question  $$\dfrac{x+7}{x}\times 4 = 16 $$

$$\Rightarrow 12x = 28 $$

$$\Rightarrow x = \dfrac{28}{12} = \dfrac{7}{3}$$

total work =$$ (\dfrac{7}{3} + 7) \times 4 $$

then 75 % work = $$ (\dfrac{7}{3} + 7) \times 4 \dfrac{3}{4}$$ (where 75% =$$\dfrac{3}{4}$$)

$$\Rightarrow (\dfrac{7+21}{3} \times 4 \times \dfrac{3}{4}$$

$$\Rightarrow 28 $$

then A = $$\dfrac{28}{4}$$  = 7 days Ans