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?

7

Given,

                A : B : C 

Efficiency   4 : 2 : 1

According to the question,

⇒ (7 + D) x 4 = D x 16 

⇒ D = \(\frac{7}{3}\)  (Efficiency)

⇒ Total work = 16 x \(\frac{7}{3}\) = \(\frac{112}{3}\) units,

⇒ A alone can finish 75% of the work in 

⇒ \(\frac{3}{4}\) x \(\frac{112}{3}\) x \(\frac{1}{4}\) = 7 days.

Therefore, A alone can finish the total work in 7 days.