A works twice as fast as B and B works twice as fast as C. All 3 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?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. 7.
D alone finishes the work in 16 days
In 4 days D is doing 4/16 = 1/4 part of the work
Let in 1 day, C is doing w units of work
So B is doing 2w work and A is doing (2 × 2w) work in 1 day
So in 4 days A, B and C together did
⇒ (4 × 4w) + (4 × 2w) + (4 × w) work = 28w work
⇒ 28w = (1- 1/4) part of the work
⇒ w = (3/4) × (1/28) = 3/(4 × 28) part of the full work .
In 1 day A is doing 4w work = (4 × 3)/(4 × 28) = 3/28 part of the whole work.
⇒ 3/28 part is done in 1 day
⇒ 1 part is done in 28/3 days
⇒ 75% = 3/4 part is doing in (28/3) × (3/4) = 7 days
A will do 75% of the work in 7 days.