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?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 7
A works twice as fast as B and B works twice as fast as C:
Let efficiency of C = 1 unit
Efficiency of B = 2 units
Efficiency of A = 4 units
Given that D alone can finish the same task in 16 days.
As total work is 100%
So, work done by D in 4 days = 25%
It is also given that all three working together can finish the total task in 4 days with the help of D.
So, work done by (A + B + C) in 4 days = 100% - 25% = 75%
Efficiency of (A + B + C) = 1 + 2 + 4 = 7 units
So, work done by (A + B + C) in 4 days = 7 × 4 units
⇒ 7 × 4 units = 75% of the work ............... (1)
Now we will find that in how many days will A alone finish 75% of the same task.
Efficiency of A = 4 units
Let A will finish 75% of the same task in x days.
So,
Number of days taken by A to finish 75% of the work = (7 × 4)/4 = 7 days