If 'A' is 6 times more efficient than 'B', 'B' takes 32 days to complete the task, then find the number of days required to finish the whole work by 'A' and 'B' working together.

Question

Question

Answer 1

Given:

A is 6 times more efficient than B, & B takes 32 days to complete the task.

Formula used:

Total work = Efficiency × Time taken

Calculation:

A is 6 times more efficient than B

Efficiency of A ∶ Efficiency of B = 7 ∶ 1

Total work = Efficiency of B × Time taken

⇒ 1 × 32 = 32 units

Number of days required to finish the whole work by (A + B) = Total work/Efficiency of (A+ B)

⇒ 32/8

⇒ 4

∴ The total number of days required to finish the whole work by (A + B) is 4 days.

There is a difference in "Efficient" and " More efficient"

A is 6 times efficient than B means if B is 1 then, A will be 6

A is 6 times more efficient than B means if B is 1 then, A will be (1 + 6) = 7

In the question, it is given that A is 6 times more efficient which means if B is 1, then A will (1 + 6) times = 7 times efficient

So, Total efficiency of A and B = (1 + 7) = 8 units/day

Time taken to complete the work together = 32/8 days

⇒ 4 days and this is the answer.