Solution:
Given:
A and B can complete a task together in 35 days.
Total time taken is 90 days.
Calculation:
Let the total amount of work be 35 units.
If A and B work together, they complete 35/35 = 1 unit of work every day.
Now,
Let the amount of work done by A by himself in one day be x units and the work done by B by himself in one day be y units.
So, x + y = 1
If A works alone and completes 5/7th of the task,
work done by A is (5/7) × 35 = 25 units.
⇒ time taken by A to complete this 25 units of work = 25/x days.
If B works alone and completes 2/7th of the task,
work done by B is (2/7) × 35 = 10 units.
⇒ Time taken by B to complete this 10 units of work = 10/y days.
According to the question,
25/x + 10/y = 90
⇒ (25 + 10)/xy = 90
⇒ 25y + 10x = 90xy
⇒ 25(1 – x) + 10x = 90xy
⇒ 25 – 25x + 10x = 90x(1 – x)
⇒ 25 – 15x = 90x – 90x2
⇒ 90x2 – 105x + 25 = 0
⇒ 18x2 – 21x + 5 = 0
⇒ 18x2 – 15x – 6x + 5 = 0
⇒ 3x(6x – 5) – 1(6x – 5) = 0
⇒ (6x – 5)(3x – 1) = 0
So, x = 5/6 (x cannot be 1/3 as A is more efficient among two)
Time taken by A alone to complete work by herself = 35/(5/6)
⇒ 42 days
∴ The required answer is 42 days.
