Two plumbers received a job. At first, one of the plumbers worked alone for 1 hour, and then they worked together for the next 4 hours. After this 40% of the job was finished. How long would it take each plumber to do the whole job by himself if it is known that it would take the first plumber 5 more hours to finish the job than it would take the second plumber?

Answer 1

If the fast plumber takes x hours, then the slow one takes x+5 hours.

1/(x+5) + 4(1/x + 1/(x+5)) = 2/5
x = 20

So, the 1st plumber takes 25 hours.

Check:
In 1 hour, the 1st plumber finishes 1/25 of the job.

During each of the next 4 hours, the two together finish 1/25 + 1/20 = 9/100 of the job.

1/25 + 36/100 = 40/100 = 40%

Answer 2

37

Answer 3

1st Plumber = 25 hrs, 2nd Plumber = 20 hrs.