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?

The first worker complete the job in 25 days himself

The second worker complete the job in 20 days himself

Explanation:

r1= rate of work done by first worker

r2=rate of work done by second worker

W= total work done

t days= time taken by the 1st worker to complete the job

r1(t)=W

r1=W/t (1)

Then the time taken by the 2nd worker to complete the job is t-5 days.

r2(t−5)=W

r2=W/(t−5) (2)

If 1st worker do the job for 1 hour and then both the worker do the job for 4 hours and 40% work is done, So

r1(1)+(r1+r2)(4)=4W/10

r1+4r1+4r2=4W/10

5r1+4r2=4W/10 (3)

Substitute equation 1 and 2 into (3)

5W/t+4{W/(t−5)}=4W/10

Multiply through by 10(t-5)

5/t+4/t−5=4/10

5(10)(t−5)+4(10)(t)=4(t−5)(t)

50t−250+40t=4t^2−20t

4t^2−110t+250=0

Solving the above quadratic equation using factorization method

4t^2−100t−10t+250=0

4t(t−25)−10(t−25)=0

(t−25)(4t−10)=0

t=25 or t=2.5

2.5 can't be the answer because if we take 2.5 days in which 1st worker completes the work, then the second worker will complete the work in -2.5days which is wrong.

The first worker completes the job in 25 days by himself and the second worker completes the job in 20 days by himself.