Question

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

Answer:

ASSUMPTIONS

The work done by the first worker be X

The work done by the second worker be Y

If Z is assumed to be the total work done

Then, the time taken for the 1st worker to complete the job is h days

X(h)=Z--------------------------------equation 1

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

Y(h-5)=Z-------------------------------equation 2

from equation 1, X=Z/h--------------- equation 3

from equation 2, Y=Z/(h-5)---------- equation 4

worker 1 do the job for 1 hour and then both worker 1&2  do the job for 4 hours and 40% work is done:

X(1) +  (X+Y)4= 4Z/10X

X+4X+4Y=4Z/10

5X+4Y=4Z/10-----------------------equation 5

From equation 3, X=Z/h

From equation 4, Y=Z/(h-5)

substituting the value of X and Y in equation 5,

5(Z/h) + 4(Z/(h-5))=4Z/10

5Z/h + 4Z/(h-5)=4Z/10

5/h + 4/(h-5)= 4/10

multiply through by the lowest common factor which is 10(h-5)

we have:  5/h×10(h-5) + 4/(h-5)×10(h-5)= 4/10×10(h-5)

50h - 250 + 40h = 4h² - 20h

4h²-110h+250=0

solving the quadratic equation,

4h²-100h-10h+250=0

4h(h-25) -10(h-25)=0

(4h-10) (h-25)=0

Therefore,

4h-10=0 or h-25=0

4h=10 or h=25

h=10/4 or h= 25

h=2.5 or h=25

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

So, the first worker completes the work in 25 days (h=25) by himself and the second worker completes thework in 20 days (h-5=25-5=20) by himself.

Explanation:

Answer 2

Answer:

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.