Question

One person can complete a task 8 hours sooner than another person. Working together, both people can perform the task in 3 hours. How many hours does it take each person to complete the tasking working alone?

Answer 1

Answer:

it takes 1st person 4 hour and second person is 8 hours

Step-by-step explanation:

Let t is the time (t >0)

We have:

• 1st person working rate per one task: 1/t-8
• 2nd person working rate per one task: 1/t
• Working rate per one task from 2 person: 1/3

<=> 1/3  = 1/t  + 1/t-8  

<=> $t^{2} -14t = -24$

<=> $(t-7)^{2} =25$

<=> t  - 7 = 5

<=> t = 12

So it takes 1st person 4 hour and second person is 8 hours

Answer 2

Answer:

Person A complete the task in 1.125 hours

Person B complete the task in 9 hours

Step-by-step explanation:

Person A complete a task 8 hours sooner than person B

Working together they perform the task in 3 hours

Let call "x" hours person A  need to do the task alone

Then person B will take 8*x hours to do the job

In 1 hour person A do 1/x  part of the task

In 1 hour person B do 1/8*x  part of the task  

Then A + B in one hour of work will do

1/x + 1/8x   =  [ 8 + 1 ]/8*x    ⇒  9/8*x

And according to problem statement that time is 1/3 (one third of the time)

Then 3* (9/8*x) =  3

9 = 8*x

x = 9/8

x =  1.125 hours

And person B will take  8*1,125  = 9 hours