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Alex takes 5 days less than Peter to do a project if they work on the project alone. If they are working together, the project would be completed 4 days earlier than Alex working alone. How long would each take to work on the job alone?

 Mar 31, 2020
 #1
avatar+112280 
+1

This is similar to the  last one

 

Let the number  of days it takes  Alex to complete the job = D

The Peter  takes D +  5  days

 

So   Alex does 1/D  of the  project  in one day

And Peter  does 1 / (D + 5)  of  the project in  one day

 

Add these

 

1/D  +  1/{D + 5)  =     [ D + 5 + D ]  / [ D ( D + 5)  ]  =   [ 2D + 5 ] / [ D^2 + 5D ]

 

The reciprocal of this is the total time for  both to completet the job  (in days )= [ D^2 + 5D ] / [ 2D + 5 ]

 

And  like  the last problem  we  know  that

 

Time working together  + 4 days  = Time  for alex working alone

 

 

[D^2 + 5D ] / [ 2D + 5 ] + 4  =  D       simplify

 

[ D^2  + 5D ]  / [ 2D + 5 ]  =  D  - 4

 

D^2  + 5D  =   [ 2D + 5] [ D - 4 ]

 

D^2  + 5D = 2D^2  + 5D  - 8D  - 20

 

D^2  - 8D  - 20   =  0      factor

 

(D  - 10 ) ( D + 2)

 

The  first factor  set to 0  results in D  = 10  days for Alex to complete the project  on his own

 

And  D +  5  = 15  days  for Peter to complete the project on his own

 

 

 

cool cool cool

 Mar 31, 2020
 #2
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+1

Here is another way:
Peter =1 / P - his work rate
Alex = 1 / [P - 5] - his work rate
1 Job = 1 / J
1 / [P - 5] = 1 / J.....................(1)
1 / P + 1/ [P - 5] = 1 / [J - 4].....(2), solve for P, J
P = 15 days for Peter to finish working by himself.
15 - 5 =10 days for Alex to finish the job by himself.
Working together: [1/15 + 1/10] =1/6 - or 6 days to finish the job. or, 10 - 6 =4 days less than Alex working alone.

 Mar 31, 2020

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