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?
+128406 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
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.
